Tutors Answer Your Questions about Trigonometry-basics (FREE)
Question 751524: cos 2x + cos x = 0 Answer by Cromlix(322) (Show Source):
You can put this solution on YOUR website!cos2x + cos x
replace cos 2x with 2cos^2-1
2cos^2-1 + cos x = 0
Rearrange
2cos^2x + cos x - 1 = 0
(2cosx - 1)(cosx + 1) = 0
cosx = 1/2 or -1
x = 60 degs, 300degs and 180degs
= pi/3 rads, 5pi/3 rads and pi
Question 751527: 2tanx + 2 = 4 Answer by Cromlix(322) (Show Source):
You can put this solution on YOUR website!Do you want this equation solved?
If so:
2tanx + 2 = 4
2tanx = 4 - 2
2tanx = 2
tanx =2/2
tanx = 1
x = 45 degrees, 225 degrees
= 0.78 rads , 3.92 rads
Question 751462: (a) prove(i have done this part already)
cos3a = 4cos^3 a - 3cosa
(b) Show that cos40 is a root of the equation 8x^3 - 6x +1 = 0
Need help with part (b) PLEASE Answer by Theo(3460) (Show Source):
You can put this solution on YOUR website!the second part sounds like it's a lot easier than the first part.
all you need to do is find the cosine of 40 degrees and replace x in the equation.
if the value of the equation comes out to be equal to 0, then cosine of 40 degrees is a root of the equation since the roots of the equation are when the value of y is equal to 0.
use your calculator to find cosine of 40 degrees.
that turns out to be equal to be equal to .766044443.
store that in your memory so as not to get rounding errors and use that stored result in place of x in the equation.
when you replace x in your equation with the stored value of .766044443, the result is 0 indicating that cosine of 40 degrees is a root of the equation.
alternately, you can graph your equation and find the roots (when the equation crosses the s-axis are the roots).
the graph of the equation actually crosses the x-axis in 3 places, and the third place (the rightmost intersection) is equal to the cosine of 40 degrees (.76604443).
the graph of the equation is shown below:
i drew a vertical line (well, almost vertical) at x = .766044443 to show you that the graph of the equation crosses the x-axis at that point.
Question 751312: Verify: (1-sin(x)/1+sin(x))=(sec(x)-tan(x))^2 Answer by lwsshak3(6494) (Show Source):
Question 751386: is there a shortcut to solve sin(x+45) or do I need to just apply sin(a+b)? Answer by josgarithmetic(1507) (Show Source):
You can put this solution on YOUR website!No. Actually, the formula for sin(a+b) IS the shortcut. For a quick reference see the Angle Sum and Difference Identities in http://en.wikipedia.org/wiki/Angle_sum_formula#Angle_sum_and_difference_identities
Question 751293: If sin(theta)= (sqrt2)/5, and pi/2 < theta < pi, then find the exact value of tan(2theta) using identities (NO inverse trig, or SOHCAHTOA) Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!If sin(theta)= (sqrt2)/5, and pi/2 < theta < pi, then find the exact value of tan(2theta) using identities
***
Identity: 
..

Identity:

..
Check with calculator:
sin(x)=√2/5
x≈16.43º
2x≈32.86º
tan(2x)=tan(32.86º)≈0.6459
exact value=  ≈0.6459
Question 751296: Find the Polar form of the Cartesian equation y=(1/2x).
Express answer as r=f(theta) Answer by tommyt3rd(528) (Show Source):
Question 751310: Verify: (sec(x)/sin(x))-(sin(x)/cos(x))=cot(x) Answer by stanbon(57327) (Show Source):
You can put this solution on YOUR website!Verify: (sec(x)/sin(x))-(sin(x)/cos(x))=cot(x)
-----
(1/(cos*sin)) - (sin/cos) = (cos/sin)
----
Multiply thru by cos*sin to get:
1 - sin^2 = cos^2
---
cos^2 = cos^2
=================
Cheers,
Stan H.
===============
Question 751285: simplify: sin^2(x)+cot^2(x)+cos^2(x) Found 2 solutions by Edwin McCravy, tommyt3rd: Answer by Edwin McCravy(8908) (Show Source): Answer by tommyt3rd(528) (Show Source):
Question 751291: Simplify: ((cos(x)/sin(x))+csc(x) Answer by jim_thompson5910(28593) (Show Source):
Question 751124: How do I find the height of the building if the adjacent=100mm and the angle opposite the right angle equal 38,7
Answer by Cromlix(322) (Show Source):
You can put this solution on YOUR website!Trigonometric ratio.
A right angled triangle.
Opposite - height of building
Adjacent - 100mm
Angle of elevation - 38.7 degrees
Tan 38.7 = opposite/100mm
Tan 38.7 x 100mm = opposite
Height of building = 80.1mm
Don't you mean metres? mm - millimetres
If it was metres then building would
be 80.1 metres high.
Hope this helps
:-)
Question 751018: A=[1 -1 0]
[1 0 -1]
[6 -2 -3]
B=[-2 -3 1]
[-3 -3 1]
[-2 -4 1]
Solve for x: 3X+4B=A
If I could get some quick help it would be greatly appreciated. :) Answer by Edwin McCravy(8908) (Show Source):
Question 751017: Find the exact value of cos .
Some quick help would be greatly appreciated. :) Found 2 solutions by stanbon, kathydots21: Answer by stanbon(57327) (Show Source):
You can put this solution on YOUR website!Find the exact value of cos{11pi/6}.
-----
(11/6)pi is the same as -pi/6
----
So, cos(11pi/6) = cos(-pi/6) = cos(pi/6) = sqrt(3)/2
=========================
Cheers,
Stan H.
=========================
Answer by kathydots21(7) (Show Source):
You can put this solution on YOUR website!http://www.mathway.com/answer.aspx?p=trig?p=cos%2811SMB19SMB106%29?p=358?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=FindSMB15theSMB15ExactSMB15Value
go here
Question 751016: Find the exact value of sec .
Some quick help would be greatly appreciated. :) Answer by kathydots21(7) (Show Source):
Question 751000: At what point does the minimum value of the quadratic function occur?
Thanks. Answer by stanbon(57327) (Show Source):
You can put this solution on YOUR website!At what point does the minimum value of the quadratic function  occur?
----
Minimum occurs when x = -b/(2a) = -2/(2*1) = -1
---
f(-1) = (-1)^2+2(-1) = 1 - 2 = -1
--------
Min: (-1,-1)
===============
Cheers,
Stan H.
Question 750959: a = 6in , b = 8 in , c = ?
I trust you are offering two sides of a triangle.
6^2 = 36
8^2 - 64
36+64 = 100
Sqr(100) = 10
c=10 in Answer by JoeTaxpayer(112) (Show Source):
Question 750918: log3x-log5=1 Answer by lwsshak3(6494) (Show Source):
Question 750672: tan^2(theta)-3sec(theta)+3=0
Solve for values on the unit circle in value of pi Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!tan^2(theta)-3sec(theta)+3=0
Solve for values on the unit circle in value of pi.
***

LCD:cos^2(x)

2cos^2(x)-3cos(x)+1=0
(2cos(x)-1)(cos(x)-1)=0
..
2cosx-1=0
cosx=1/2
x=π/3, 5π/3 (in Q1 and Q4 where cos>0)
..
cosx-1=0
cosx=1
x=0
Question 750741: What is the phase shift of y = -csc(3x – pi)? Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!What is the phase shift of y = -csc(3x – pi)?
form of equation for csc function:y=csc(Bx-C)
For given function:
B=3
C=π
phase shift=C/B=π/3 (shift to the right)
Question 750761: (-3,3) is on the terminal side of theta, what is the cosine of theta Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!(-3,3) is on the terminal side of theta, what is the cosine of theta.
reference angle in quadrant II
hypotenuse of reference right triangle=√(3^2+3^2)=√18=3√2
cos theta=-3/(3√2)=-√2
Question 750782: What are the x's of: sec^2(x)+tan(x)-1 Answer by tommyt3rd(528) (Show Source):
Question 750777: What are the x's of: csc^2(x)=csc(x)+2 Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!What are the x's of:
0 ≤ cos(x) ≤ 2π
csc^2(x)=csc(x)+2
csc^2(x)-csc(x)-2=0
(csc(x)-2)(csc(x)+1)=0
csc(x)-2=0
csc(x)=2
sin(x)=1/2
x=π/6, 5π/6
(csc(x)+1=0
csc(x)=-1
sin(x)=-1
x=π
x's are π/6, 5π/6 and π
Question 750779: What are the x's of: 2cos^2(x)+3cos(x)=0 Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!What are the x's of:
2cos^2(x)+3cos(x)=0
0 ≤ cos(x) ≤ 2π
2cosx(cosx+3)=0
2cosx=0
x=π/2, 3π/2
cosx+3)=0
cosx=-3 (reject, not in domain)
Question 750800: Use algebra to solve the equation 4cos3x=-2/sin3x over the domain 0 greater than or equal to x and less than or equal to pi/3. Give exact answers. Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!Use algebra to solve the equation 4cos3x=-2/sin3x over the domain 0 greater than or equal to x and less than or equal to pi/3. Give exact answers.
***
0 ≤ x ≤ π/3
4cos3x=-2/sin3x
4cos3xsin3x=-2
divide by 2
2cos3xsin3x=-1
sin6x=-1
6x=3π/2
x=3π/12=π/4
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(528) (Show Source):
Question 749962: Find the exact value of csc(pi/8) Found 2 solutions by Alan3354, lwsshak3: Answer by Alan3354(30993) (Show Source): Answer by lwsshak3(6494) (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:

csc(pi/8)=1/sin(pi/8)= 
..
Check with calculator (use sin function)
sin(pi/8)≈0.3827..
exact value=  ≈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(34) (Show Source):
You can put this solution on YOUR website!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(6494) (Show Source):
You can put this solution on YOUR website!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(1423) (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  = Sin 60 Cos 45 – Cos 60 Sin 45
Length of hypotenuse =  , or  (in simplest radical form)
You can do the check!!
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
Answer by Alan3354(30993) (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.
----------------
c is the hypotenuse
sin(15) = 2/c
c = 2/sin(15)
------
Use the Half-Angle formula to get sin(15)
Answer by JoeTaxpayer(112) (Show Source):
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(6494) (Show Source):
You can put this solution on YOUR website!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(30993) (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(112) (Show Source):
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(1507) (Show Source): Answer by lwsshak3(6494) (Show Source):
You can put this solution on YOUR website!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(528) (Show Source):
You can put this solution on YOUR website!√2cosx=-1
The principle value is
but our angle is negative so that puts our angle in quadrants II and III, meaning our solutions are
 or
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(528) (Show Source):
Question 749967: How to graph y=sin2(x-π/4)+1
Answer by stanbon(57327) (Show Source):
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(57327) (Show Source): Answer by josmiceli(9671) (Show Source):
You can put this solution on YOUR website!It's a pure sine wave that is phase-shifted by 
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 
phase-shifted sine wave.
Here's a plot of both
 and
Question 749958: if 15x-7+6y=20 what is the value of 10x+4y-5?
Answer by reviewermath(518) (Show Source):
You can put this solution on YOUR website!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 =
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