Prove that Cos(2x) = -Sin(x)

Question

Took a quiz in precal and this question completely stumped me to the point of writing it down to try and figure it out later. I know you would start with the left side but that’s as far as I have gotten.

Answer 1

for x=0,  cos(2*0) =1 while -sin(0) = 0

 

for x=pi, cos(2*pi) = 1 while -sin(pi) = 0

 

for x = pi/2, the cosine is 1 while the sine is -1

 

So no, they are not EXACTLY the same.

 

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But we can find where they are equal:

 

cos(2x) = - sin(x)

1 - 2*(sinX)^2 = -sinX

Let z = sinX

1 - 2z^2 = -z

1 + z - 2z^2 = 0


2z^2 - z - 1 = 0

( 2z + 1)( z - 1 ) = 0

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2z + 1 = 0 ---> z = -1/2

sin(x) = -1/2

x = 210, 330

x = (7/6)*pi, or x = (11/6)*pi
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z - 1 = 0 ---> z =1

sin(x) = 1

x = 90 ---> x = (pi/2)