Evaluate:(i) cos 20° + cos 100° + cos 140°(ii) sin 50° – sin 70�

1.	Evaluate:(i) cos 20° + cos 100° + cos 140°(ii) sin 50° – sin 70° + sin 10°
Answer» (i) LHS = (cos 20° + cos 100°) + cos 140° = 2 cos (\(\frac{20°+100°}{2}\)) + cos (\(\frac{20°-100°}{2}\))140° [∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))] = 2 cos 60° cos(-40°) + cos 140° = 2 x \(\frac{1}{2}\) x cos(-40°) + cos(180° – 140°) [∵ cos(-θ) = cos θ, cos 60° = \(\frac{1}{2}\) = cos 40° – cos 40° = 0 Hence Proved. (ii) LHS = (sin 50° – sin 70°) + sin 10° = 2 cos (\(\frac{50+70}{2}\)) sin (\(\frac{50+70}{2}\)) + sin 10° [∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))] = 2 cos 60° sin(-10°) + sin 10° = 2 x \(\frac{1}{2}\)(-sin 10°) + sin 10° [∵ sin(-θ) = -sin θ] = -sin 10° + sin 10° = 0 = RHS

Evaluate:(i) cos 20° + cos 100° + cos 140°(ii) sin 50° – sin 70° + sin 10°

Answer»

(i) LHS = (cos 20° + cos 100°) + cos 140°

= 2 cos (\(\frac{20°+100°}{2}\)) + cos (\(\frac{20°-100°}{2}\))140°

[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]

= 2 cos 60° cos(-40°) + cos 140°

= 2 x \(\frac{1}{2}\) x cos(-40°) + cos(180° – 140°)

[∵ cos(-θ) = cos θ, cos 60° = \(\frac{1}{2}\)

= cos 40° – cos 40°

= 0

Hence Proved.

(ii) LHS = (sin 50° – sin 70°) + sin 10°

= 2 cos (\(\frac{50+70}{2}\)) sin (\(\frac{50+70}{2}\)) + sin 10°

[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]

= 2 cos 60° sin(-10°) + sin 10°

= 2 x \(\frac{1}{2}\)(-sin 10°) + sin 10° [∵ sin(-θ) = -sin θ]

= -sin 10° + sin 10°

= 0

= RHS