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Prove that:

(i) sin 65° + cos 65° = √2 cos 20°

(ii) sin 47° + cos 77° = cos 17°

1 Answer

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(i) Let us consider the LHS

sin 65° + cos 65° = sin 65° + sin (90° – 65°)

= sin 65° + sin 25°

On using the formula,

sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2

sin 65° + sin 25° = 2 sin (65° + 25°)/2 cos (65° – 25°)/2

= 2 sin 90°/2 cos 40°/2

= 2 sin 45° cos 20°

= 2 × 1/√2 × cos 20°

= √2 cos 20°

= RHS

Thus proved.

(ii) Let us consider the LHS

sin 47° + cos 77° = sin 47° + sin (90° – 77°)

= sin 47° + sin 13°

On using the formula,

sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2

sin 47° + sin 13° = 2 sin (47° + 13°)/2 cos (47° – 13°)/2

= 2 sin 60°/2 cos 34°/2

= 2 sin 30° cos 17°

= 2 × 1/2 × cos 17°

= cos 17°

= RHS

Thus proved.

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