1. ## Double argument property?

I'm not quite sure if I'm answering the question...?

"Use the double argument property, $\displaystyle cos 2x = 1 - 2 sin2x$ to express cos 60° in terms of sin 30°."

My response:

$\displaystyle x=30° cos(2*30°)=1-2sin^2(30°) cos(60°)=1-2sin^2(30°) (cos(60°)-1)/-2)=sin^2(30°) 0.25=sin^2(30°) 0.5=sin(30°)$

2. The next question asks the exact same thing, except in reverse:

"Use the double argument property: $\displaystyle cos 2x = 1 - 2sin2 x$ to express cos 60° in terms of sin 30°."

Would I just take my work and swap everything over the equal sign? Assuming I approached this problem correctly...

3. Originally Posted by wiseguy
I'm not quite sure if I'm answering the question...?

"Use the double argument property, $\displaystyle cos 2x = 1 - 2 sin2x$
This is incorrect- probably a typo. You mean $\displaystyle cos(2x)= 1- 2 sin^2(x)$

My response:

$\displaystyle x=30° cos(2*30°)=1-2sin^2(30°) cos(60°)=1-2sin^2(30°)p$
Very good! And you are finished! You have "expressed cos(60) in terms of sin(30)".

$\displaystyle (cos(60°)-1)/-2)=sin^2(30°) 0.25=sin^2(30°) 0.5=sin(30°)$
You were not asked to evaluate cos(60) or sin(30).