Using the equation cos(a+b) = cos(a)cos(b) - sin(a)sin(b) or otherwise, show that cos(2x) = 2cos^2(x) - 1.

First let a = b = x such that:          

          cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

becomes:

          cos(x + x) = cos(x)cos(x) - sin(x)sin(x)

Leading to:

          cos(2x) = cos2(x) - sin2(x)

Using the fact that sin2(y) + cos2(y) = 1 or rearranged sin2(y) = 1 - cos2(y):

          cos(2x) = cos2(x) - (1 - cos2(y)) = 2cos2(x) - 1, as required.

Another suitable approach may involve the Maclaurin series of cos(2x) and cos2(x) to arrive at the required relation, although this is more involved.

BH

Related Maths A Level answers

All answers ▸

If y = 1/(x^2) + 4x, find dy/dx


Find the turning point(s) of the following function f(x) = x^2-2x+4. Determine whether the turning point is a minimum or maximum.


How do I find the maxima and minima of f(x) = e^(x^2)?


Solve for x (where 0<x<360) 2sin^2(x) - sin(x) - 1 = 0