Question

1. Prove the following identities :
sin 10°sin 50° + sin 50°sin 250° + sin 250° sin 10° = -3/4​

Answer 1

Answer:

Step-by-step explanation:

Given,

sin 10°sin 50° + sin 50°sin 250° + sin 250° sin 10° = -3/4​

To Find :-

Prove the above equation = -3/4

Solution :-

Taking L.H.S :-

sin 10°sin 50° + sin 50°sin 250° + sin 250° sin 10°

Multiplying and dividing with 2 :-

$\sf\dfrac{2}{2}sin10\degree sin50\degree+ \dfrac{2}{2}sin50\degree sin250\degree+ \dfrac{2}{2}sin250\degree sin10\degree$

Taking 1/2 common :-

$\sf\dfrac{1}{2}(2sin10\degree sin50\degree + 2sin50\degree sin250\degree + 2sin250\degree sin10\degree)$

We know that, 2sinAsinB = cos(A-B) - cos(A + B) :-

$\sf\dfrac{1}{2}(cos(10 - 50) - cos(10 + 50) + cos(50 - 250) - cos(50 + 250) + cos(250 - 10) - cos(250 + 10))$

$\sf\dfrac{1}{2}(cos(-40\degree) - cos(60\degree) + cos(-200\degree) - cos(300\degree) + cos(240\degree) - cos(260\degree)$

Since, as cos(-A) = cosA :-

$\sf\dfrac{1}{2}(cos40\degree - cos60\degree + cos200\degree - cos300\degree + cos240\degree - cos260\degree)$

$\sf\dfrac{1}{2}(cos40\degree - cos60\degree + cos(180 + 20) - cos(360 - 60) + cos(180 + 60) - cos(180 + 80)$

$\sf\dfrac{1}{2}(cos40\degree - cos60\degree - cos20\degree - cos60\degree - cos60\degree +cos80\degree)$

$\sf\dfrac{1}{2}( -3cos60\degree + cos40\degree  - cos20\degree + cos80\degree)$

$\sf\dfrac{1}{2}( -3cos60\degree  + cos40\degree cos80\degree  - cos20\degree)$

$cosC + cosD = 2cos\bigg(\dfrac{C + D}{2}\bigg)cos\bigg(\dfrac{C - D}{2}\bigg)$

$\sf\dfrac{1}{2}( -3cos60\degree + 2cos\bigg(\dfrac{40 + 80}{2}\bigg)cos\bigg(\dfrac{40 - 80}{2}\bigg) - cos20\degree)$

$\sf\dfrac{1}{2}( -3cos60\degree + 2cos60\degree cos20\degree - cos20\degree)$

$\sf\dfrac{1}{2}( -3\times \dfrac{1}{2} + 2\dfrac{1}{2} cos20\degree - cos20\degree)$

$\sf\dfrac{1}{2}( \dfrac{-3}{2} + cos20\degree - cos20\degree)$

$\dfrac{1}{2}\times \dfrac{-3}{2}\\\\= \dfrac{-3}{4}$

Hence , Proved