Using induction, prove that `cos theta · cos 2theta · cos 2^2 theta. ... cos 2^(n-1) theta =(sin2^n theta)/(2^n sin theta)`

Using induction, prove that `cos theta · cos 2theta · cos 2^2 theta.

1.	Using induction, prove that `cos theta · cos 2theta · cos 2^2 theta. ... cos 2^(n-1) theta =(sin2^n theta)/(2^n sin theta)`
Answer» Let P(n) : `costhetacos2theta. . . cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sintheta)` Step I For n=1,P(1): `costhetacos2theta. . . cos2^(n-1)theta=(sin2^(1)theta)/(2^(n)sintheta)` `=(sin2theta)/(2sintheta)=(2sinthetacostheta)/(2sintheta)=costheta` which is true. Step II Assume that P(n) is true, for n=k. `P(k):costhetacos2thetacos2^(2)theta. . .cos2^(k-1)theta=(sin2^(k)theta)/(2^(k)sintheta)` is true. Step III To prove P(k+1) is true. `P(k=1):costhetacos2thetacos2^(2)theta. . .cos2^(k-1)thetacos2^(k)theta` `(sin2^(k)theta)/(2^(k)sintheta)cos2^(k)theta` `=(2sin2^(k)thetacos2^(k)theta)/(22^(k)sintheta)` `=(sin2*2^(k)theta)/(2^(k+1)sintheta)=(sin2^((k+1))theta)/(2^(k+1)sintheta)` which is true. So, P(k+1) is true. Hence, P (n) is true.