2

Question :

$\sin\beta\sin\left(\dfrac\pi3-\beta\right)\sin\left(\dfrac\pi3+\beta\right)=\frac {1}{4}\sin 3\beta$

I tried using the sine expansion but could not get it. Please help me.

Akash Yadav
  • 1,167
Myst1cal
  • 267
  • https://math.stackexchange.com/questions/1145406/prove-that-cos-x-cdot-cosx-60-circ-cdot-cosx60-circ-frac14 – lab bhattacharjee Apr 09 '20 at 03:46
  • Try using $\sin x=(e^{ix}-e^{-ix})/(2i)$. – Gerry Myerson Apr 09 '20 at 03:56
  • There is a generalised version. See the long postscript (under the trigonometric proof and above the geometric proof) of this answer: https://math.stackexchange.com/questions/3523754/show-that-sin220-circ-sin40-circ-sin10-circ-sin30-circ-sin60-circ/3523768#3523768. A proof is as Gerry Myerson suggests in the comment above me. – Batominovski Apr 09 '20 at 04:09

1 Answers1

1

Firstly I will give you an easy exercise :

$$\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$$

Hint for the exercise :

Use the formula : $\cos(A)-\cos(B)=-2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$.

Now, for the question :

$\sin\beta\sin\left(\dfrac\pi3-\beta\right)\sin\left(\dfrac\pi3+\beta\right)$

$=\sin\beta\left(\sin^2\left(\dfrac\pi3\right)-\sin^2\left(\beta\right)\right)$

$=\sin\beta\left(\frac{3}{4}-\sin^2\left(\beta\right)\right)$

$=\frac{\sin 3\beta }{4}$

Akash Yadav
  • 1,167