Question

This is a trigonometric proof of a generalized case,question is in the details box?

Answer 1

Proof by induction is below.

Explanation:

Let's prove this identity by induction.

A. For `n=1` we have to check that

`(2cos(2theta)+1)/(2cos(theta)+1) = 2cos(theta)-1`

Indeed, using identity `cos(2theta)=2cos^2(theta)-1`, we see that

`2cos(2theta)+1 = 2(2cos^2(theta)-1)+1 = 4cos^2(theta)-1 =`
`=(2cos(theta)-1)*(2cos(theta)+1)`

from which follows that

`(2cos(2theta)+1)/(2cos(theta)+1) = 2cos(theta)-1`

So, for `n=1` our identity holds true.

B. Assume that the identity is true for `n`

So, we assume that

`(2cos(2^ntheta)+1)/(2cos(theta)+1) = Pi _(j in [0,n-1])[2cos(2^jtheta)-1]`

(symbol `Pi` is used for product)

C. Using assumption B above, let's prove the identity for `n+1`

We have to prove that from assumption B follows

`(2cos(2^(n+1)theta)+1)/(2cos(theta)+1) = Pi _(j in [0,n])[2cos(2^jtheta)-1]`

(notice that the right boundary for an index of multiplication is `n` now).

PROOF

Using an identity `cos(2x)=2cos^2(x)-1` for `x=2^ntheta`,

`2cos(2^(n+1)theta)+1 = 2cos(2*(2^n*theta))+1 =`

`= 2[2cos^2(2^ntheta)-1]+1 =`

`=4cos^2(2^ntheta)-1 =`

`=[2cos(2^ntheta)-1]*[2cos(2^ntheta)+1]`

Divide beginning and ending expressions by `[2cos(theta)+1]`, getting

`[2cos(2^(n+1)theta)+1]/[2cos(theta)+1] =`

`= [2cos(2^ntheta)-1]*[2cos(2^ntheta)+1] /[2cos(theta)+1]`

Now we use assumption B getting

`[2cos(2^(n+1)theta)+1]/[2cos(theta)+1] =`

`= [2cos(2^ntheta)-1]*Pi _(j in [0,n-1])[2cos(2^jtheta)-1] =`

`= Pi _(j in [0,n])[2cos(2^jtheta)-1]`

(notice the range of an index now is extended to `n`).

The last formula is exactly the same for `n+1` as original is for `n`. That completes the proof by induction that our formula is true for any `n`.

Answer 2

See the Proof in Explanation Section below.

Explanation:

This is equivalent to prove that,

#(2cosx+1)(2cosx-1)(2cos2x-1)(2cos4x-1)...(2cos2^(n-1)x-1) =(2cos2^nx+1)#

`"The L.H.S."={(2cosx+1)(2cosx-1)}(2cos2x-1)(2cos4x-1)...(2cos2^(n-1)x-1)`

`={4cos^2x-1}(2cos2x-1)(2cos4x-1)...(2cos2^(n-1)x-1)`

`={4((1+cos2x)/2)-1}(2cos2x-1)(2cos4x-1)....(2cos2^(n-1)x-1)`

`=(2cos2x+1)(2cos2x-1)(2cos4x-1)...(2cos2^(n-1)x-1)`

`=(4cos^2(2x)-1)(2cos4x-1)...(2cos2^(n-1)x-1)`

`=(2cos(2*2x)+1)(2cos4x-1)...(2cos2^(n-1)x-1)`

`=(2cos4x+1)(2cos4x-1)...(2cos2^(n-1)x-1)`

`=(2cos8x+1)...(2cos2^(n-1)x-1)`

`vdots`

`={2cos(2*2^(n-1)x)+1)}`

`=(2cos2^nx+1)`

`="the R.H.S."`

Enjoy Maths.!