Question

Given that $\sin \alpha = \dfrac{1}{2}$ and $\cos \beta = \dfrac{1}{2}$, then the value of $(\alpha + \beta )$ is
A) ${0^ \circ }$
B) ${30^ \circ }$
C) ${60^ \circ }$
D) ${90^ \circ }$

Answer 1

Hint: Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions. It is also called arcus functions. Taking the arcus function of the same function always gives the value of the angle. For example, taking the arcsine function of the sine function gives ${\sin ^{ - 1}}\sin \theta = \theta $.

Complete step by step answer:
We are given that $\sin \alpha = \dfrac{1}{2}$ and $\cos \beta = \dfrac{1}{2}$, and the objective is to determine the value of $(\alpha + \beta )$
To determine the value of $(\alpha + \beta )$, we first need to determine the value of $\alpha $ and $\beta $.
First take,
$\sin \alpha = \dfrac{1}{2}$
Taking the inverse function, that is ${\sin ^{ - 1}}$ on both sides of the given equation ,
${\sin ^{ - 1}}(\sin \alpha ) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Since ${\sin ^{ - 1}}(\sin \alpha ) = \alpha $
$\alpha = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
For ${0^ \circ } \leqslant \alpha \leqslant {90^ \circ }$, we need to determine the value of ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Since we know that $\sin {30^ \circ } = \dfrac{1}{2}$, take the inverse function, that is ${\sin ^{ - 1}}$ on both sides of the given equation
${\sin ^{ - 1}}\dfrac{1}{2} = {\sin ^{ - 1}}(\sin {30^ \circ })$
Since ${\sin ^{ - 1}}(\sin {30^ \circ }) = {30^ \circ }$
${\sin ^{ - 1}}\dfrac{1}{2} = {30^ \circ }$
So $\alpha = {30^ \circ }$
Now take ,
$\cos \beta = \dfrac{1}{2}$
Taking the arcus function , that is ${\cos ^{ - 1}}$ on both sides of the given equation ,
${\cos ^{ - 1}}(\cos \beta ) = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Since ${\cos ^{ - 1}}(\cos \beta ) = \beta $
$\beta = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
For ${0^ \circ } \leqslant \beta \leqslant {90^ \circ }$, we need to determine the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$.
Since we know that $\cos {60^ \circ } = \dfrac{1}{2}$, take the inverse function, that is ${\cos ^{ - 1}}$ on both sides of the given equation
${\cos ^{ - 1}}\dfrac{1}{2} = {\cos ^{ - 1}}(\cos {60^ \circ })$
Since ${\cos ^{ - 1}}(\cos {60^ \circ }) = {60^ \circ }$
${\cos ^{ - 1}}\dfrac{1}{2} = {60^ \circ }$
So $\beta = {60^ \circ }$
Now, we need to determine the value of $(\alpha + \beta )$ by substituting the values of $\alpha = {30^ \circ }$ and $\beta = {60^ \circ }$.
$\alpha + \beta = {30^ \circ } + {60^ \circ }$
$\alpha + \beta = {90^ \circ }$
So, the value of $(\alpha + \beta )$ is (D)${90^ \circ }$.

So, the correct answer is “Option D”.

Note:
Trigonometric functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant.
For a right-angled triangle, the values of the basic trigonometric functions are given by
$
  \sin \theta = \dfrac{{Perpendicular}}{{Hypotenuse}} \\
  \cos \theta = \dfrac{{Base}}{{Hypotenuse}} \\
  \tan \theta = \dfrac{{Perpendicular}}{{Base}} \\
 $