Math, asked by sanya55, 1 year ago

find sin x/2 Cos x/2 and tan x/2 in the following

1. Tan x = -4/3 x in quad. 2

2. Sin x =1/4 x in quad.2

3. Cos x = -1/3 in quad.3


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Answers

Answered by Deepsbhargav
41
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 |1| \\ \\ in \: the \: {2}^{nd} \: quad. \: only \: sin \: of \: angle \: is \: positive.... \\ \\ = > tanx = \frac{2tan \frac{x}{2} }{1 - {tan}^{2} \frac{x}{2} } = - \frac{4}{3} \\ \\ = > 4 {tan}^{2} \frac{x}{2} - 6tan \frac{x}{2} - 4 = 0 \\ \\ = > 4 {tan}^{2} \frac{x}{2} - 8tan \frac{x}{2} + 2tan \frac{x}{2} - 4 = 0 \\ \\ = > 4tan \frac{x}{2} (tan \frac{x}{2} - 2) + 2(tan \frac{x}{2} - 2) = 0 \\ \\ = > tan \frac{x}{2 } = 2 \: \: \: (not \: a \: possible \: value) \\ \\ = > tan \frac{x}{2} = - \frac{1}{2} \: \: \: \: ...eq _{1} \\ \\ = > 1 + {tan}^{2} \frac{x}{2} = {sec}^{2} \frac{x}{2} \\ \\ > {sec}^{2} \frac{x}{2} = 1 + \frac{1}{4} = \frac{5}{4} \\ \\ = > {cos}^{} \frac{x}{2} = \binom{ + }{ - } \sqrt{ \frac{4}{5} } \\ \\ = > cos \frac{x}{2} = - \sqrt{ \frac{4}{5} } = - \frac{2}{ \sqrt{5} } \: \: \: ......eq _{2} \\ \\ = > sin \frac{x}{2} = tan \frac{x}{2} \times cos \frac{x}{2} = - \frac{1}{2} \times ( - \frac{2}{ \sqrt{5} } ) = \frac{1}{ \sqrt{5} }

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{2}

sinx = \frac{1}{4} .......x \: in \: {2}^{nd} quad \\ \\ x \: is \: lies \: in \: quad \: {2}^{nd} than \: sinx \: is \: positive \: \\ and \: cosx \: and \: tanx \: is \: negative \\ \\ now \\ \\ = > {sin}^{2} x + {cos}^{2} x = 1 \\ \\ = > cosx = \sqrt{1 - {sin}^{2} x} = \sqrt{1 - \frac{1}{16} } = \frac{ \sqrt{15} }{4} \\ \\ = > cos \frac{x}{2} = \sqrt{ \frac{1 + cosx}{2} } = \sqrt{ \frac{1 + \frac{ \sqrt{15} }{4} }{2} } = - \sqrt{ \frac{8 + \sqrt{15} }{8} } \\ \\ = > sin \frac{x}{2} = \sqrt{ \frac{1 - \sqrt{15} }{8} } \\ \\ = > tan \frac{x}{2} = -\frac{ \sqrt{1 - \sqrt{15} } }{ \sqrt{1 + \sqrt{15} } }

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[3]

coax = -1/3 .... in the 3rd quadrant

now use this formula...

=> cosx = 2cos²(x/2)-1

=> -1/3 = 2cos²(x/2)-1

=> 2cos²(x/2) = 1-1/3 = 2/3

=> cos²(x/2) = 1/3

=> cos(x/2) = ±root(1/3)

=> cos(x/2) = +root(1/3)
(not a possible value)

=> cos(x/2) = -root(1/3) ...eq(1)

now

=> sin²(x/2) + cos²(x/2) = 1

=> sin²(x/2) = 1 - 1/3 = 2/3

=> sin(x/2) = root(2/3) ...eq(2)

=> tan(x/2) = -root2. ___eq(3)


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Answered by wikremsinghp4v470
4

Answer:

easy question

Step-by-step explanation:

use formula for tanx=2tanx/1-tan^2x

1+tan^2x=sec^2x

secx+1/cosx

and then sin^2x+cos^2x=1

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