SOLVED Double argument property?

Jul 2010
101
0
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°)\)
 
Jul 2010
101
0
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
Apr 2005
20,249
7,909
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).
 
  • Like
Reactions: wiseguy