# Questions on Algebra: Trigonometry answered by real tutors!

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 Algebra: Trigonometry Solvers Lessons Answers archive Quiz In Depth

Question 817237: What are the values that satisfy the trigonometric equation for 0 <= q <= 2? sin(theta) + tan(-theta)=0
a.)0,pie
b.)pie/4,5 pie/4
c.)0,pie/2, 3pie/2
d.)pie/2,3pie/2

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"0 <= q <= 2? sin(theta) + tan(-theta)=0"?? I suspect no one has responded to this because they don't understand what this means.

Question 817278: tanx=1
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What are you/we supposed to do with this?

Please include the instructions for the problem when you post.

Question 817280: Simplify csc(x)-cos^2(x)csc(x)
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Here's one way...
Factor out csc(x):

We should recognize that :

Since csc and sin are reciprocals of each other we can replace csc with 1/sin:

The sin in the denominator cancels out with one of the factors of sin in leaving:
sin(x)

Question 817228: Question:
Solve over the interval [0 degrees, 360 degrees):
3cos^2x-4cosx-2=0
I have absolutely no idea how to do this problem. I tried using the quadratic formula but I knew that wasn't right once I did it that way. I then tried this way of doing it:
3cos^2x-4cosx-2=0
3cos^2x-4cosx=2
cosx(3cosx-4)=2
cosx=2/3cosx-4

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You have a good idea with the factoring. But factoring an expression that is not equal to zero does us no good. So adding the 2 at the start was a mistake. Finding out that cos(x)(3cos(x)-4) = 2 does not help us at all.

The key is to factor the original equation (which has a zero on one side). It may not be easy to see how to factor this. It is in what is called quadratic from. This means the equation has the same structure as a regular quadratic, , and can be solved using the same techniques.

Let q = cos(x). Replacing the cos(x)'s with q's we get:

It should be obvious that we now have a quadratic equation. It does not factor but we can use the quadratic formula:

Simplifying...

which is short for:
or

Of course we do not care about what q equals. We want to know what x equals. So we replace the q's:
or
The right sides of these equations are not special value numbers for cos. So we get out our calculators (and rounding to 2 places):
cos(x) = 1.7 or cos(x) = -0.39

Since cos is always between -1 and 1, the first equation is impossible. So we will not get any solutions from it. But the second equation is possible. First we find the reference angle. Using our calculators to find we should get 67.05 for the reference angle. (Note that we did not use the "-". Never use minus signs when looking for a reference angle.) Since -0.3874 is negative (here is where the negative gets used) and since cos is negative in the 2nd and 3rd quadrants we should get the following general solution equations:
These simplify to:

Now we use these and various integers for n to find the x's that are in the specified interval.
From ...
if n = 0 then x = 112.95
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval
From ...
if n = 0 then x = 247.05
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval

So the only solutions that are in the specified interval are 112.95 and 247.05 degrees.

Question 817271: sqrt (3)*tg * (3*x) +1 = 0
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• What are you/we supposed to do with this? Please include the instructions for the problem, not just the equation/expression.
• What is "tg"??
• Is it supposed to be "tan" (since you did post this in the Trigonometry category)? If yes, then there is no multiplication between the function name and its argument. For example: tan(3x-2) not tan*(3x-2).
• Or is "tg" something else? If so then more information will be needed to solve the problem.

Question 817265: 2*sin^2(x)+3*sin(x)+1=0
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```
Hi,
2*sin^2(x)+3*sin(x)+1=0
(2sin(x)+1)(sin(x) + 1) = 0
(2sin(x)+1) = 0 , sin(x) = -1/2
(sin(x) + 1) = 0, sin(x) = -1
See(cos(x), sin(x)) reference below:

```

Question 816649: Find the exact value of the sin cos and tan of the angle -225 and 13pie/12
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Find the exact value of the sin cos and tan of the angle -225 and 13pie/12
***
reference angle for -225˚ is 45˚ in quadrant II where sin>0,cos<0,tan<0.
tan(-225=tan(45)=-1
sin(-225)=sin(45)=√2/2
cos(-225)=cos(45)=-√2/2
..
(13π/12)=(9π/12+4π/12)=(3π/4+π/3)
tan(3π/4)=-1
tan(π/3)=√3
Identity:tan(3π/4+π/3)=(tan(3π/4)+tan(π/3))/(1+tan(3π/4)*tan(π/3)
=(-1+√3)/(1+(-1)*(√3)=(-1+√3)/(1+√3)
tan(13π/12)=2-√3
..
sin(3π/4)=√2/2
sin(π/3)=√3/2
cos(3π/4)=-√2/2
cos(π/3)=1/2
..
Identity:sin(3π/4+π/3)=sin(3π/4)*cos(π/3)+cos(3π/4)*sin(π/3)
=√2/2*(1/2)+(-√2/2)*√3/2=(√2/4)-√6/4=-√6+√2)/4
..
Identity: cos(3π/4+π/3)=cos(3π/4)*cos(π/3)-sin(3π/4)*sin(π/3)
=-√2/2*(1/2)-(√2/2)*√3/2=(-√2/4)-√6/4=-√2-√6)/4=-(√6+√2)/4
..
Calculator check:
tan(13π/12)≈0.2679..
exact value as calculated=2-√3≈0.2679.
..
sin(13π/12)≈-0.2588..
exact value as calculated=-√6+√2)/4≈-0.2588
..
cos(13π/12)≈-0.9659..
exact value as calculated=-(√6+√2)/4≈-0.9659..

Question 816927: for all values of theta, the expression 1-cos²(θ) simplifies to...
Answer by Edwin McCravy(9716)   (Show Source):

Question 816758: Solve the equation cosA/(1+sinA)+1+sinA/cosA=4, giving your answers for A in the interval 0
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First of all, I assume that the second fraction's numerator is (1+sin(A)). If so, then please put multiple-term numerators (and denominators) in parentheses. If I am wrong then please re-post your question.

Here's a solution that may be faster than most. For reasons that will become clear soon, multiply the first fraction by (1-sin(A)):

Simplifying...

The factor of cos(A) in the numerator cancels one of the two factors of cos(A) in the denominator. (Do you see now why we multiplied by 1-sin(A)?)

The denominators are the same so we can add. The sin's cancel:

Multiplying both sides by cos(A):

Dividing by 4:

We should recognize that 1/2 is a special angle value for cos. It tells us that the reference angle is 60 degrees. Since the 1/2 is positive and since cos is positive in the first and fourth quadrants we should get general solution equations of:
A = 60 + 360n (for the first quadrant)
A = -60 + 360n (for the fourth quadrant)

Now we try various integers for n as we look for specific solutions within the given interval.
From A = 60 + 360n
if n = 0 then A = 60
if n = 1 (or larger) then A is too large for the interval
if n = -1 (or smaller) then A is too small for the interval
From A = -60 + 360n
if n = 0 (or smaller) then A is too small for the interval
if n = 1 then A = 300
if n = 2 (or larger) then A is too large for the interval

So the only solutions within the given interval are: 60 and 300.

Question 816780: Use inverse trigonometric functions to find the solutions of the equation that are in the interval [0, 2π).You may enter an exact answer or round your solutions to four decimal places (this is best for this problem). (Enter your 4 answers in a comma separated list.)
cos(x)(8cos(x) + 4) = 3

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The left side is factored. But the right side is not zero. So having the left side factored does not help. Multiplying out the left side:

Now we get a zero on the right by subtracting 3:

Now we factor, if we can. But this will not factor. It is a quadratic however. So we can use the quadratic formula with an "a" of 8, a "b" of 4 and a "c" of -3:

Simplifying...

which is short for:
or

Using our calculator to get decimal approximations (and rounding them to 4 places):
or
Simplifying:
or

Using the inverse cos on our calculator on these decimals...
For , is 1.1467. So the reference angle is 1.1467. And since the 0.4115 is positive and since cos is positive in the 1st and 4th quadrants we should get general solution equations of:

For , is 0.4239. (Note: Do not enter the minus of -0.9115 when looking for a reference angle! With a minus on the decimal, , we will get second quadrant angle, not a reference angle!) So the reference angle is 0.4239. And since the -0.9115 is negative (here is where the minus gets used) and since cos is negative in the 2nd and 3rd quadrants we should get general solution equations of:
Replacing first 's with 3.1416:
which simplifies to:

So the general solution equations are:

From these we will try various integer values for n looking for x's that are in the given interval.
From ...
if n = 0 then x = 1.1467
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval
From ...
if n = 0 (or smaller) then x is too small for the interval
if n = 1 then x = 1.1467 + = 5.1365
if n = 2 (or larger) then x is too large for the interval
From ...
if n = 0 then x = 2.7177
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval
From ...
if n = 0 then x = 3.5655
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval

So the specific solutions within the given interval are:
1.1467, 5.1365, 2.7177 and 3.5655

Question 816773: find the indicated value without using a calculator csc(sin^-1 3/5)
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represents an angle which has a sin ratio of 3/5. Since sin and csc are reciprocals of each other, the csc of an angle whose sin is 3/5 will be the reciprocal of 3/5: 5/3.

Question 816769: Show all steps necessary to verify the trigonometric identity:
1+tan^2x
-------- = csc^2x
tan^2x

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First of all, please do not try to use fraction bars when you post. Most of the time they look so bad they are hard to understand. Instead, put parentheses around the numerator and around the denominator and separate them with a slash:
(1+tan^2(x))/tan^2(x) = csc^2(x)

When verifying identities and you're not sure what to do, try changing sec's, csc's, tan's and cot's into expressions of sin and/or cos. This often makes the path clearer. Changing the tan's to sin/cos:

Now we can multiply the numerator and denominator by (to eliminate the fractions within the larger fraction:

Using the Distributive Property on top we get:

We should recognize that the numerator is 1:

And since sin and csc are reciprocals of each other, the left side is equal to the right:

tan13degrees= y/x Tan19 degrees= y/25-x
What is x and y?
Thank you!

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tan(13) = y/x = .23
tan(19) = y/(25 - x) = .34
from equation 1 we have x = y/.23
substitute for x in second equation
y / (25 - y/.23) = .34
y = .34 * (25 - y/.23)
y = 8.5 - 1.48y
2.48y = 8.5
y = 3.43
x = 3.43 /.23 = 14.91
check our answers by substituting in the second equation
3.43 / (25 - 14.91) = .34
.339 = .34
.34 = .34
x = 14.91, y = 3.43

Question 816297: Please Prove the following identities
1.) Tan (x/2) = tan (x)/sec (x) + 1
2.) sin(2x) - tan(x) = tan(x) cos (2x)

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1.) Tan (x/2) = tan (x)/sec (x) + 1
tanx/(secx+1)=(sin/cos)/((1/cos)+1)=(sin/cos)/((1+cos)/cos)
=
verified: right side=left side
..
2.) sin(2x) - tan(x) = tan(x) cos (2x)
tanxcos2x=(sin/cos)(2cos^2-1)=2sincos-sin/cos=sin(2x)-tan(x)
verified: right side=left side

Question 816597: I have a trig problem involving vector operations.
[ An airplane has an airspeed of 150 km/h. It is to make a flight in a direction of 70 degrees while there is a 25 km/h wind [TO] 340 degrees. What will the airplane's actual heading be? ]
NOTE: the original problem stated that the wind was FROM 340 degrees, but that is impossible assuming that the given result of 60 degrees is the solution to the problem. To obtain the result of 60 degrees, the wind must be TO 340 degrees, not FROM 340 degrees.

how did my instructor figure it out?

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a = magnitude 150 kph direction to 70 degrees
w = magnitude 25 kph direction to 340 degrees
---
convert angles from true north (aviation format), to polar coordinates:
70 degrees true north = +20 degrees polar
340 degrees true north = +110 degrees polar
---
calculate the rectangular coordinates of each vector:
---
a vector:
ay:
sin( 20 ) = opp/hyp
sin( 20 ) = ay/150
ay = 150 * sin( 20 )
---
ay = 51.303
---
ax:
cos( 20 ) = ax/150
ax = 150 * cos( 20 )
---
ax = 140.953
---
w vector:
wy:
sin( 110 ) = opp/hyp
sin( 110 ) = wy/25
wy = 25 * sin( 110 )
---
wy = 23.492
---
wx:
cos( 110 ) = wx/25
wx = 25 * cos( 110 )
---
wx = -8.551
---
calculate the result vector, from vector addition in rectangular coordinates
r vector:
rx:
rx = ax + wx
rx = 140.953 - 8.551
---
rx = 132.402
---
ry:
ry = ay + wy
ry = 51.303 + 23.492
---
ry = 74.795
---
convert r vector to polar coordinates:
rmag = sqrt( 132.402^2 + 74.795^2 )
rmag = 152.067 kph
rang = arctan( 74.795 / 132.402 )
rang = 29.462 degrees
convert polar angle to aviation format (true north)
rang = 60.538 degrees
---
the airplane's actual heading after wind correction is 60.538 degrees, and its ground speed accounting for wind is 152.067 kph

Question 816392: Use the given information to find tan2x.
Then I am given a simple graph with a triangle drawn in the first quadrant. Angle is given as x, the hypotenuse is labeled 5, and the x-axis on the triangle is labeled 3. It is a right triangle.

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Use the given information to find tan2x.
Then I am given a simple graph with a triangle drawn in the first quadrant. Angle is given as x, the hypotenuse is labeled 5, and the x-axis on the triangle is labeled 3.
***
Reference angle (x) is in quadrant I
You are working with a 3-4-5 reference right triangle in quadrant I
sinx=4/5
cosx=3/5
tanx=sinx/cosx=4/3
Identity: tan(2x)=2tan(x)/1-tan^2(x)=(8/3)/(1-16/9)=(8/3)/(-7/9)=(8/3)*(-9/7)=-72/21
tan(2x)=-72/21
..
Calculator check:
tanx=4/3
x≈53.13˚
2x≈106.26
tan(2x)≈tan(106.26)≈-3.4286..
exact value as calculated=-72/21≈-3.4286..

Question 816554: If sin t= -3/5 and cos t>0, Find A) Cos t=____ and B) Tan t=____
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If sin t= -3/5 and cos t>0, Find A) Cos t=____ and B) Tan t=____
***
You are working with a 3-4-5 reference right triangle in quadrant IV where sin>0 and cos>0
..
Cos(t)=4/5
Tan(t)=sin(t)/cos(t)=-3/4

Question 816603: Solve for B; cos(3B - pi/4) =1/2
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Solve for B; cos(3B - pi/4) =1/2
assume(3B-π/4) is an acute angle
cos(3B-π/4)=1/2=cos(π/3)
3B-π/4=π/3
3B=π/3+π/4=4π/12+3π/12=7π/12
3B=7π/12
B=7π/36

Question 816620: Use Substitution to solve the system:
-x+y=2
x^2+y^2=20

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-x+y=2
x^2+y^2=20
-x+y=2
y=x+2

x^2+y^2=20
substitute y
x^2+(x+2)^2=20
x^2+x^2+4x+4=20
2x^2+4x-16=0
/2
x^2+2x-8=0
x^2+4x-2x-8=0
x(x+4)-2(x+4)=0
(x+4)(x-2)=0
x=-4 or x=2
y=x+2
when x=-4
y=-4+2=-2
(-4,-2)
when x=2
y=2+2=4
(2,4)

Question 816479: Can you help me prove the identity:
1+tan^2x=1/cos^2x
I don't understand how to solve this.

Found 2 solutions by Alan3354, lwsshak3:
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Can you help me prove the identity:
1+tan^2x=1/cos^2x
-------------
1+tan^2x = sec^2
QED

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Can you help me prove the identity:
1+tan^2x=1/cos^2x
***

..

..

..

..
verified:
left side=right side

Question 816468: If the cos x is -(1/2) and the x is in the 2nd quadrant, find the cos 2x. (no calculator)
I have been working on this problem for 30 mins. I think I am missing one thing for it to make sense. I think that sin x= 2/sq.rt. 5...
Show that this is a trig identity.
csc^2 = cot^2x+1

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If the cos x is -(1/2) and the x is in the 2nd quadrant, find the cos 2x.
cos(2x)=cos^2(x)-sin^2(x)
..
cos(x)=(-1/2)
cos^2(x)=1/4
..
sin(x)=√3/2
sin^2(x)=3/4
..
cos(2x)=(1/4)-(12/16)=(4/16)-(12/16)=-8/16=-1/2
...
Show that this is a trig identity.
csc^2x = cot^2x+1
cot^2x+1=cos^2/sin^2x+1
(sin^2(x)+cos^2(x))/sin^2(x)
1/sin^2(x)=csc^2(x)
verified:
right side=left side

Question 816469: Use inverse trigonometric functions to find the solutions of the equation that are in the interval [0, 2π).You may enter an exact answer or round your solutions to four decimal places (this is best for this problem). (Enter your 4 answers in a comma separated list.)
cos(x)(15cos(x) + 7) = 3

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cos(x)(15cos(x) + 7) = 3
15cos^2(x)=3-7=-4
cos^2(x)=-4/15
no solution:
cos^2(x)≥0

0 If cosx= 8/17 and tany= 5/12, find sin(x+y)

Answer by Edwin McCravy(9716)   (Show Source):
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```since the cosine is the adjacent over the hypotenuse,
and we are given that the cosine of x is 8 over 17, we
will draw a right triangle with an adjacent side of 8
and a hypotenuse of 17.  Then the angle in that
triangle that has 8 for its adjacent side will be
angle x.

Since the tangent is the opposite over the adjacent,
and we are given that the tangent of y is 5 over 12,
we will draw a right triangle with an opposite side
of 5 and an adjacent side of 12.  Then the angle in
that triangle that has 5 for its opposite side and
12 for its adjacent side will be angle y.

Now we will find the missing side in each of those
triangles using the Pythagorean theorem:

c² = a² + b²         c² = a² + b²
17² = 8² + b²         c² = 12² + 5²
289 = 64 + b²         c² = 144 + 25
225 = b²              c² = 169
15 = b                c = 13

Now we use the identity

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

and use the sides of those triangles to substitute
the trig ratios:

sin(x+y) =  =  =

Edwin```

Question 816236: Find the function that will make the given equation an identity:
(1+ cos(2X))/ (sin(2X)) = f(x)

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When you learned about the cos(2x) I hope you learned all three variations:
• cos(2x) = cos^2(x)-sin^2(x)
• cos(2x) = 2cos^2(x)-1
• cos(2x) = 1-2sin^2(x)
While all of them will work in this problem, the one in the middle, with the -1, will work well in our numerator with its +1:

The +1 and -1 cancel:

The factors of 2 (in front) will cancel:

The cos(x) in the denominator will cancel with one of the two factors of cos(x) in the numerator:

which is equal to:

Question 816119: solve tanx- 1=0 for all values of x.
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tan(x) - 1 = 0
tab(x) = 1
We should recognize that 1 is a special angle value for tan. It tells us that the reference angle is . Since the 1 is positive and since tan is positive in the 1st and 3rd quadrants, we should get the following general solution equations:
The second equation simplifies:

Since the problem asks for all values of x, the general solution is the answer:

P.S. Since the period of tan (and cot) is . The general solution can be expressed in a single equation:

(Note that the "2" is gone.)

Question 816116: solve tanx cosx= 1/2 for all values of x.
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One way to solve this is to start by replacing tan with sin/cos:

The cos's cancel:

We should recognize that 1/2 is a special angle value for sin. It tells us that the reference angle is . Since the 1/2 is positive and sin is positive in the 1st and 2nd quadrants, we should get the following general solution equations:
The second equation will simplify:

The problem asks for all values of x. This is what the general solution is. So all the solutions are described by:

Question 816106: Find all solutions for 2sinx+2=0 on the interval [0, 2pi).
I tried to factor it and got 2(sinx)=0, but I think that's wrong.

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When you factor it you get:
2(sin(x)+1) = 0
not what you got. Now you should be able to solve this.

Question 816020: So, the instructions on the problem say:
'Use the identities to rewrite each expression in terms of a single trigonometric function. Check your result by graphing.

I recognized that it has a similar form to sin2t= 2sintcost but the sin and cos are squared in this problem. What else can I do? Maybe transform the sin^2 or cos^2 to a different identity? Thank you so much, this is such a great site for help.

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Recognizing that the sin(2x) identity is buried in there is more than half the battle! Very good!

Now we just have to see how we can use it. Yes, the sin and cos in your expression are squared. But isn't 4 also the square of the 2? So isn't your expression just the square of sin(2x)?:

Now we can use the sin(2x) formula with 7t in the place of x:

The argument simplifies:

which can also be written as:

Question 815924: Find the values of all the trigonometric functions from the given information: cotT=3, pi < or equal to T< or equal to 3 pi over 2
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Find the values of all the trigonometric functions from the given information: cotT=3, pi < or equal to T< or equal to 3 pi over 2
***
hypotenuse of reference right triangle in quadrant III=√(3^2)+(1)^2)=√(9+1)=√10
sinT=-1/√10=-√10/10
cosT=-3/√10=-3√10/10
tanT=1/3
cscT=-√10
secT=-√10/3
cotT=3 (given)

Question 816073: Given that cotx=-12/5 with x in the IV quadrant and secy=-17/15 with y in the III quadrant find csc(x+y)? I have found everything else for the triangle but I just do not know which identity to use. I know that the csc is 1/sin so do I use the sin(u+v) identity?
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Given that cotx=-12/5 with x in the IV quadrant and secy=-17/15 with y in the III quadrant find csc(x+y)?
***
csc(x+y)=1/sin(x+y)
The problem then is to solve for sin(x+y)then take its reciprocal.
sin(x+y)=sinxcosy+cosxsiny
hypotenuse of reference right triangle in quadrant IV=√(12^2+5^2)=√144+25)=√169=13
cotx=-12/5 (given)
sinx=-5/13
cosx=12/13
..
secy=-17/15 (given)
siny=-15/17
cosy=-√(1-sin^2y)=-√(1-(15/17)^2)=-√(1-225/289)=-√(64/289)=-8/17
..
sin(x+y)=(-5/13)*(-8/17)+(12/13)*(-15/17)
=40/221-180/221=-140/221
csc(x+y)=-221/140
..
calculator check:
tanx=-5/12
x=337.38˚
..
cosy=8/17
y=241.93˚
..
x+y=579.31˚
..
sin(x+y)=sin(579.31˚)≈-0.6335
Exact value as calculated=-140/221≈-0.6335
..

Question 816104: solve 12sin^2(x)+2sin(x)-2=0 for all solutions where x is in radians between 0 and 2pi
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solve 12sin^2(x)+2sin(x)-2=0 for all solutions where x is in radians between 0 and 2pi
***

divide by 2

solve for sin(x)
(3sin(x)-1)(2sin(x)+1)=0
..
3sin(x)-1=0
sin(x)=1/3
x≈0.34, 2.80
or
2sin(x)+1=0
sin(x)=-1/2
x=7π/6,11π/6

Question 816078: verify the identity by showing that the left equals right sec^2x/1-tan^2=sec2x
do I use 1/cos^2x/1-tan^2x or do I use 1+tan^2x/1-tan^2x? either way I do not know where to go from either one of those this is really confusing to me and I just cannot get it.

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verify the identity by showing that the left equals right
sec^2x/1-tan^2=sec2x
***

..

..
verified: left side=right side

Question 816069: If cscA=2, find the value of sinA.
I got sinA= 1/2 because csc is the inverse of sin, but I'm not sure that is right.

You can put this solution on YOUR website!
If cscA=2, find the value of sinA.
I got sinA= 1/2 because csc is the inverse of sin, but I'm not sure that is right.
-----------
That is right.

Question 816076: Simplify cscx- cosxcotx
I rewrote the problem using reciprocals and quotients to get . Then I multiplied the reciprocals of cosx and cotx to get I'm not sure what to do after that.

You can put this solution on YOUR website!
Simplify cscx- cosxcotx
I rewrote the problem using reciprocals and quotients to get . Then I multiplied the reciprocals of cosx and cotx to get I'm not sure what to do after that.
===============
cscx- cosxcotx
---
= 1/sin - cos^2/sin
= (1 - cos^2)/sin
= sin^2/sin
= sin(x)

Question 815755: Solve for X:
log to the base of 3(x^2+4x+13)-log to the base of 3(x+3)=2

You can put this solution on YOUR website!

Solving equations like this usually starts with using algebra and/or properties of logs to rewrite the equation in one of the following forms:
log(expression) = number
or
log(expression) = log(other_expression)
With the "non-log" term of 2 on the right side, it would seem that the "all-log" second form would be harder to achieve. So we will aim for the first form.

To achieve the first form all we need to do is find a way to combine the logs. Fortunately there is a property of logs, , which will do exactly what we need:

We now have the first form.

The next stage is to eliminate the logs. With the first form, you rewrite the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:

which simplifies to:

The next stage is to solve the equation (now that the logs are gone and the variables are exposed). First we'll eliminate the fraction by multiplying each side by x+3:

This is a quadratic equation so we want one side to be zero. Subtracting 9x and 27 from each side we get:

Now we factor:
(x-7)(x+2) = 0
Next the Zero Product Property:
x-7 = 0 or x+2 = 0
Solving these we get:
x = 7 or x = -2

The last stage is to check. This is not optional! A check must be made to ensure that the "solutions" make all the bases and arguments valid. (Valid bases are positive but not 1 and valid arguments are positive.) Any "solution" that makes any base or argument invalid must be rejected.

Use the original equation to check:

Checking x = 7:

We should be able to see already that the arguments are going to be positive and the bases are all 3's. (If you can't see this then keep simplifying.) So this solution checks.
Checking x = 7:

Simplifying...

And the arguments are positive (and the bases are still 3's) so this solution checks out, too. (Note: As this problem shows, negative numbers can be solutions. It is the arguments and bases which cannot be negative.

So there are two solutions to the equation: 7 and -2.

Question 815899: How can I solve for the remaining sides and angles if in a triangle C=95 degrees, a=1/2, and b=1/3?
And in another triangle if A=pi/6 b=35.6, and c=41.4? Thanks for any help!

You can put this solution on YOUR website!
How can I solve for the remaining sides and angles if in a triangle C=95 degrees, a=1/2, and b=1/3?
Find side c using the Cosine Law.
The use the Law of Sines for the 2 angles.
-------------
And in another triangle if A=pi/6 b=35.6, and c=41.4?
Same as above.

Question 815886: Find all values of t in the interval [0,2pi] that satisfy the given equation: 2sin2t- square root of 2 ( not under radical but next to the square root) tan2t=0
You can put this solution on YOUR website!

If we would factor the left side we could solve this. At first glance it does not appear to factor. But if we replace tan with sin/cos we will notice something:

We should now notice a common factor of sin(2t). Factoring this out we get:

Now we can use the Zero Product Property:
or

Now we solve these two equations. I'm going to start with the second one because it is more complex:

Subtracting 2:

Dividing by :

The minuses cancel on the right side:

Finding the reciprocal of each side:

We should recognize as a special angle value for cos. It tells us that the reference angle is radians. And since is positive and cos is positive in the 1st and 4th quadrants we get general solution equations of:
Dividing both of these equations by 2 we get:

Now on to the other equation:

Again we should recognize zero as a special angle value for sin. sin is zero at zero and at radians. So the general solution equations are:

Again we divide both by 2:

So the general solution equations are:

Now we use different integers for the "n" in these equations looking for specific solutions in the desired interval.
From :
if n = 0 then
if n = 1 then
if n = 2 (or greater) then t is too large for the interval
if n = -1 (or smaller) the t is too small for the interval
From :
if n = 0 (or smaller) then t is too small for the interval
if n = 1 then
if n = 2 then
if n = 3 (or greater) then t is too large for the interval
From :
if n = 0 then
if n = 1 then
if n = 2 then
if n - 3 (or greater) then t is too large for the interval
From :
if n = 0 then
if n = 1 then
if n = 2 (or greater) then t is too large for the interval
if n = -1 (or smaller) the t is too small for the interval

So the specific solutions within the interval [0, ] are:
, , , , 0, , , and

Question 815618: solve 12sin^2(w)+13cos(w)-15=0 for all solutions where w is in radians between 0 and 2pi. Give your answers accurate to 2 decimal places, as a list separated by commas
You can put this solution on YOUR website!
NOTE: My original posting had an error. This posting has corrected this.

If we could factor this we might be able to solve it. But with both sin and cos in the equation it will not factor easily. Since the sin is squared, we can use to substitute in for the sin squared:

which simplifies as follows:

Now we can factor. Since factoring with a positive leading coefficient is easier, I'll start by multiplying (or dividing) both sides by -1:

Factoring we get:
(3cos(w)-1)(4cos(w)-3) = 0
From the Zero Product Property:
3cos(w) - 1 = 0 or 4cos(w)-3 = 0
Solving 3cos(w) - 1 = 0:
3cos(w) = 1
cos(w) = 1/3
Using inverse cos we get a reference angle of (approximately) 1.23. Since the 1/3 is positive and since cos is positive in the 1st and 4th quadrants we get general solution equations of:

Solving 4cos(w)-3 = 0:
4cos(w) = 3
cos(w) = 3/4
Using inverse cos we get a reference angle of (approximately) 0.72. Since the 1/3 is positive and since cos is positive in the 1st and 4th quadrants we get general solution equations of:

Now we use the general solution equations to find the specific solutions which are in the given interval.
From we get:
if n = 0 then w = 1.23
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From we get:
if n = 0 or smaller then w is too small for the interval
if n = 1 then w = -1.23 + 6.28 = 5.05
if n = 2 (or larger) then w is too large for the interval
From we get:
if n = 0 then w = 0.72
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From we get:
if n = 0 or smaller then w is too small for the interval
if n = 1 then w = -0.72 + 6.28 = 5.56
if n = 2 (or larger) then w is too large for the interval

So the only solutions in the given interval are: 1.23, 5.05, 0.72, 5.56

Question 815687: help the step by step process for this.
establish the identity of sec theta over tan theta-sin theta = 1+ cos theta over sin^3 theta.

You can put this solution on YOUR website!
Please put multiple-term numerators and denominators in parentheses. What you posted meant:

But maybe you meant:

or

etc.
or

etc.

Please re-post your question using parentheses to make the expression clear and unambiguous.

Question 815617: solve 3sin^2(w)-17sin(w)+10=0 for all solutions where w is in radians between 0 and 2pi
You can put this solution on YOUR website!
Note: When I first posted a solution, I had neglected to set my calculator to radian mode. So there were errors in the solution. This posting has these errors corrected.

First we factor:

If you cannot see how the equation factored this way, then try using a temporary variable. Let q = sin(w). Then the equation becomes . It should not be too difficult to see that it factors into (3q-2)(q-5). Then just replace the q's.

Next we use the Zero Product Property:
3sin(w)-2 = 0 or sin(w)-5=0
Solving 3sin(w)-2 = 0 ...
3sin(w) = 2
sin(w) = 2/3
Using inverse sin we get a reference angle of (approximately) 0.73. Since the 2/3 is positive and since sin is positive in the 1st and 2nd quadrants we get a general solution of:

Solving sin(w) - 5 - 0 ...
sin(w) = 5
But sin is never larger than 1 (or less than -1). So there is no solution for this.

Last we use the general solution equations to find the specific solutions in the specified interval.
From
if n = 0 then w = 0.73
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval

if n = 0 then w = 2.41
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval

So the only solutions in the given interval are 0.73 and 2.41.

Question 815656: Prove the following is an identity: cos5y+cos3y/cos5y-cos3y=-cot4ycoty
You can put this solution on YOUR website!
• Put multiple-term numerators and denominators in parentheses. What you posted meant:

when I suspect you meant:
• Put arguments of functions in parentheses. cos(5y) is clearer than cos5y (which could mean (cos(5))*y)

One of the keys to proving identities is to get the arguments to match. On the left side of our equation we have arguments of 5y and 3y. On the right side we have arguments of 4y and y. One way to match these arguments is to rewrite the arguments on the left side in terms of the arguments on the right. With a little effort we should be able to find that 5y = 4y + y and 3y = 4y - y. Rewriting the left side using these arguments we get:

Now we can use the cos(A+B) and cos(A-B) formulas on the left side:

In the numerator the terms with sin in them cancel each other out. In the denominator it is the terms with cos that cancel out. (Be careful with the "-" in front of the parentheses in the denominator!). So this simplifies to:

The two's cancel:

The path to the end should be clear now. Moving the minus out in front and "un-multiplying" fractions we get:

Since cot = cos/sin we get:

Question 815665: Tan^4(k)-sec^4(k)=1-sec^2(k)
You can put this solution on YOUR website!

P.S.
• If the problem is to solve for k or to prove the identity, then here are some steps to try:
1. Factor the left side as a difference of squares.
2. Use the identity to replace the tan^2's with
• If the problem is to prove the identity, then I think the right side should be , not what you posted.

Question 815619: solve sin^2(t)=-5cos(t) for all solutions where t is in radians between 0 and 2pi. Give your answers accurate to 2 decimal places, as a list separated by commas