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Question: Evaluate \[\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{3}{7}} \right)} \right)\]....

Evaluate cos⁡(sin⁡−1(37))\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{3}{7}} \right)} \right).

Explanation

Solution

In the given problem, we are required to calculate the cosine of an angle whose sine is given to us. Such problems require basic knowledge of trigonometric ratios and formulae. Besides this, knowledge of concepts of inverse trigonometry is extremely essential to answer these questions correctly.

Formula used:

sin⁡θ=Opposite sideHypotenuse\sin \theta = \dfrac{\text{Opposite side}}{\text{Hypotenuse}}
cos⁡θ=Adjacent sideHypotenuse\cos \theta = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}}

Complete step by step solution:
In the given problem, we have to find the value ofcos⁡(sin⁡−1(37))\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{3}{7}} \right)} \right).
Hence, we have to find the cosine of the angle whose sine is given to us as (37)\left( {\dfrac{3}{7}} \right).
Let us assume θ\theta to be the concerned angle.
Then, θ=sin⁡−1(37)\theta = {\sin ^{ - 1}}\left( {\dfrac{3}{7}} \right)
Taking sine on both sides of the equation, we get
=sin⁡θ=37= \sin \theta = \dfrac{3}{7}
To evaluate the value of the required expression, we must keep in mind the formulae of basic trigonometric ratios.
We know that, sin⁡θ=PerpendicularHypotenuse\sin \theta = \dfrac{{Perpendicular}}{{Hypotenuse}}and cos⁡θ=BaseHypotenuse\cos \theta = \dfrac{{Base}}{{Hypotenuse}}.
So, sin⁡θ=PerpendicularHypotenuse=37\sin \theta = \dfrac{{Perpendicular}}{{Hypotenuse}} = \dfrac{3}{7}
Let the length of the perpendicular be3x3x.
Then, length of hypotenuse =7x = 7x.
Now, applying Pythagoras Theorem,
(Hypotenuse)2=(Base)2+(Perpendicular)2{\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}
=(7x)2=(Base)2+(3x)2= {\left( {7x} \right)^2} = {\left( {Base} \right)^2} + {\left( {3x} \right)^2}
Computing the squares, we get,
=49x2=(Base)2+9x2= 49{x^2} = {\left( {Base} \right)^2} + 9{x^2}
=(Base)2=49x2−9x2= {\left( {Base} \right)^2} = 49{x^2} - 9{x^2}
=(Base)=40x2= \left( {Base} \right) = \sqrt {40{x^2}}
Simplifying the expression by taking the factor appearing twice in the factorisation of the number outside the square root. So, we get,
=(Base)=210x= \left( {Base} \right) = 2\sqrt {10} x
So, we get Base=210xBase = 2\sqrt {10} x
Hence, cos⁡θ=210x7x=2107\cos \theta = \dfrac{{2\sqrt {10} x}}{{7x}} = \dfrac{{2\sqrt {10} }}{7}
Therefore, the value ofcos⁡(sin⁡−1(37))\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{3}{7}} \right)} \right) is 2107\dfrac{{2\sqrt {10} }}{7}.

Note:
The given problem can also be solved by use of some basic trigonometric identities such as sin⁡2(θ)+cos⁡2(θ)=1{\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = 1. This method also provides exposure to the applications of trigonometric identities in various mathematical questions.