# SOLVEDDouble argument property?

#### wiseguy

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:

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

#### wiseguy

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...

#### HallsofIvy

MHF Helper
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).

wiseguy