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

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2 Answers
Aug 26, 2016

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#.

Aug 26, 2016

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.!