Question

PROVE➠
L.H.S=R.H.S
(N.C.E.R.T Class 12th)

$4cosxcos( \dfrac{\pi}{3} + x) cos( \dfrac{\pi}{3} - x) = cos3x$
​

Answer 1

Answer:

Since, cos(α−β)=1

⇒α−β=nπ,

But −2π≤α−β≤2π [ as, α,β∈(−π,π)]

∴α−β={0,−2π,2π} ...(i)

And α+β={2α,2α±2π}

Given, cos(α+β)=

e

1

⇒cos2α=

e

1

<1, which is true for four values of α. as −2π<2α<2π

Answer 2

$\star\:\:\:\sf\large\underline\blue{Question:-}$

• Prove that $\sf 4\cos(x) \cos( \frac{\pi}{3} + x) \cos( \frac{\pi}{3} - x ) = \cos(3x)$

$\star\:\:\:\sf\large\underline\blue{Proof:-}$

While solving this problem, we shall use these formulae.

• $\sf \cos(180 \degree - x) = - \cos(x)$
• $\sf2 \cos(x) \cos(y) = \cos(x + y) + \cos(x - y)$
• $\sf \cos( - x) = \cos(x)$

So, here the proof comes.

Taking LHS,

$\sf 4\cos(x) \cos( \frac{\pi}{3} + x) \cos( \frac{\pi}{3} - x )$

$\sf = 2 \cos(x) \{2 \cos( \frac{\pi}{3} + x) \cos( \frac{\pi}{3} - x ) \}$

$\sf = 2 \cos(x) \{ \cos( \frac{2\pi}{3} ) + \cos(2x) \}$

$\sf = 2 \cos(x) \{ \cos( \pi - \frac{\pi}{3} ) + \cos(2x) \}$

$\sf = 2 \cos(x) \{ \cos(- \frac{\pi}{3} ) + \cos(2x) \}$

$\sf = 2 \cos(x) \{ \cos(2x) + \frac{ - 1}{2} \}$

$\sf = 2 \cos(x) \cos(2x) - \cos(x)$

$\sf = \cos(x + 2x) + \cos(x - 2x) - \cos(x)$

$\sf = \cos(3x) + \cos( - x) - \cos(x)$

$\sf = \cos(3x) + \cos( x) - \cos(x)$

$\sf = \cos(3x)$

Taking RHS,

$\sf = \cos(3x)$

Therefore, LHS=RHS.

Hence Proved.