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This is a question from transformation of sums and products solve it please

This is a question from transformation of sums and products solve it please

Question Image
Grade:11

2 Answers

Avinash
52 Points
2 years ago
4sin23×sin37×sin83;it can be written as 2[2sin23×sin37]sin83;We know that cos(A-B)-cos(A+B)=2sinA.sinBBy using this 2[cos14-cos60]sin832sin83.cos14-2cos60.sin83We know that sin(A+B)+sin(A-B)=2sinAcosBBy using this we will getSin97+sin69-2×1/2×sin83sin97=sin83Therefore we left with sin69Sin69=cos21Hence proved
Soumendu Majumdar
159 Points
2 years ago
Dear Student,
cos(A – B) – cos(A + B) = 2 sin A sin B
so 
Given equation on L.H.S is
4sin23^{\circ}sin37^{\circ}sin83^{\circ}
2sin83^{\circ}(2sin23^{\circ}sin37^{\circ})
=2sin83^{\circ}(cos 14^{\circ}-cos 60^{\circ})
2sin83^{\circ}cos 14^{\circ}-sin 83^{\circ}
sin69^{\circ}+sin 97^{\circ}-sin 83^{\circ}                           since 2 sin A cos B = sin (A + B) + sin (A – B)
cos21^{\circ}+cos 7 ^{\circ}-cos7^{\circ}
cos21^{\circ}
= R.H.S [Proved]
Hope it helps!
 

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