Derive the formula for the cos( 2t ) in terms of cos( t ) and sin( t )

Question

I want to know how to solve it because i'am having problems

Answer 1

If you don't know things like the addition identity, you can derive the formula by using the law of cosines.

 

Start with an isosceles triangle with legs of unit length and vertex angle 2t. Let x be the length of the base.

 

By the law of cosines [c²=a²+b²-2ab cos θ], you have x = √(2-2cos 2t).

 

Also, the altitude bisects the vertex angle producing two right triangles. Thus x=2 sin t.

 

Equating gives

 

2 sin t = √(2-2cos 2t)

 

4 sin² t = 2-2cos 2t

 

cos 2t = 1 - 2 sin² t.

 

We can two other formulas by using the pythagorean identity sin² t + cos² t = 1.

 

cos 2t = 1 - sin² t - (1 - cos² t) = cos² t - sin² t

 

and

 

cos 2t = 1-2(1-cos² t) = 2 cos² t - 1.

Answer 2

Addition identity:

cos (A+B) = CosAcosB - sinAsinB

 assuming A=B=t

 

cos(2t) = cos^2t - sin^2t

 

also sin^2t + cos^2t = 1

cos^2t = 1- sin^2t

 

substitite

 

cos2t = 1- sin^2t - sin^2t

cos2t = 1- 2sin^2t

 

You can also derive in terms of cos^2t

 

cos2t = 2cos^2t - 1

:)