1 - cos 56- - cos 58- - cos 66- -

1 + cos 56^∘+cos 58^∘ cos 66^∘= =2cos^228^∘+2sin 62^∘.sin 4^∘ =2cos^228^∘+2cos 28^∘.sin 4^∘ =2cos28^∘(cos 28^∘+ cos 86^∘) =2cos28^∘.2cos 57^∘.cos 29^∘ =4cos 2 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

1 + cos 56∘ +...

Question

$1 + c o s 56^{\circ} + c o s 58^{\circ} - c o s 66^{\circ} =$

$2 c o s 28^{\circ} c o s 29^{\circ} c o s 33^{\circ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$4 c o s 28^{\circ} c o s 29^{\circ} c o s 33^{\circ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$2 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$

$1 + c o s 56^{\circ} + c o s 58^{\circ} - c o s 66^{\circ} =$
= $2 c o s^{2} 28^{\circ} + 2 s i n 62^{\circ} . s i n 4^{\circ}$
= $2 c o s^{2} 28^{\circ} + 2 c o s 28^{\circ} . s i n 4^{\circ}$
= $2 c o s 28^{\circ} (c o s 28^{\circ} + c o s 86^{\circ})$
= $2 c o s 28^{\circ} .2 c o s 57^{\circ} . c o s 29^{\circ}$
= $4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$
Aliter: Apply the condition identify

$c o s A + c o s B - c o s C = - 1 + 4 c o s \frac{A}{2} c o s \frac{B}{2} s i n \frac{c}{2}$
$[∴ 56^{\circ} + 58^{\circ} + 66^{\circ} = 180^{\circ}]$
We get the value of required expression equal to $4 c o s 28^{\circ} c o s 29^{\circ} s i n 33^{\circ}$ .

Suggest Corrections

1+cos 56∘+cos 58∘−cos 66∘=

$1 + c o s 56^{\circ} + c o s 58^{\circ} - c o s 66^{\circ} =$