Answer to Question #88936 in Trigonometry for Manu Chemjong

Question

cos^3a.cos 3a+sin^3a.sin3a=cos^3 2a

Answer 1

You need the following identities: 

$$1. cos^2 X + sin^2 X = 1$$ $$2. sin(X+Y) = sin X cosY + sin Y cos X$$ $$3. cos(X-Y) = cos X cos Y + sin X sin Y$$ $$4. sin(2X)= 2*sin X cos X$$

Solution:

$$cos^3 (A) cos (3A) + sin^3 (A) sin (3A) =$$ $$cos^2 (A) (cos A cos (3A)) + sin^2 (A)(sin A sin (3A)) =$$

Formula 1 gives:

$$cos^2 (A) = 1 - sin^2 (A)$$

and

$$sin^2 (A) = 1 - cos^2 (A)$$

$$(1-sin^2 (A)) (cos (A) cos (3A)) + (1-cos^2 (A))(sin (A) sin (3A)) =$$ $$cos (A) cos (3A) + sin (A) sin (3A) - sin^2 (A) cos (A) cos (3A) - cos^2 (A) sin (A) sin (3A) =$$ $$cos (A) cos (3A) + sin (A) sin( 3A) - sin( A )cos (A)(sin (A) cos (3A) + sin (3A) cos(A))=$$

Formua 2 gives:

$$sin (A) cos (3A) + sin (3A) cos (A) = sin (4A)$$

Formua 3 gives:

$$cos (A) cos (3A) + sin (A) sin (3A) = cos (2A)$$ $$cos (2A) - sin (A) cos (A) (sin (4A)) =$$

Formua 4 gives:

$$sin (4A) = 2*sin (2A) cos (2A)$$ $$2*sin (A) cos (A) = sin (2A)$$ $$cos (2A) - 2*sin (A) cos (A) (sin (2A) cos (2A)) =$$ $$cos (2A) - sin (2A) sin (2A) cos( 2A) =$$ $$cos (2A) (1 - sin^2 (2A)) =$$

Using Formua 1 again :

$$cos( 2A) (cos^2 (2A)) =$$

$$cos^3 (2A)$$