Question

cosA cos2A cos4A Cos 8A = Sin2^4A/2^4 sinA

Answer 1

1.Multiply And Divide The LHS by sinA.

 2. Then use the formula sinA.cosA=(sin2A)/2.

3. Your LHS will be = (sin2A.cos2A.cos4A.cos8A)/(2.sinA)

4. Now use the same formula as used in step 2 but this time for sin2A.cos2A=(sin4A)/(2)
Therefore now the LHS will be = (sin4A.cos4A.cos8A)/(2.2.sinA)

5. Now follow the same procedure as in above steps till u reach cos8A.

6. The Result I Got Is= (sin16A)/(2^4)*(sinA)

P.S. The Question u posted has (sin2A)^4 in the Numerator.
Hope The Answer Helped. :-)

Answer 2

Answer:

cos A cos2A cos4A cos8A=(1)/(2^(4)sin A)*sin2^(4)A Prove that : cos A cos2A cos4A cos8A=(1)/(2^(4)sin A)*sin2^(4)A