Question

Class 11

Q1 Find the values of other five trigonometric functions

1. cos x = -1/2, x lies in third quadrant.

Q2 Prove that

$\cos ^{2} x + \cos ^{2} (x + \frac{\pi}{3} ) + \cos ^{2} (x - \frac{\pi}{3} ) = \frac{3}{2}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

x lies in third quadrant.

$\rm \: cosx = - \dfrac{1}{2}$

We know,

$\rm \: {sin}^{2}x + {cos}^{2}x = 1$

On substituting the value of cosx, we get

$\rm \: {sin}^{2}x + {\bigg[ - \dfrac{1}{2} \bigg]}^{2} = 1$

$\rm \: {sin}^{2}x + \dfrac{1}{4} = 1$

$\rm \: {sin}^{2}x = 1 - \dfrac{1}{4}$

$\rm \: {sin}^{2}x = \dfrac{3}{4}$

$\rm \: {sin}x = \pm \: \dfrac{ \sqrt{3} }{2}$

As x lies in third quadrant, so sinx < 0

$\rm\implies \:\rm \: {sin}x \: = \: - \: \dfrac{ \sqrt{3} }{2}$

Now, Other 5 Trigonometric ratios are

$\rm \: sinx = - \dfrac{ \sqrt{3} }{2}$

$\rm \: cosx = - \dfrac{1}{2}$

$\rm \: tanx = \dfrac{sinx}{cosx} = \sqrt{3}$

$\rm \: cosecx = - \dfrac{2}{ \sqrt{3} }$

$\rm \: secx = - 2$

$\rm \: cotx \: = \: \dfrac{1}{ \sqrt{3} }$

$\large\underline{\sf{Solution-2}}$

$\rm \: \cos ^{2} x + \cos ^{2}\bigg (x + \dfrac{\pi}{3} \bigg) + \cos ^{2} \bigg(x - \dfrac{\pi}{3} \bigg)$

can be rewritten as

$\rm \: = \: \dfrac{1}{2}\bigg[2 \cos ^{2} x + 2\cos ^{2}\bigg (x + \dfrac{\pi}{3} \bigg) + 2\cos ^{2} \bigg(x - \dfrac{\pi}{3} \bigg)\bigg]$

We know

$\boxed{\tt{ \: {2cos}^{2}x = 1 + cos2x \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{1}{2}\bigg[1 + \cos2x + 1 +\cos\bigg (2x + \dfrac{2\pi}{3} \bigg) + 1 + \cos \bigg(2x - \dfrac{2\pi}{3} \bigg)\bigg]$

$\rm \: = \: \dfrac{1}{2}\bigg[3 + \cos2x +\cos\bigg (2x + \dfrac{2\pi}{3} \bigg) + \cos \bigg(2x - \dfrac{2\pi}{3} \bigg)\bigg]$

We know,

$\boxed{\tt{ \: cos(x + y) + cos(x - y) = 2 \: cosx \: cosy \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{1}{2}\bigg[3 + \cos2x +2 \: \cos2x \: cos\dfrac{2\pi}{3} \bigg]$

$\rm \: = \: \dfrac{1}{2}\bigg[3 + \cos2x +2 \: \cos2x \: cos\bigg(\pi - \dfrac{\pi}{3}\bigg)\bigg]$

We know,

$\boxed{\tt{ \: \: cos(\pi - x) \: = \: - \: cosx \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{1}{2}\bigg[3 + \cos2x - 2 \: \cos2x \: cos\bigg(\dfrac{\pi}{3}\bigg)\bigg]$

$\rm \: = \: \dfrac{1}{2}\bigg[3 + \cos2x - 2 \: \cos2x \: \times \dfrac{1}{2}\bigg]$

$\rm \: = \: \dfrac{1}{2}\bigg[3 + \cos2x - \: \cos2x \:\bigg]$

$\rm \: = \: \dfrac{3}{2}$

Hence,

$\boxed{\tt{ \rm \: \cos ^{2} x + \cos ^{2}\bigg (x + \dfrac{\pi}{3} \bigg) + \cos ^{2} \bigg(x - \dfrac{\pi}{3} \bigg) = \frac{3}{2} \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \: sin2x = 2sinxcosx = \frac{2 \: tanx}{1 + {tan}^{2} x} \: }}$

$\boxed{\tt{ \: cos2x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1 \: }}$

$\boxed{\tt{ \: cos2x = {cos}^{2}x - {sin}^{2}x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x} \: }}$

$\boxed{\tt{ \: tan2x = \frac{2tanx}{1 - {tan}^{2} x} \: }}$

$\boxed{\tt{ sin3x = 3sinx \: - \: {4sin}^{3}x \: }}$

$\boxed{\tt{ cos3x = \: {4cos}^{3}x \: - \: 3cosx }}$

$\boxed{\tt{ \: tan3x = \frac{3tanx - {tan}^{3} x}{1 - 3 {tan}^{2}x } \: }}$

Answer 2

Refer the Given Attachments.