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Question

Prove that:
(i) sin 38° + sin 22° = sin 82°
(ii) cos 100° + cos 20° = cos 40°
(iii) sin 50° + sin 10° = cos 20°
(iv) sin 23° + sin 37° = cos 7°
(v) sin 105° + cos 105° = cos 45°
(vi) sin 40° + sin 20° = cos 10°

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Solution

(i)

Consider LHS:sin 38° + sin 22°= 2sin 38° + 22°2 cos 38° - 22°2 sin A + sin B = 2sin A + B2 cos A - B2= 2sin 30° cos 8°= 2×12cos(90°-8°)= sin 82°= RHSHence, LHS=RHS.

(ii)

Consider LHS:cos 100° + cos 20°= 2cos 100° + 20°2 cos 100° - 20°2 cosA+cosB=2cosA+B2cosA-B2= 2cos 60° cos 40°= 2×12cos 40°= cos 40°Hence, LHS=RHS.

(iii)

Consider LHS:sin 50° + sin 10°= 2sin 50° + 10°2 cos 50° - 10°2 sin A + sin B = 2sin A + B2 cos A - B2= 2sin 30° cos 20°=2×12cos 20°= cos 20°Hence, LHS = RHS.


(iv)

Consider LHS:sin 23° + sin 37°= 2sin 23° + 37°2 cos 23°- 37°2 sin A + sin B = 2sin A + B2 cos A - B2= 2sin 30° cos -7°= 2sin 30°cos 7°= 2×12cos 7°= cos 7°Hence, LHS=RHS.

(v)

Consider LHS:sin 105° + cos 105°= sin 105° + cos 90° + 15°= sin 105° - sin 15°= 2sin 105° - 15°2 cos 105° + 15°2 sinA+sinB=2sinA-B2cosA+B2= 2sin 45°cos 60°= 2sin 90° - 45° cos 60°= 2×12cos45°=cos 45°Hence, LHS=RHS.

(vi)

Consider LHS:sin 40° + sin 20°= 2sin 40° + 20°2 cos 40° - 20°2 sinA+sinB=2sinA+B2cosA-B2= 2sin 30° cos 10°= 2×12cos 10°= cos10°Hence, LHS = RHS.

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