#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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!