NCERT Solutions For Class 11 Math Chapter – 3 Miscellaneous

Miscellaneous Exercise page: 81

Prove that:

1.

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Solution:

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We get

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= 0

= RHS

2. (sin 3+ sin x) sin + (cos 3– cos x) cos =

0

Solution:

Consider

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

By further calculation

= sin 3x sin x + sin2 x + cos 3x cos x – cos2 x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos2 x – sin2 x)

Using the formula

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

So we get

= cos 2x – cos 2x

= 0

= RHS

3.

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Solution:

Consider

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using formula we get

= cos2 x + cos2 y + 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y

Grouping the terms

= (cos2 x + sin2 x) + (cos2 y + sin2 y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

By further calculation

= 2 + 2 cos (x + y)

Taking 2 as common

= 2 [1 + cos (x + y)]

From the formula cos 2A = 2 cos2 A – 1

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4.

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Solution:

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using formula

= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y

Grouping the terms

= (cos2 x + sin2 x) + (cos2 y + sin2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]

By further calculation

= 2 [1 – cos (x – y)]

From formula cos 2A = 1 – 2 sin2 A

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5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

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6.

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Solution:

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7.

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Solution:

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8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

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Solution:

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cos x = -3/5

From the formula

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9. cos x = -1/3, x in quadrant III

Solution:

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10. sin x = 1/4, x in quadrant II

Solution:

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