sinA . sin(60-A) . sin(60+A) = 1/4 sin3A
Answers
Answered by
1
hope you like it. zzzzz
Attachments:
![](https://hi-static.z-dn.net/files/dde/81cd067e40dfdea27224ca01937851db.jpg)
Answered by
9
Answer:
1/4 sin3A
Step-by-step explanation:
LHS = sinA . sin(60° + A) . sin(60° - A)
= 1/2 sinA.[2sin(60° + A).sin(60° - A)]
= 1/2 sinA.[cos{(60° + A) - (60° - A)} - cos{( 60° + A) + (60° - A)}]
= 1/2 sinA.[cos(60° + A - 60° + A) - cos120° ]
= 1/2 sinA.[cos2A - {- 1/2}]
= 1/2 sinA.[cos2A + 1/2]
= 1/2 sinA. cos2A + 1/4 sinA
= 1/4 ( 2cos2A.sinA)+1/4 sinA
= 1/4[sin(2A+A) - sin(2A - A)] + 1/4 sinA
= 1/4(sin3A-sinA) + 1/4 sinA
= 1/4 sin3A - 1/4 sinA + 1/4 sinA
= 1/4 sin3A
Similar questions