1-cos 56-cos 58-cos 66-

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

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Byju's Answer

Standard XI

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

1+cos 56∘+cos...

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}

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

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

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

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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 conditional identity
$c o s A + c o s B - c o s C = - 1 + 4 c o s \frac{A}{2} c o s \frac{B}{2} s i n \frac{C}{2}$
$[∵ 56^{\circ} + 58^{\circ} + 66^{\circ} = 180^{\circ}]$
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∘=

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