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# Find sin x/2, cos x/2, and tan x/2 in each of the following: sin x = 1/4, x in quadrant II

**Solution:**

Since x lies in quadrant II i.e. π/2 < x < π

Therefore, π/4 < x/2 < π/2

Hence, sin x/2, cos x/2, and tan x/2 all are positive as x/2 lies in quadrant I.

It is given that sin x = - 1/4

By one of the trigonometric identities,

cos^{2}x = 1 - sin^{2}x

= 1 - (-1/4)^{2}

= 1 - 1/16

= 15/16

cos x = ±√(15/16)

Since, x lies in quadrant II so cos x is negative

So, cos x = -√(15/16)

cos 2(x/2) = -√(15/16)

2cos^{2}(x/2) - 1 = -√(15/16)

2cos^{2}x/2 = 1 - (√15 / 4)

2cos^{2}x/2 = [4 -√15] / 4

cos^{2}x/2 = [4 -√15] / 8

cos x/2 = ±√[{4 -√15} / 8]

Since cos x/2 lies in quadrant I and positive so,

cos x/2 = √[{4 -√15} / 8]

multiplying and dividing the fraction inside the square root by 2,

cos x/2 = √[8 -2√15] / 4

Now, sin^{2}x/2 = 1 - cos^{2}x/2 [Because sin^{2}A + cos ^{2}A = 1]

= 1 - (√[8 -2√15] / 4)^{2}

= 1 - [8 -2√15] / 16

= (16 - [8 + 2√15]) / 16

= [8 + 2√15] / 16

sin x/2 = ± √([8 + 2√15] / 16)

Since sin x/2 lies in quadrant I and positive so, sin x/2 = √[8 + 2√15] / 4

Now, tan x/2 = [sin x/2] / [cos x/2]

= (√[8 + 2√15] / 4) / (√[8 -2√15] / 4)

= √[(4 + √15) / (4 - √15)]

Rationalizing the denominator,

= √[(4 + √15) / (4 - √15)] × √[(4 + √15) / (4 + √15)]

= √[(4 + √15)^{2} / 16 - 15]

= 4 + √15

NCERT Solutions Class 11 Maths Chapter 3 Exercise ME Question 10

## Find sin x/2, cos x/2, and tan x/2 in each of the following: sin x = 1/4, x in quadrant II

**Summary:**

When sin x = 1/4, x in quadrant II, sin x/2 = √[8 + 2√15] / 4, cos x/2 = √[8 - 2√15] / 4, and tan x/2 = 4 + √15.

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