Question

tRIGONOMETRY

Simplify
i) cos 100°.cos 40° + sin 100°. sin 40°
ii) cos 340°.cos 40° + sin 200°. sin 140°​

Answer 1

Answer:

The trigonometric identities

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

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

should be used here.

Now

cos 340° cos 40°

= (1/2)[cos(340°+40°) + cos(340°-40°)]

= (1/2)[cos 380° + cos 300°]

= (1/2)[cos(360°+20°) + cos(360°-60°)]

Now, we have to know that

cos (360°+A) = cos A

cos (360°-A) = cos(-A) = cos A [Since, cos x is an even function].

Therefore,

cos 340° cos 40° = (1/2)[ cos 20° + cos 60° ]

Now,

sin 200° sin 140°

= (1/2)[cos(200°-140°) - cos(200°+140°)]

= (1/2)[cos 60° - cos 340°]

= (1/2)[cos 60° - cos (360°-20°)]

sin 200° sin 140° = (1/2)[cos 60° - cos 20°]

Therefore,

cos 340° cos 40° + sin 200° sin 140°

= (1/2)[cos 20° + cos 60°] + (1/2)[cos 60° - cos 20°]

Taking 1/2 in common,

= (1/2)[cos 20° + cos 60° + cos 60° - cos 20°]

Cancel out cos 20°.

= (1/2)×[2cos 60°]

We know cos 60° = 1/2

= (1/2)×[2×(1/2)] = (1/2)×1

= 1/2