# By the method of Induction prove that cos^2 A +cos^2 2A+cos^2 3A+....+cos^2 nA= 1/2((n+(cos n+1 A . sin nA)/sinA)) .First we prove it for n=1, it is ok. Then we take it to be true for...

By the method of Induction prove that

cos^2 A +cos^2 2A+cos^2 3A+....+cos^2 nA=

1/2((n+(cos n+1 A . sin nA)/sinA)) .

First we prove it for n=1, it is ok.

Then we take it to be true for n=k,(LHS=RHS),

We have to prove it to be true when n=k+1; that is RHS of P(k+1)=

1/2((k+1) +(cos k+2 A . sin k+1 A )/2))

### 1 Answer | Add Yours

`cos^2 A +cos^2 2A+cos^2 3A+....+cos^2 nA= 1/2(n+(cos (n+1)A*sin(nA))/sinA)`

When n = 1

LHS = `cos^2A`

RHS

`= 1/2(1+cos2A*sinA/sinA) `

`= 1/2(1+cos2A)`

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

`= cos^2A`

LHS = RHS

So for n=1 the result is true.

Let us assume for n = p where p is a positive integer the result is true.

Then;

`cos^2 A +cos^2 2A+cos^2 3A+....+cos^2 pA= 1/2(p+(cos (p+1)A*sin(pA))/sinA)`

Lets add cos^2(p+1)A to both sides of the above equation.

LHS = `cos^2 A +cos^2 2A+cos^2 3A+....+cos^2 pA+cos^2(p+1)A`

RHS

`= 1/2(p+cos(p+1)A*sin(pA)/sinA)+cos^2(p+1)A`

`= 1/2(p+cos(p+1)A*sin(pA)/sinA)+cos^2(p+1)A`

`= 1/2(p+cos(p+1)A*(sin(pA)/sinA+2cos(p+1)A)`

`= 1/2(p+cos(p+1)A*(sinpA+2cos(p+1)A*sinA)/sinA)`

We know that

`2sin((x-y)/2)*cos((x+y)/2) = sinx-siny`

`2cos(p+1)A*sinA`

`= sin(p+2)A-sinpA`

RHS

`= 1/2(p+cos(p+1)A*(sinpA+2cos(p+1)A*sinA)/sinA)`

`= 1/2(p+cos(p+1)A*(sinpA+sin(p+2)A-sinpA)/sinA)`

`= 1/2(p+cos(p+1)A*sin((p+2)A)/sinA)`

`= 1/2(p+1/2(sin(2p+3)-sin(-A))/sinA)`

`= 1/2(p+1/2(sin(2p+3)+sinA)/sinA)`

`= 1/2(p+1/2(sin(2p+3)-sinA+2sinA)/sinA)`

`= 1/2(p+1/2((sin(2p+3)-sinA)/sinA)+1)`

`= 1/2(p+1+1/2*2cos(p+2)A*sin((p+1)A)/sinA)`

`= 1/2(p+1+cos(p+1+1)A*sin((p+1)A)/sinA`

This represent the answer of n = p+1

So from mathematical induction for all `n in Z^+` the result is true.

**Sources:**