cos⁡22∘+cos⁡780+cos⁡80∘=

1 + 4 sin ⁡ 11 ∘ sin ⁡ 39 ∘ sin ⁡ 40 ∘

4 sin ⁡ 11 ∘ sin ⁡ 39 ∘ sin ⁡ 40 ∘

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\cos 22^{\circ} + \cos 78^{0} + \cos 80^{\circ} =$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$1 + 4 \sin 11^{\circ} \sin 39^{\circ} \sin 40^{\circ}$

$4 \sin 11^{\circ} \sin 39^{\circ} \sin 40^{\circ}$

$4 \cos 11^{0} \cos 39^{0} \cos 40^{0}$

$1 + 4 \cos 11^{0} \cos 39^{0} \cos 40^{0}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$(\cos 78^{0} + \cos 22^{\circ}) + \cos 80^{\circ} = 2 \cos \frac{78^{0} + 22^{\circ}}{2} \cos \frac{78^{0} - 22^{\circ}}{2} + (1 - 2 \sin^{2} 40^{\circ}) = 1 + 2 \cos 50 ° \cos 28 ° - 2 \cos^{2} 50 ° = 1 - 2 \cos 50 ° (\cos 50 ° - \cos 28 °) = 1 + 2 \cos 50 ° (2 \sin \frac{50 ° + 28 °}{2} \sin \frac{50 ° - 28 °}{2}) = 1 + 4 \sin 40 ° \sin 39 ° \sin 11 °$