1-cos 56-cos 58-cos 66-

The correct option is C 1 + cos 56^∘+cos 58^∘ nbsp; cos 66^∘= =2cos^228^∘+2sin 62^∘.sin 4^∘ =2cos^228^∘+2cos 28^∘.sin 4^∘ =2cos28^∘(cos 28^∘+ cos 86^∘) =2c ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

1+cos 56∘+cos...

Question

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

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Solution

The correct option is C

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

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Similar questions

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

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

[IIT 1964]

Q. Prove that

1 + cos 56^{\circ} + cos 58^{\circ} - cos 66^{\circ} = 4 cos 28^{\circ} cos 29^{\circ} sin 33^{\circ}

1 + cos 56^{\circ} + cos 58^{\circ} - cos 66^{\circ} = m cos 28^{\circ} cos 29^{\circ} sin 33^{\circ}

. Find m

Q. If

- 1 + cos 56^{\circ} + cos 58^{\circ} + cos 66^{\circ} = k sin 28^{\circ} sin 29^{\circ} sin 33^{\circ}

, then the value of k is

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1+cos 56∘+cos 58∘−cos 66∘=

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