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

Explanation

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):costheta*cos2theta*cos2^(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):costheta*cos2theta*cos2^(2)theta. . .cos2^(k-1)theta*cos2^(k)theta`
`(sin2^(k)theta)/(2^(k)sintheta)*cos2^(k)theta`
`=(2sin2^(k)theta*cos2^(k)theta)/(2*2^(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.