Question

Find the values of the five trigonometric functions if Cos x = - 1/2, x lies in third quadrant.​

Answer 1

Answer:

cos==1/2 =log =cos1/2-1/2

Step-by-step explanation:

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

$\mathsf {\cos \: x = - \frac{1}{2}}$

$\therefore \mathsf {\sec \: x = \frac{1}{ \cos \: x } = \frac{1}{( - \frac{1}{2}) } = - 2}$

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

$\implies \mathsf {\sin {}^{2} \: x = 1 - \cos {}^{2} \: x}$

$\implies \mathsf {\sin {}^{2} \: x = 1 - ( - \frac{1}{2} ) {}^{2}}$

$\implies \mathsf {\sin {}^{2} \: x = 1 - \frac{1}{4} = \frac{3}{4}}$

$\implies \mathsf {\sin \: x = \pm \: \frac{ \sqrt{3} }{2}}$

• Since, x lies in the 3rd quadrant, the value of sin x will be negative.

$\therefore \mathsf {\sin \: x = - \frac{ \sqrt{3} }{2}}$

$\mathsf {\cos \: ecx = \frac{1}{ \sin \: x } = \frac{1}{( - \frac{ \sqrt{3} }{2}) } = - \frac{2}{ \sqrt{3} } }$

$\mathsf {\tan \: x = \frac{ \sin \: x }{ \cos \: x} = \frac{( - \frac{ \sqrt{3} }{2}) }{( - \frac{1}{2}) } = \sqrt{3}}$

$\mathsf \blue {\cos \: x = \frac{1}{ \tan \: x } = \frac{1}{ \sqrt{3} }}$