prove sin 38 + sin 22 = sin 82
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Answered by
84
HELLO DEAR,
sin38° + sin22° = sin82°
Now,
from , L.H.S,
sin38° + sin22° = 2sin(38 + 22)/2 * cos(38 - 22)/2
∴ [ sinA + sinB = 2sin(A + B)/2 * cos(A - B)/2 ]
⇒2sin(60/2) * cos(16/2)
⇒2sin30° * cos8°
⇒2 * 1/2 * cos8°
∴ [ sin30° = 1/2 ]
⇒ 2̶ * 1/ 2̶ * cos8°
⇒cos8°
we know that:-
cos(90 - Ф) = sinФ
now using here,
we get,
cos8° = cos(90 - 82)°
⇒sin82°
hence,
sin38° + sin22° = sin82°
I HOPE ITS HELP YOU DEAR,
THANKS
sin38° + sin22° = sin82°
Now,
from , L.H.S,
sin38° + sin22° = 2sin(38 + 22)/2 * cos(38 - 22)/2
∴ [ sinA + sinB = 2sin(A + B)/2 * cos(A - B)/2 ]
⇒2sin(60/2) * cos(16/2)
⇒2sin30° * cos8°
⇒2 * 1/2 * cos8°
∴ [ sin30° = 1/2 ]
⇒ 2̶ * 1/ 2̶ * cos8°
⇒cos8°
we know that:-
cos(90 - Ф) = sinФ
now using here,
we get,
cos8° = cos(90 - 82)°
⇒sin82°
hence,
sin38° + sin22° = sin82°
I HOPE ITS HELP YOU DEAR,
THANKS
Answered by
27
HEY!!
___________________________
✴Consider LHS:
✔sin 38° + sin 22°
✔2sin (38° + 22°/2) cos (38° − 22°/2)
✔{∵ sin A + sin B = 2sin (A + B/2) cos (A − B/2)}
✔2sin 30° cos 8°= 2×1/2cos(90°-8°)
✔sin 82°= RHS
▶▶Hence, LHS=RHS.
___________________________
✴Consider LHS:
✔sin 38° + sin 22°
✔2sin (38° + 22°/2) cos (38° − 22°/2)
✔{∵ sin A + sin B = 2sin (A + B/2) cos (A − B/2)}
✔2sin 30° cos 8°= 2×1/2cos(90°-8°)
✔sin 82°= RHS
▶▶Hence, LHS=RHS.
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