Question

If 13a =
$\pi$
, then prove that
cos a cos 2a cos 3a cos 4a cos 5a cos 6a =1/64​

Answer 1

Consider the LHS.

$\begin{aligned}&\cos\alpha\ \cos(2\alpha)\ \cos(3\alpha)\ \cos(4\alpha)\ \cos(5\alpha)\ \cos(6\alpha)\end{aligned}$

We can rewrite this as,

$\cos(13\alpha-12\alpha)\ \cos(2\alpha)\ \cos(13\alpha-10\alpha)\ \cos(4\alpha)\ \cos(13\alpha-8\alpha)\ \cos(6\alpha)$

Since $13\alpha=\pi,$ $cos(13\alpha-\theta)=-\cos\theta.$ So it'll be,

$-\cos(2\alpha)\ \cos(4\alpha)\ \cos(6\alpha)\ \cos(8\alpha)\ \cos(10\alpha)\ \cos(12\alpha)$

Rearranging it as follows:

$-\cos(2\alpha)\ \cos(4\alpha)\ \cos(8\alpha)\ \cos(6\alpha)\ \cos(12\alpha)\ \cos(10\alpha)$

Now, on applying the formula,

$\cos A\ \cos(2A)\ \cos(4A)\ \cos(8A)\ \dots\ \cos(2^{n-1}A)=\dfrac{\sin(2^nA)}{2^n\sin A}$

we get,

$-\left[\cos(2\alpha)\ \cos(2\cdot2\alpha)\ \cos(2^2\cdot2\alpha)\right]\ \left[\cos(6\alpha)\ \cos(2\cdot6\alpha)\right]\ \cos(10\alpha)$

as,

$-\ \dfrac{\sin(2^3\cdot2\alpha)}{2^3\sin(2\alpha)}\cdot\dfrac{\sin(2^2\cdot6\alpha)}{2^2\sin(6\alpha)}\cdot\cos(10\alpha)$

or,

$-\ \dfrac{\sin(16\alpha)\ \sin(24\alpha)\ \cos(10\alpha)}{8\sin(2\alpha)\cdot4\sin(6\alpha)}\dfrac{}{}$

or,

$-\ \dfrac{\sin(26\alpha-10\alpha)\ \sin(26\alpha-2\alpha)\ \cos(10\alpha)}{32\ \sin(2\alpha)\ \sin(6\alpha)}$

But we should notice that $26\alpha=2\pi.$

Thus we can say that $\sin(26\alpha-\theta)=-\sin\theta.$

Then the modified LHS becomes,

$-\ \dfrac{-\sin(10\alpha)\cdot -\sin(2\alpha)\cdot \cos(10\alpha)}{32\ \sin(2\alpha)\ \sin(6\alpha)}$

or,

$-\ \dfrac{\sin(10\alpha)\ \cos(10\alpha)}{32\ \sin(6\alpha)}$

Multiplying and dividing both the numerator and the denominator by 2, we get it as,

$-\ \dfrac{2\ \sin(10\alpha)\ \cos(10\alpha)}{64\ \sin(6\alpha)}$

Since $2\sin A\cos A=\sin(2A),$ it'll be,

$-\ \dfrac{\sin(20\alpha)}{64\ \sin(6\alpha)}$

or,

$-\ \dfrac{\sin(26\alpha-6\alpha)}{64\ \sin(6\alpha)}$

Since it is known that $\sin(26\alpha-\theta)=-\sin\theta,$ the modified LHS will be,

$-\ \dfrac{-\sin(6\alpha)}{64\ \sin(6\alpha)}$

or simply,

$\dfrac{1}{64}$

This is nothing but the RHS.

Hence Proved!

Answer 2

Answer:

Takng LHS...

cos9A + cos3A + cos7A + cos5A

{ 2 cos ( 9A + 3A / 2 ) cos ( 9A - 3A / 2 ) } + { 2 cos ( 7A+ 5A / 2) cos ( 7A - 5A / 2 ) }

( 2 cos6A cos3A ) + ( 2 cos6A cosA )

2 cos6A ( cos3A + cosA )

2 cos6A { 2 cos ( 3A + A / 2) cos ( 3A - A / 2 ) }

2 cos6A ( 2 cos2A cosA )

4 cosA cos2A cos6A = RHS.....