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Question: Given that \(\sin \alpha = \dfrac{1}{2}\) and \(\cos \beta = \dfrac{1}{2}\), then the value of \((\a...

Given that sin⁡α=12\sin \alpha = \dfrac{1}{2} and cos⁡β=12\cos \beta = \dfrac{1}{2}, then the value of (α+β)(\alpha + \beta ) is
A) 0∘{0^ \circ }
B) 30∘{30^ \circ }
C) 60∘{60^ \circ }
D) 90∘{90^ \circ }

Explanation

Solution

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⁡−1sin⁡θ=θ{\sin ^{ - 1}}\sin \theta = \theta .

Complete step by step answer:
We are given that sin⁡α=12\sin \alpha = \dfrac{1}{2} and cos⁡β=12\cos \beta = \dfrac{1}{2}, and the objective is to determine the value of (α+β)(\alpha + \beta )
To determine the value of (α+β)(\alpha + \beta ), we first need to determine the value of α\alpha and β\beta .
First take,
sin⁡α=12\sin \alpha = \dfrac{1}{2}
Taking the inverse function, that is sin⁡−1{\sin ^{ - 1}} on both sides of the given equation ,
sin⁡−1(sin⁡α)=sin⁡−1(12){\sin ^{ - 1}}(\sin \alpha ) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
Since sin⁡−1(sin⁡α)=α{\sin ^{ - 1}}(\sin \alpha ) = \alpha
α=sin⁡−1(12)\alpha = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
For 0∘⩽α⩽90∘{0^ \circ } \leqslant \alpha \leqslant {90^ \circ }, we need to determine the value of sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
Since we know that sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2}, take the inverse function, that is sin⁡−1{\sin ^{ - 1}} on both sides of the given equation
sin⁡−112=sin⁡−1(sin⁡30∘){\sin ^{ - 1}}\dfrac{1}{2} = {\sin ^{ - 1}}(\sin {30^ \circ })
Since sin⁡−1(sin⁡30∘)=30∘{\sin ^{ - 1}}(\sin {30^ \circ }) = {30^ \circ }
sin⁡−112=30∘{\sin ^{ - 1}}\dfrac{1}{2} = {30^ \circ }
So α=30∘\alpha = {30^ \circ }
Now take ,
cos⁡β=12\cos \beta = \dfrac{1}{2}
Taking the arcus function , that is cos⁡−1{\cos ^{ - 1}} on both sides of the given equation ,
cos⁡−1(cos⁡β)=cos⁡−1(12){\cos ^{ - 1}}(\cos \beta ) = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)
Since cos⁡−1(cos⁡β)=β{\cos ^{ - 1}}(\cos \beta ) = \beta
β=cos⁡−1(12)\beta = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)
For 0∘⩽β⩽90∘{0^ \circ } \leqslant \beta \leqslant {90^ \circ }, we need to determine the value of cos⁡−1(12){\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right).
Since we know that cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}, take the inverse function, that is cos⁡−1{\cos ^{ - 1}} on both sides of the given equation
cos⁡−112=cos⁡−1(cos⁡60∘){\cos ^{ - 1}}\dfrac{1}{2} = {\cos ^{ - 1}}(\cos {60^ \circ })
Since cos⁡−1(cos⁡60∘)=60∘{\cos ^{ - 1}}(\cos {60^ \circ }) = {60^ \circ }
cos⁡−112=60∘{\cos ^{ - 1}}\dfrac{1}{2} = {60^ \circ }
So β=60∘\beta = {60^ \circ }
Now, we need to determine the value of (α+β)(\alpha + \beta ) by substituting the values of α=30∘\alpha = {30^ \circ } and β=60∘\beta = {60^ \circ }.
α+β=30∘+60∘\alpha + \beta = {30^ \circ } + {60^ \circ }
α+β=90∘\alpha + \beta = {90^ \circ }
So, the value of (α+β)(\alpha + \beta ) is (D)90∘{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⁡θ=PerpendicularHypotenuse cos⁡θ=BaseHypotenuse tan⁡θ=PerpendicularBase  \sin \theta = \dfrac{{Perpendicular}}{{Hypotenuse}} \\\ \cos \theta = \dfrac{{Base}}{{Hypotenuse}} \\\ \tan \theta = \dfrac{{Perpendicular}}{{Base}} \\\