Using Demoivre's Theorem prove that $ {\cos5 \theta} = 16{\cos^5 \theta} - 20{\cos^3 \theta} + 5{\cos \theta} $ .

Question

$ {\cos5 \theta} = 16{\cos^5 \theta} - 20{\cos^3 \theta} + 5{\cos \theta} $ .

Demoivre's Theorem

$$ \{\cos \theta + i \sin \theta \}^n = \cos n\theta + i\sin n\theta $$

Where n is an integer .

I tried

$$ \{\cos \theta + i \sin \theta \}^5 - i\sin 5\theta = \cos 5\theta $$

But then I cant eliminate sines .

Answer 1

You don't need to eliminate anything. Just equate the real parts on the LHS and the RHS.

Answer 2

Hint: $(\cos \theta \pm i\sin \theta)^5=\cos 5\theta \pm i\sin 5\theta$. Add them and simplify.

Answer 3

Notice, $$\cos5\theta=(\cos\theta+i\sin\theta)^5-i\sin 5\theta$$ $$=(\cos\theta+i\sin\theta)^2(\cos\theta+i\sin\theta)^3-i\sin 5\theta$$ $$=(\cos^2\theta-\sin^2\theta+2i\sin\theta\cos\theta)(\cos^3\theta-3\sin^2\theta\cos\theta-i\sin^3\theta+3i\sin\theta\cos^2\theta)-i\sin 5\theta$$ comparing the real parts on both the sides, one should get $$\cos5\theta=(\cos^2\theta-\sin^2\theta)(\cos^3\theta-3\sin^2\theta\cos\theta)$$$$+2\sin\theta\cos\theta(\sin^3\theta-3\sin\theta\cos^2\theta)$$

$$\cos5\theta=(2\cos^2\theta-1)(4\cos^3\theta-3\cos\theta)+2\cos\theta(1-\cos^2\theta)(1-4\cos^2\theta)$$ $$\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$