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

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2 Answers

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.
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
:)