Find Radians: How to Use Sin to Solve for X

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To find sin(3x) when sin(x) = 1/3, use the sine of the sum of angles formula: sin(3x) = sin(2x + x). This requires applying the double-angle formulas for sin and cos to express sin(2x) and cos(2x) in terms of sin(x) and cos(x). Given sin(x) = 1/3 and knowing x is in the first quadrant, you can derive cos(x) using the Pythagorean identity. For the case of cos(2θ) = 1/3, apply the inverse cosine function to find θ, which allows you to subsequently determine cos(θ). Understanding these formulas and their applications is crucial for solving these types of problems.
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I do not need you to give me an answer just a formula

I have some problems to work and I cannot find anything through searching, I can make up a problem, but for example

find sin3x if sinx=1/3 and 0 < or equal to sin greater than or equal to pie/2

How would I find x? I now how to plug in a radian into my calculator to get the answer the opposite way but not this way.
 
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Put your calculator down. You don't need it for any of this. If you want the exact answer, you don't need it at all.

You need the formula for the sine of the sum of two angles.
sin(3x) = sin(2x + x) = ??

You'll get an expression involving sin(2x), sin(x), cos(2x), and cos(x).

Convert the sin(2x) and cos(2x) factors by using the double-angle formulas for sin and cos.

After you have done all that, you should have an expression that involves only sinx and cosx. You're given that sinx = 1/3, and I believe you are given than 0 < x < pi/2 (not pie/2). If you know that sinx = 1/3 and that x is in the first quadrant, what must cosx be?
 
How would I find it if the roles were reversed, let's say

cos2theta = 1/3 and find costheta?

I understand the double angle formula, but I do not understand howto apply it in this case.
 
pooker said:
How would I find it if the roles were reversed, let's say

cos2theta = 1/3 and find costheta?

I understand the double angle formula, but I do not understand howto apply it in this case.

The double-angle formula doesn't apply here. Instead, you need to use the inverse of the cosine function.

Given that cos(2\theta) = 1/3, then 2\theta = cos-1(1/3), so \theta = 1/2 cos-1(1/3).

Hopefully there are some restrictions on \theta, but once you know it, you can determine cos(\theta).
 

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