1. ## Double argument property?

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

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

My response:

$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: $cos 2x = 1 - 2sin2 x$ 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, $cos 2x = 1 - 2 sin2x$
This is incorrect- probably a typo. You mean $cos(2x)= 1- 2 sin^2(x)$

My response:

$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)".

$(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).