Question

How do you find the value of $\cos 120?$

Answer 1

Hint: The given question is to find out the value of the given trigonometric ratio. Since there are a total six trigonometric ratios in trigonometry. Which are $\sin \theta ,{\text{ }}\cos \theta ,{\text{ }}\tan \theta ,{\text{ }}\cos ec\theta ,{\text{ }}\sec \theta ,{\text{ }}\cot \theta $ with an angle$\theta $. Also these six trigonometric ratios have specific or fixed values at some fixed angles, which are ${0^0},{\text{ }}{30^0},{\text{ }}{45^0},{\text{ }}{60^0}{\text{ }}and{\text{ }}{90^0}$ . Value of Trigonometric ratios at these angles is fixed. Also these angles are in degrees and we are also have angles whose values for ${30^0},{\text{ }}{45^0},{\text{ }}{60^0}{\text{ }}and{\text{ }}{90^0}$ are $\dfrac{\pi }{6},{\text{ }}\dfrac{\pi }{4},{\text{ }}\dfrac{\pi }{3}and{\text{ }}\dfrac{\pi }{2}$ respectively where $\pi = {180^0}(in{\text{ }}radians)$

Complete Step by step Solution:
The given question is to find out the value of the given trigonometric ratio. Since-there are six trigonometric ratios whose value of r some specific angle is fixed
Since angle is in degree as well as in radians.
and the fixed values of the trigonometric ratios at certain angles which are ${30^0},{\text{ }}{45^0},{\text{ }}{60^0}{\text{ }}and{\text{ }}{90^0}$ were angle in degree
Also angles in radians in radians is symbolized in terms of ${30^0},{\text{ }}{45^0},{\text{ }}{60^0}{\text{ }}and{\text{ }}{90^0}\pi $ where value of in radians is ${180^0}$
Therefore angles in degrees and radians are
${0^0},{\text{ }}{30^0},{\text{ }}{45^0},{\text{ }}{60^0}{\text{ }}and{\text{ }}{90^0}$ is equivalent to \[0,\dfrac{\pi }{6},{\text{ }}\dfrac{\pi }{4},{\text{ }}\dfrac{\pi }{3},{\text{ }}\dfrac{\pi }{2}{\text{ where }}\pi = {180^0}\]are given as
And the fixed values for $\sin \theta {\text{ and }}\cos \theta $ are given as

angles	${0^0}$	${\text{ }}{30^0}$	\[{\text{ }}{45^0}\]	\[{60^0}\]	\[{90^0}\]
$\sin \theta $	\[\sqrt {\dfrac{0}{4}} = 0\]	\[\sqrt {\dfrac{1}{4}} = \dfrac{1}{2}\]	\[\sqrt {\dfrac{1}{4}} = \sqrt {\dfrac{1}{2}} = \dfrac{1}{{\sqrt 2 }}\]	\[\sqrt {\dfrac{3}{4}} = \dfrac{{\sqrt 2 }}{2}\]	\[\sqrt {\dfrac{4}{4}} = 1\]
$\cos \theta $	$1$	\[\sqrt {\dfrac{3}{2}} \]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	$0$

Value of \[\sin \theta \] are derived as talking \[0{\text{ }},{\text{ }}1{\text{ }},{\text{ }}2{\text{ }},{\text{ }}3{\text{ }},4\]
Dividing all by \[4\] and taking square roots and the values obtained are the values of \[\sin \theta \] at angle ${0^0},{\text{ 3}}{{\text{0}}^0},{\text{ }}{60^0}{\text{, 4}}{{\text{5}}^0}{\text{, }}{90^0}$
and value of \[\cos \theta \] are derived by taking the inverse of all the values of \[\sin \theta \] for angles ${0^0},{\text{ 3}}{{\text{0}}^0},{\text{ }}{60^0}{\text{, 4}}{{\text{5}}^0}{\text{ and }}{90^0}$
Means \[\sin {0^0} = \cos {90^0},{\text{ }}\sin {30^0} = \cos {60^0},{\text{ }}\sin {45^0} = \cos {45^0},{\text{ }}\]
\[\sin {60^0} = \cos {30^0}{\text{ and }}\sin {90^0} = \cos {0^0}\]

 Since we want to find out the value of $\cos {120^0}$ But the value of ${120^0}$ angle is not known to us.
Also we can write ${120^0}$as \[\left( {{{90}^0} + {{30}^0}} \right)\]
Therefore $\cos {120^0}$ becomes $\cos \left( {{{90}^0} + {{30}^0}} \right)$
Which becomes – $\sin {30^0}$
By using identity $\cos \left( {{{90}^0} + {{30}^0}} \right) = - \sin \theta $
So value of \[\sin {30^0}{\text{ is }}\dfrac{1}{2}\]
Therefore $ - \sin {30^0} = \dfrac{{ - 1}}{2}$
So, value of $\cos {120^0}{\text{ is }}\dfrac{{ - 1}}{2}$

Note: The angles in degree as well as in radians are having fixed value for the fixed trigonometric ratio where angle in degree is denoted by degree after the angle and in radians is denoted by using.
$\pi $ where $\pi = {180^0}$ which is the relation to convert degree to radiation in trigonometry