Questions on Algebra: Trigonometry answered by real tutors!

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Tutors Answer Your Questions about Trigonometry-basics (FREE)


Question 750383: Suppose sin a=((1)/(3)), with a in quadrant 1, and cos b=((3)/(4)), with b in quadrant 4 Find sin(a+b)
Answer by tommyt3rd(449) About Me  (Show Source):
You can put this solution on YOUR website!

All trig functions are positive in quad 1:
cos%28a%29=sqrt%281-%281%2F3%29%5E2%29=sqrt%288%29%2F3

In quadrant 4 sine is negative:
sin%28b%29=-sqrt%281-%283%2F4%29%5E2%29=-sqrt%287%29%2F4



%0D%0Asin%28a%2Bb%29=sin%28a%29cos%28b%29%2Bcos%28a%29sin%28b%29%0D%0A





Question 749962: Find the exact value of csc(pi/8)
Found 2 solutions by Alan3354, lwsshak3:
Answer by Alan3354(30983) About Me  (Show Source):
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csc(pi/8)=1/sin(pi/8)=2%2Fsqrt%282-sqrt%282%29%29
------------
csc^2 = 4%2F%282+-+sqrt%282%29%29
= 4%282+%2B+sqrt%282%29%29%2F2
= 2%282+%2B+sqrt%282%29%29
= 4+%2B+sqrt%288%29
-----
csc%28pi%2F8%29+=+sqrt%284+%2B+sqrt%288%29%29

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
Find the exact value of csc(pi/8)
csc(pi/8)=1/sin(pi/8)
half-angle formula for sin: sin%28x%2F2%29=sqrt%28%281-cos%28x%29%29%2F2%29

csc(pi/8)=1/sin(pi/8)=2%2Fsqrt%282-sqrt%282%29%29
..
Check with calculator (use sin function)
sin(pi/8)≈0.3827..
exact value=sqrt%282-sqrt%282%29%29%2F2≈0.3827..


Question 750324: pls i need help
At an airport, a flight controller was in a tower 40m above the ground. When she first observed a plane it was at an angle of elevation of 12 (degrees) from her line of sight and flying at a constant altitude (height)
The plane had a constant speed of 360km/h
Eight seconds later the plane flew directly overhead.
On reaching the tower, the plane climbed for the next 15 seconds, without changing speed, to reach a new altitude of 1500 meters above the ground.
What was the plane's angle of ascent during this time?
i dont know if i have to convert the velocity to m/s. i got the final answer and it's 59degrees im really not sure about my answer
can u please clarify the solutions u get thanks very much

Answer by dkppathak(23) About Me  (Show Source):
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speed 360km/h
convert into m/s=100m/s
air craft travels for 8 sec to come overhead
distance traveled in 8 sec=800 m
it means aircraft was 800 m away at 12 degree from tower to altitude of aircraft
using trigonometry
we can say tan 12 = altitude of air craft/800
altitude of air craft from tower = 800x0.212 =169.6 m
as we know tan 12 =0.212
altitude of aircraft from ground= 169.6+40 =209.6 m
from over head of tower let elevation ang is x degree
it travel 15 sec at x degree to gain altitude upto 1500 m
difference of altitude gain by aircraft in 15 sec after tower =1500-209.6=1290.4
sin x=1290.4/1500=0.926
sin x=0.926
x= sin inverse of 0.926 =67.81 degree
answer 67.81 degree


Question 750106: For the equation y=2cos5(theda)
find the: range (which I found as [-2, 2])
the period: (which I found as 2pi/5)
the frequency: this is what I am having a problem finding
Please help! Thank you.

Answer by lwsshak3(6469) About Me  (Show Source):
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To determine frequency you need to know the time it takes to complete one period.
For example, if it takes one second to complete a period, the frequency is one cycle/second.
If it takes 2 seconds to complete a period, the frequency is 1/2 cycle/second, etc.


Question 750137: I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
I doubled the 15 degree angle and created a 30-60-90 triangle so I know the new hypotenuse is twice the new short side, and that the common long side is root three times the short side. I also have the Pythagorean relations for the two right triangles, so I have a system of equations, but my system resolves to an identity. I need to introduce some new information.
Thanks, John

Found 3 solutions by MathTherapy, Alan3354, JoeTaxpayer:
Answer by MathTherapy(1421) About Me  (Show Source):
You can put this solution on YOUR website!
I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
I doubled the 15 degree angle and created a 30-60-90 triangle so I know the new hypotenuse is twice the new short side, and that the common long side is root three times the short side. I also have the Pythagorean relations for the two right triangles, so I have a system of equations, but my system resolves to an identity. I need to introduce some new information.
Thanks, John

The easiest method is to use the difference of angles formula.

sin (A – B) = Sin A Cos B – Cos A Sin B

sin (60 - 45) = Sin 60 Cos 45 – Cos 60 Sin 45

sin 15%5Eo = Sin 60 Cos 45 – Cos 60 Sin 45

Length of hypotenuse = highlight_green%282%28sqrt%286%29+%2B+sqrt%282%29%29%29, or highlight_green%282sqrt%286%29+%2B+2sqrt%282%29%29 (in simplest radical form)

You can do the check!!

Send comments and “thank-yous” to “D” at MathMadEzy@aol.com

Answer by Alan3354(30983) About Me  (Show Source):
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I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
----------------
c is the hypotenuse
sin(15) = 2/c
c = 2/sin(15)
------
Use the Half-Angle formula to get sin(15)
sin%2815%29+=+%28sqr%282+-+sqr%283%29%29%2F2%29
c+=+2%2F%28sqrt%282+-+sqr%283%29%29%2F2%29
c+=+4%2F%28sqrt%282+-+sqr%283%29%29%29
c%5E2+=+16%2F%282+-+sqrt%283%29%29
c%5E2+=+16%2A%282+%2B+sqrt%283%29%29
c+=+4sqrt%282+%2B+sqrt%283%29%29

Answer by JoeTaxpayer(93) About Me  (Show Source):
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Remember SOHCAHTOA
Sin15=2/hypotenuse
%28Sqr%282-Sqr%283%29%29%29%2F2+=+2%2Fhypotenuse

hypotenuse=4%2F%28Sqr%282-Sqr%283%29%29%29
this is ~ 7.727


Question 750083: state the amplitude, period, phase shift, and vertical shift for the function. Then graph the function. y=4cos[3theta+(3/2)pi]+2
Answer by lwsshak3(6469) About Me  (Show Source):
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state the amplitude, period, phase shift, and vertical shift for the function. Then graph the function. y=4cos[3theta+(3/2)pi]+2
***
Equation for cos function: y=Acos(Bx-C), A=amplitude, period=2π/B, phase shift=C/B
For given equation: y=4cos(3x+3π/2)+2
A=4
B=3
period=2π/B=2π/3
1/4 period=2π/12=π/6
C=3π/2
phase shift=C/B=(3π/2)/3=π/2 (shift to the left)=3π/6
curve is bumped up 2 units
..
Graphing function:
On an x-y coordinate system with the x-axis scaled in radians, start with coordinates of the basic cos functon with period=2π/3,amplitude =1, phase shift=0, vertical shift=0, for one period.
coordinates of basic cos function:
(0,1), (π/6,0), (π/3,-1), (π/2,0), (2π/3,1), (5π/6,0)
..
shift curve π/2 to the left:
(-π/2,1), (-π/3,0), (-π/6,-1), (0,0), (π/6,1), (π/3,0)
..
change amplitude to 4:
(-π/2,4), (-π/3,0), (-π/6,-4), (0,0), (π/6,4), (π/3,0)
..
vertical shift up 2 units (final configuration)
(-π/2,6), (-π/3,2), (-π/6,-2), (0,2), (π/6,6), (π/3,2)
set x=0
y-intercept: 4cos(3x+3π/2)+2=4cos(3π/2)+2=-4+2=2

..
It is easy to make a mistake here. Please check my calculations.


Question 750206: Find the sec ɵ and tan ɵ if sin ɵ =1/5 and ɵ is in the 1st quadrant
Answer by Alan3354(30983) About Me  (Show Source):
You can put this solution on YOUR website!
Find the sec ɵ and tan ɵ if sin ɵ =1/5 and ɵ is in the 1st quadrant
-------------
cos = sqrt(1 - sin^2) = sqrt(24/25) = sqrt(24)/5
sec = 1/cos = 1/sqrt(24)/5 = 5/sqrt(24)
sec = 5*sqrt(24)/24
sec = 5*sqrt(6)/12
------------
tan = sin/cos = sin*sec
tan = sqrt(6)/12


Question 750191: Find all real solutions. Cos ɵ = 0.5 for 0 ≤ ɵ ≤ 2π
Answer by JoeTaxpayer(93) About Me  (Show Source):
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The Cos of 60 degrees gives you .5, as would the Cos of -60 degrees which is the same as 300 degrees.


Question 750034: Please help me
Solve the following equation for x, if 0 ≤ x ≤ 2π
a) 2cos²θ + cosθ - 1 = 0
b) sin²θ - sinθ = 0
c) √58cos(θ + 0.78) = -6
Thanks

Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(1480) About Me  (Show Source):
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Look at #a.


First, let w=cos%28theta%29, then 2w%5E2%2Bw-1=0
w=%28-1%2B-+sqrt%281-4%2A2%2A%28-1%29%29%29%2F%282%2A2%29
w=%28-1%2B-+3%29%2F4
w=-1 or w=1%2F2

Reverse the substitution.
cos%28theta%29=-1 or cos%28theta%29=1%2F2

highlight%28theta=pi%29 or highlight%28theta=pi%2F3%29, or also 5%2Api%2F3

Answer by lwsshak3(6469) About Me  (Show Source):
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Solve the following equation for x, if 0≤ x ≤ 2π
a) 2cos²θ + cosθ - 1 = 0
(2cosx-1)(cosx+1)=0
cosx=1/2
x=π/3, 5π/3
or cosx=-1
x=π
..
b) sin²θ - sinθ = 0
sinx(sinx-1)=0
sinx=0
x=0,π
sinx=1
x=π/2
..
c) √58cos(θ + 0.78) = -6
cos(x+.78)=-6/√58≈-0.79
cos^-1(-0.79)≈2.48 in Q2
x+.78=2.48
x=2.48-.78=1.7


Question 750048: Please Help
Please help me
Solve the following equation for x, if 0 ≤ x ≤ 2π
√2cosx=-1

Thanks

Answer by tommyt3rd(449) About Me  (Show Source):
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√2cosx=-1
cosx=-1%2Fsqrt%282%29=-sqrt%282%2F2%29

The principle value is pi%2F4
but our angle is negative so that puts our angle in quadrants II and III, meaning our solutions are

%283pi%29%2F4 or %285pi%29%2F4


Question 750028: Hello, Please help me with this question
Solve the following equation for x and write answer in form of a + bi with simplified surds.
2x^2 - 4x + 3 = 0
Thank You,

Answer by tommyt3rd(449) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-4x%2B3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-4%29%5E2-4%2A2%2A3=-8.

The discriminant -8 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -8 is + or - sqrt%28+8%29+=+2.82842712474619.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-4%2Ax%2B3+%29


Question 749967: How to graph y=sin2(x-π/4)+1

Answer by stanbon(57246) About Me  (Show Source):
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How to graph y=sin2(x-π/4)+1
-----
graph%28400%2C400%2C-7%2C7%2C-3%2C3%2Csin%5E2%28x-%28pi%2F4%29%29%2B1%29
--------------
cheers,
Stan H.
============


Question 749968: How to graph y=-sin(x+π/4) and what the looks like graphed
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(57246) About Me  (Show Source):
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How to graph y=-sin(x+π/4) and what the graph looks like.
graph%28400%2C400%2C-7%2C7%2C-2%2C2%2C-sin%28x%2B%28pi%2F4%29%29%29
-------------------
Cheers,
Stan H.

Answer by josmiceli(9660) About Me  (Show Source):
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It's a pure sine wave that is phase-shifted by +pi%2F4+
Make a chart with 3 headings:
x --------- x + pi/4 ---------- sin( x + pi/4 )
------------------------------------------
0 --------- 0 + pi/4 ---------- sin( pi/4 )
pi/4 ------ pi/4 + pi/4 -------- sin( pi/2 )
pi/2 ------ pi/2 + pi/4 -------- sin( ( 3*pi)/4 )
Keep on going and calculate the sines
You will end up with the +pi%2F4+
phase-shifted sine wave.
Here's a plot of both
+sin%28+x+%29+ and +sin%28+x+%2B+pi%2F4+%29+
+graph%28+600%2C+300%2C+-10%2C+10%2C+-1%2C+1%2C+sin%28x%29+%2C+sin%28+x+%2B+pi%2F4+%29+%29+


Question 749958: if 15x-7+6y=20 what is the value of 10x+4y-5?

Answer by reviewermath(507) About Me  (Show Source):
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15x - 7 + 6y = 20, add 7 to each side of the equation
15x + 6y = 27, multiply both sides by 2/3
10x + 4y = 18, subtract 5 from both sides of the equation
10x + 4y - 5 = highlight%2813%29


Question 749617: Simplifying trig identities.
2tanx sinx + 2cosx-cscx
Please show steps

Answer by josgarithmetic(1480) About Me  (Show Source):
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Given, shown in abbreviated form: 2tan*sin+2cos-csc

2(sin/cos*)sin+2*cos-1/sin
2(sin/cos)*sin*(sin/sin)+2*cos*(cos/cos)*(sin/sin)-(1/sin)*(cos/cos)
(2sin^3+2(cos^2)sin-cos)/(sin*cos)
(2sin*(sin^2+cos^2)-cos)/(sin*cos)
Notice the identity expression sin^2+cos^2=1;
2sin*(1-cos)/(sin*cos)
.
2/cos-2cos/cos
highlight%282sec-2%29 or highlight%282%28sec-1%29%29

For some unknown reason, rendering was not working, so I removed the triple brackets tags on most of the steps. I know this makes reading more difficult.


Question 749317: 1/(sec(θ)-tan(θ))=sec(θ)+tan(θ) how do I prove this?
Answer by JoeTaxpayer(93) About Me  (Show Source):
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1/(sec(θ)-tan(θ))=sec(θ)+tan(θ)

(sec(θ)-tan(θ)) (sec(θ)+tan(θ)) = 1 (we cross multiplied)

(sec^2(θ)-tan^2(θ)) = 1 ( (a+b) * (a-b) = a^2 - b^2 )

tan^2(θ) + 1 = sec^2(θ) (rearranged)

sin^2(θ)/cos^2(θ) + cos^2(θ)/cos^2(θ)= 1/cos^2(θ)

[Here, Tan = Sin/Cos, Substitute Cos/Cos for 1 and 1/Cos for Sec]

sin^2(θ)+cos^2(θ)^2=1 (multiply both sides by cos^2)

Above equation is the pythagorean theorum. Solved.



Question 749613: Please help me prove the following identities:
a) cot^2σsin^2σ + sin^2σ=1
b) cosσsecσ/1+tan^2σ=cos^2σ
c) sin2x/1+cos2x=tanx
d) csc2A = cot2A= cotA
Thank you,
-Nie

Answer by lwsshak3(6469) About Me  (Show Source):
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Please help me prove the following identities:
a) cot^2σsin^2σ + sin^2σ=1

..
b) cosσsecσ/1+tan^2σ=cos^2σ

..
c) sin2x/1+cos2x=tanx

d) csc2A = cot2A= cotA
??


Question 749590: Simplifying Basic Trig identities.
tanx cotx + sinx secx
please show work.

Answer by tommyt3rd(449) About Me  (Show Source):

Question 749600: please help me on this: Solve the following equations for x, if 0 ≤ x ≤ 2pi
a) sinx=(-1/2)
b)sqrt(2)x= -1
c) 3tan + sqrt(3)= 0
d) 8cos^2= 4
I'd really appreciate it if someone can help me on this, Thank You so much!

Answer by lwsshak3(6469) About Me  (Show Source):
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please help me on this: Solve the following equations for x, if 0 ≤ x ≤ 2pi
a) sinx=(-1/2)
x=7π/6, 11π/6 (Q3 and Q4 where sin<0)
..
b)sqrt(2)x= -1
??
..
c) 3tan + sqrt(3)= 0
tanx=-√3/3
x=5π/6, 11π/6 (Q2 and Q4 where tan<0)
..
d) 8cos^2= 4
cos^2x=4/8=1/2
cosx=1/√2
x=π/4, 7π/4 (Q1 and Q4 where cos>0)


Question 749596: Verifying Trig identities
+%28sinx%2Bcosx%29%5E2+=+%28+2%2Bsecx+cscx+%2F+secx+cscx+%29+ The 2 is on the numerator.

Answer by lwsshak3(6469) About Me  (Show Source):
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+%28sinx%2Bcosx%29%5E2+=+%28+2%2Bsecx+cscx%29+%2F%28+secx+cscx+%29+
start with left side
+%28sinx%2Bcosx%29%5E2

verified: left side=right side


Question 749602: cosy/(1-siny)= (1+siny)/cosy
Answer by stanbon(57246) About Me  (Show Source):
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cosy/(1-siny)= (1+siny)/cosy
------------
Cross-multiply to get:
cos^2(y) = 1-sin^2(y)
------
cos^2(y) = cos^2(y)
----
Solution y can have any value that does not make cos(y) = 0
and sin(y) = 1
=======================
Cheers,
Stan H.


Question 749575: Please help me find the exact value of this trig. function: 6sin75-6sin15
Answer by stanbon(57246) About Me  (Show Source):
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6sin75-6sin15
-----
6*sin(30+45) - 6*sin(60-45)
-----
= 6[sin(30)(cos(45))+(cos(30)(sin(45)] - 6[sin(60)*cos(45)-cos(60)sin(45)]
-------
= 6[(1/2)(sqrt(2)/2)+(sqrt(3)/2)(sqrt(2)/2)]- 6[(sqrt(3)/2)(sqrt(2)/2)-(1/2)(sqrt(2)/2]
---------
= 6[sqrt(2)/4 + sqrt(6)/4] - 6[sqrt(6)/4 - sqrt(2/4]
---------
= 6[(sqrt(2)+sqrt(6))/4] - 6[(sqrt(6)-sqrt(2))/4]
---------
= (3/2)[sqrt(2)+sqrt(6)] - (3/2)[sqrt(6)-sqrt(2)]
----
= 3*sqrt(2)
===========
Cheers,
Stan H.
===========


Question 749530: Simplifying Basic Trig identities
(1+sin^2 x ) (1-sin^2 x )

Answer by Alan3354(30983) About Me  (Show Source):
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(1+sin^2 x ) (1-sin^2 x )
-------
= 1 - sin^4(x)


Question 749495: 1) If tan equals 0.4, then what is sec equal to?

Answer by lwsshak3(6469) About Me  (Show Source):
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If tan equals 0.4, then what is sec equal to?
≈1.077


Question 749375: cos x- 5 sin x=0 (x is between 0 and 360 degrees)
The answers I got are 0,11.3,180,191.3 and 360 degrees.
May i please know why 0, 180 and 360 degrees are invalid?

Answer by KMST(1868) About Me  (Show Source):
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cos%280%5Eo%29=cos%28360%5Eo%29=1 and
sin%280%5Eo%29=sin%28360%5Eo%29=0 so
cos%280%5Eo%29-5sin%280%5Eo%29=1-5%2A0=1 and cos%28360%5Eo%29-5sin%28360%5Eo%29=1-5%2A0=1

cos%28180%5Eo%29=-1
sin%28180%5Eo%29=0 so
cos%28180%5Eo%29-5sin%28180%5Eo%29=-1-5%2A0=-1


Question 749108: Just started summer Trig review for pre-calc/ Calc. Could be going better.
I understand that S = R where S = 1, but they are using real numbers now in terms of 't' and I'm unsure. The question asks...
"Sketch the oriented arc on the unit circle corresponding to each of the following real numbers."
1.) t = 3pi/4 (which I got; 135 degrees counter clockwise, Quadrant II angle)
2.) t = -2pi (also good; 1 full rotation clockwise, quadrilateral angle)
3.) t = -2 (lost me)
4.) t = 117 (same issue as the last).
-How do I handle regular old numbers in these scenarios?

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
1.) t = 3pi/4 (which I got; 135 degrees counter clockwise, Quadrant II angle)
2.) t = -2pi (also good; 1 full rotation clockwise, quadrilateral angle)
3.) t = -2 (lost me)
4.) t = 117 (same issue as the last).
***
3π/4, -2π, -2, and 117 are in radians which are real numbers as you said.
To illustrate:
1.) t = 3pi/4 =2.3532..radians=135º
2.) t = -2pi=-6.28 radians=0
3.) t = -2=-2 radians=-2/π*180≈-114.59º
4.) t = 117=117 radians=117/π*180≈6703.61º
note:special angles like 30º, 60º, 45º, etc are written in terms of π for ease of calculation and explanation but angles in-between are expressed as real numbers without π.
Hope this helps.


Question 749158: sin[tan^-1(5/12)-sin^-1(-1)]
I need to solve this equation for 0 I am familiar with the Product to sum formulas but I do not see where that would play in here because of the arctan(5/12). I know that tan is sin/cos. I tried to see if I can maybe solve it knowing that much, but I cannot seem to figure it out.
Any help or feedback would be great! Thank you!
Mayra

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
sin[tan^-1(5/12)-sin^-1(-1)]
let x=the angle whose tan=5/12
tan(x)=5/12
hypotenuse=13 (5-12-13 right triangle)
sin(x)=5/13
cos(x)=12/13
..
let y=the angle whose sin=-1
sin(y)=-1
cos(y)=√(1-sin^2(y))=0
...
sin[tan^-1(5/12)-sin^-1(-1)]=sin(x-y)=sin(x)cos(y)-cos(x)sin(y)=5/13*0-12/13*(-1)=12/13
..
Check: (with calculator)
tan(x)=5/12
x=22.62º
sin(y) =-1
y=270º
x-y=(22.62-270)=-247.38 (in Q3)
reference angle:67.38º
sin(x+y)=sin(67.38º)≈0.923..
exact value=12/13≈0.923..


Question 748964: X Y
-3 0
-2 -4
-1 -7
0 -6
1 -4
2 0
3 6
What are the zeros of this function?

Answer by MathLover1(6622) About Me  (Show Source):
You can put this solution on YOUR website!

X|%09Y
-3|%090
-2| -4
-1|%09-7
0|%09-6
1|%09-4
2|%090
3|%096
What are the zeros of this function? A value of x which makes a function y equal 0. The zeros of a function are the x coordinates of the x intercepts of the graph of f%28x%29 or y.
so, your zeros are:
-3|%090 and 2|%090

if x=-3 => y=0
if x=2 => y=0



Question 748920: Solve the equation
Solve tan θ = 1/√3 for θ, where 0 ≤ θ ≤ π/2

Answer by Cromlix(269) About Me  (Show Source):
You can put this solution on YOUR website!
tan θ = 1/√3 for θ, where 0 ≤ θ ≤ π/2
tan theta = pi/6


Question 748850: how do you solve 1/cot(pi/4)- 2/csc(pi/6) without using a calculator
Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
how do you solve 1/cot(pi/4)- 2/csc(pi/6) without using a calculator
cot(π/4)=1
csc(π/6)=1/sin(π/6)=1/(1/2)=2
1/cot(pi/4)- 2/csc(pi/6)=1/1-2/2=0


Question 748758: Find the exact values of sin 2a, cos 2a, and tan 2a.
tan a=2, pi < a < 3pi/2

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
Find the exact values of sin 2a, cos 2a, and tan 2a.
tan a=2, pi < a < 3pi/2
***
tan(a)=2
hypotenuse=√(2^2+1)=√5
sin(a)=-2/√5
cos(a)=-1/√5
..
sin(2a)=2sin(a)cos(a)=2*-2/√5*-1/√5=4/5
cos(2a)=cos^2(a)-sin^2(a)=1/5-4/5=-3/5
tan(2a)=(2tan(a))/(1-tan^2(a))=4/(1-4)=-4/3
..
computer check:
tan(a)=2
a=243.13º (in quadrant III)
2a=486.86º-360=126.86º(in quadrant II)
reference angle=126.86-180=53.13º
..
sin(2a)=sin(53.1º)≈0.799...
exact value=4/5=0.800
..
cos(2a)=cos(53.1º)≈-0.600...
exact value=-3/5=-0.600
..
tan(2a)=tan(53.1º)≈-1.331...
exact value=-4/3=-1.333...


Question 748654: Show that the value of tan 3 α cot α cannot lie between 1/3and 3?
Answer by AnlytcPhil(1276) About Me  (Show Source):
You can put this solution on YOUR website!
tan3α·cotα;

It's easy if you're taking calculus, and can use
L'Hopital's rule, but not if you aren't.  Tell me 
in the thank-you note form if you are taking 
calculus or just taking trigonometry.  Then I'll
help you with it.  Is this familiar?

matrix%282%2C1%2C%0D%0A%0D%0Alim%2C%0D%0A%27x-%3En%27pi%29%28tan%283x%29%2Ftan%28x%29%29

Numerator and denominator both approach 0, so we
can use L'hopital's rule:

matrix%282%2C1%2C%0D%0A%0D%0Alim%2C%0D%0A%27x-%3En%27pi%29%283sec%5E2%283x%29%2F%28sec%5E2%28x%29%29%29

Using reciplocal identity

matrix%282%2C1%2C%0D%0A%0D%0Alim%2C%0D%0A%27x-%3En%27pi%29%283cos%5E2%28x%29%2F%28cos%5E2%283x%29%29%29 = 3

That's the start of the way to do it by calculus.
I'll finish it if you tell me whether you're 
taking calculus or just trig.

Edwin








Question 748244: what is the exact value if 0 < x < pi/2 and 0 < y < pi/2, of the cos (x+y) if the tan x = 5/3 and the sin y= 1/3

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
what is the exact value if 0 < x < pi/2 and 0 < y < pi/2, of the cos (x+y) if the tan x = 5/3 and the sin y= 1/3
***
tanx=5/3

cos(x)=1/sec(x)=3/√34)
sin(x)=5/√34)
..
sin(y)=1/3
cos%28y%29=sqrt%281-sin%5E2%28y%29%29=sqrt%281-1%2F9%29=sqrt%288%29%2F3
...
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)=[3/√34*√8/3]-[5/√34*/3)]









Question 748516: Which of these statements are true??
(i) The domain of the inverse cosine function (y = cos–1 x) is –1 ≤ x ≤ 1.
(ii) The range of the inverse sine function (y = sin–1 x) is 0 ≤ y ≤ p.
(iii) The range of the inverse tangent function (y = tan–1 x) is –p/2 < y < p/2.

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
Which of these statements are true??
(i) The domain of the inverse cosine function (y = cos–1 x) is –1 ≤ x ≤ 1.
(ii) The range of the inverse sine function (y = sin–1 x) is 0 ≤ y ≤ p.
(iii) The range of the inverse tangent function (y = tan–1 x) is –p/2 < y < p/2.
***
(iii) is true


Question 748448: What are the x's of: ((csc^2)x)= csc(x) +2
Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
What are the x's of: ((csc^2)x)= csc(x) +2
For[0,360º]
((csc^2)x)= csc(x) +2
csc^2x-cscx-2=0
(csc+2)(csc-1)=0
..
csc+2=0
cscx=-2=1/cosx
cosx=-1/2
x=240º, 300º
..
csc-1=0
cscx=1=1/cos
cosx=1
x=0


Question 748635: cos 2x – 3sin x cos 2x = 0 for the principal value(s) to
two decimal places

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
cos 2x – 3sin x cos 2x = 0 for the principal value(s) to
two decimal places
***
cos 2x–3sinxcos2x = 0
factor out cos2x
1-3sinx=0
3sinx=1
sinx=1/3
x=19.47º


Question 748251: Use the given information to find sin 2theta, cos 2theta, and tan 2theta.
#1) cos theta = 4/5 , 0 deg < theta < 90 deg
#2) sin theta = 1/3, 0 < theta < pi/2

Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
Use the given information to find sin 2theta, cos 2theta, and tan 2theta.
#1) cos theta = 4/5 , 0 deg < theta < 90 deg
#2) sin theta = 1/3, 0 < theta < pi/2
***
#1
cos(x)=4/5
sin(x)=3/5 (3-4-5) right triangle
tan(x)=sin(x)/cos(x)=3/4
...
sin(2x)=2sin(x)cos(x)=2*3/5*4/5=24/25
cos(2x)=cos^2(x)-sin^2(x)=16/25-9/25=7/25
tan(2x)=(2tan(x))/(1-tan^2(x)=(6/4)/(1-(9/16))=(24/16)/(7/16)=24/7
..
#2
sin(x)=1/3
adjacent side=√(3^2)-1)=√8
cos(x)=√8/3
tan(x)=1/√8=√8/8=√2/4
...
sin(2x)=2sin(x)cos(x)=2*1/3*√8/3=2√8/9
cos(2x)=cos^2(x)-sin^2(x)=8/9-1/9=7/9
tan(2x)=(2tan(x))/(1-tan^2(x))=(√2/2)/(1-(1/8))=(√2/2)/(7/8)=4√2/7
..


Question 748430: i need help trying to figure out this problem. sin^(2)x= cos^(2)x for 0 degrees is less than or equal to x is less than or 360 degrees
Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
sin^(2)x= cos^(2)x for 0 degrees is less than or equal to x is less than or 360 degrees
cos^2x-sin^2x=0
cos(2x)=0
2x=0
x=0


Question 748452: What are the x's of: (2(cos^2)x)+(3(cos x))=0
Answer by lwsshak3(6469) About Me  (Show Source):
You can put this solution on YOUR website!
What are the x's of: (2(cos^2)x)+(3(cos x))=0
2cos%5E2%28x%29%2B3cos%28x%29=0
divide by cos(x)
2cos%28x%29%2B3=0
2cos(x)=-3
cos(x)=-3/2
no solution: (-1 ≤ cosx ≤ 1)


Question 748450: What are the x's of: sin(x)+sin(-x)=2
Answer by josgarithmetic(1480) About Me  (Show Source):
You can put this solution on YOUR website!
Look at the unit circle to see how sine of x and sine of negative x are related. If you rotate a ray of 1 unit in the clockwise direction from the positive horizontal axis, then you have a -x. The sine of -x is the opposite of sine of positive x.

Your equation means,
sine%28x%29%2Bsine%28-x%29=2
sine%28x%29%2B%28-sine%28x%29%29=2
sine%28x%29-sine%28x%29=2
highlight%280=2%29
FALSE.


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