Question

Prove that :Sin^3A +sin^3(2pi/3+A) + sin^3(4pi/3+A)= -3/4 sin3A

Answer 1

ANSWER:

To Prove:

$\:\:\bullet\:\:\sin^3A+\sin^3\left(\dfrac{2\pi}{3}+A\right)+\sin^3\left(\dfrac{4\pi}{3}+A\right)=-\dfrac{3}{4}\sin3A$

Proof:

We are given that,

$\implies\sin^3A+\sin^3\left(\dfrac{2\pi}{3}+A\right)+\sin^3\left(\dfrac{4\pi}{3}+A\right)=-\dfrac{3}{4}\sin3A$

Taking and Solving LHS,

$\implies\sin^3A+\sin^3\left(\dfrac{2\pi}{3}+A\right)+\sin^3\left(\dfrac{4\pi}{3}+A\right)$

We can re-write,

• 2π/3 as (π - π/3), and
• 4π/3 as (π + π/3)

So,

$\implies\sin^3A+\sin^3\left[\left(\pi-\dfrac{\pi}{3}\right)+A\right]+\sin^3\left[\left(\pi+\dfrac{\pi}{3}\right)+A\right]$

So,

$\implies\sin^3A+\sin^3\left(\pi-\dfrac{\pi}{3}+A\right)+\sin^3\left(\pi+\dfrac{\pi}{3}+A\right)$

We know that,

⇒ sin(π + ө) = -sin ө

So,

$\implies\sin^3A-\sin^3\left(-\dfrac{\pi}{3}+A\right)-\sin^3\left(+\dfrac{\pi}{3}+A\right)$

$\implies\sin^3A-\sin^3\left(A-\dfrac{\pi}{3}\right)-\sin^3\left(A+\dfrac{\pi}{3}\right)$

We know that,

$\hookrightarrow\sin3\theta=3\sin\theta-4\sin^3\theta$

So,

$\hookrightarrow\sin^3\theta=\dfrac{1}{4}(3\sin\theta-\sin3\theta)$

We had,

$\implies\sin^3A-\sin^3\left(A-\dfrac{\pi}{3}\right)-\sin^3\left(A+\dfrac{\pi}{3}\right)$

So,

$\implies\bigg[\dfrac{1}{4}(3\sin A-\sin3A)\bigg]-\bigg[\dfrac{1}{4}\left(3\sin\left(A-\dfrac{\pi}{3}\right)-\sin3\left(A-\dfrac{\pi}{3}\right)\right)\bigg]-\\\bigg[\dfrac{1}{4}\left(3\sin\left(A+\dfrac{\pi}{3}\right)-\sin3\left(A+\dfrac{\pi}{3}\right)\right)\bigg]$

Taking 1/4 common,

$\implies\dfrac{1}{4}\bigg[(3\sin A-\sin3A)-\left(3\sin\left(A-\dfrac{\pi}{3}\right)-\sin(3A-\pi)\right)-\left(3\sin\left(A+\dfrac{\pi}{3}\right)-\sin(3A+\pi)\right)\bigg]$

$\implies\dfrac{1}{4}\bigg[(3\sin A-\sin3A)-\left(3\sin\left(A-\dfrac{\pi}{3}\right)+\sin(\pi-3A)\right)-\left(3\sin\left(A+\dfrac{\pi}{3}\right)-\sin(\pi+3A)\right)\bigg]$

We know that,

⇒ sin(π + ө) = -sin ө, and

⇒ sin(π - ө) = sin ө

So,

$\implies\dfrac{1}{4}\bigg[(3\sin A-\sin3A)-\left(3\sin\left(A-\dfrac{\pi}{3}\right)+\sin3A\right)-\left(3\sin\left(A+\dfrac{\pi}{3}\right)+\sin3A\right)\bigg]$

So,

$\implies\dfrac{1}{4}\bigg[3\sin A-\sin3A-3\sin\left(A-\dfrac{\pi}{3}\right)-\sin3A-3\sin\left(A+\dfrac{\pi}{3}\right)-\sin3A\bigg]$

$\implies\dfrac{1}{4}\bigg[3\sin A-3\sin3A-3\sin\left(A-\dfrac{\pi}{3}\right)-3\sin\left(A+\dfrac{\pi}{3}\right)\bigg]$

Taking 3 common,

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-\sin\left(A-\dfrac{\pi}{3}\right)-\sin\left(A+\dfrac{\pi}{3}\right)\bigg]$

We know that,

⇒ sin C ± sin D = sin C cos D ± cos D sin C

So,

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-\left(\sin A\cos\left(\dfrac{\pi}{3}\right)-\cos A\sin\left(\dfrac{\pi}{3}\right)\right)-\left(\sin A\cos\left(\dfrac{\pi}{3}\right)+\cos A\sin\left(\dfrac{\pi}{3}\right)\right)\bigg]$

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-\sin A\cos\left(\dfrac{\pi}{3}\right)+\cos A\sin\left(\dfrac{\pi}{3}\right)-\sin A\cos\left(\dfrac{\pi}{3}\right)-\cos A\sin\left(\dfrac{\pi}{3}\right)\bigg]$

Cancelling cos A sin(π/3),

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-\sin A\cos\left(\dfrac{\pi}{3}\right)-\sin A\cos\left(\dfrac{\pi}{3}\right)\bigg]$

So,

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-2\sin A\cos\left(\dfrac{\pi}{3}\right)\bigg]$

We know that,

⇒ cos(π/3) = 1/2

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-2\!\!\!/\:\sin A\left(\dfrac{1}{2\!\!\!/}\right)\bigg]$

$\implies\dfrac{3}{4}\bigg[\sin A-\sin3A-\sin A\bigg]$

Cancelling sin A,

$\implies\dfrac{3}{4}(-\sin3A)$

So,

$\bf\implies-\dfrac{3}{4}sin3A =RHS$

As, LHS = RHS

Hence Proved!!