Prove the identity:

cos(3t)=cos^3(t)-3sin^2(t)cos(t)

Solutions 

To solve the given equation lets use double-angle theorem. Double-angle formulas allow the expression of trigonometric functions of angles equal to 2α in terms of α, which can simplify the functions and make it easier to perform more complex calculations, such as integration, on them.

cos(3t)=cos^3(t)-3sin^2(t)cos(t)

cos(A+B) = cosAcosB - sinAsinB 

Now cos(3t) = cos(2t+t) 

⇒cos(2t)cos(t) - sin(2t)sin(t) 

⇒(2cos^2(t)-1)cos(t) - 2sin(t)cos(t)sin(t) 

⇒ 2cos^(3)t - cos(t) - 2sin^2(t)cos(t) 

⇒ 2cos^3(t) - cos(t) - 2(1-cos^2(t))cos(t) 

⇒2cos^3(t) - cos(t) - 2cos(t) + 2cos^3(t) 

⇒4cos^3(t) - 3cos(t)