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

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

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.
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
:)