Solveeit Logo
Login

Question

Question: How do you use \( \sin \theta = \dfrac{1}{3} \) to find the value of \( \cos \theta \) ?...

How do you use sin⁡θ=13\sin \theta = \dfrac{1}{3} to find the value of cos⁡θ\cos \theta ?

Explanation

Solution

Hint : In the given problem, we are required to calculate 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.

Complete step by step solution:
So, In the given problem, we have to find the value of cosine of an angle .
Hence, we have to find the cosine of the angle whose sine is given to us as 13\dfrac{1}{3} .
Let us assume θ\theta to be the concerned angle.
Then, θ=sin⁡−1(13)\theta = {\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right)
Taking sine on both sides of the equation, we get
sin⁡θ=13\sin \theta = \dfrac{1}{3} which was already given to us in the question itself.
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=13\sin \theta = \dfrac{{Perpendicular}}{{Hypotenuse}} = \dfrac{1}{3}
Let length of the perpendicular be xx .
Then, length of hypotenuse =3x= 3x .

Now, applying Pythagoras Theorem,
(Hypotenuse)2=(Base)2+(Perpendicular)2{\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}
=(3x)2=(Base)2+(x)2= {\left( {3x} \right)^2} = {\left( {Base} \right)^2} + {\left( x \right)^2}
=9x2=(Base)2+x2= 9{x^2} = {\left( {Base} \right)^2} + {x^2}
=(Base)2=8x2= {\left( {Base} \right)^2} = 8{x^2}
=(Base)=8x2= \left( {Base} \right) = \sqrt {8{x^2}}
=(Base)=8x= \left( {Base} \right) = \sqrt 8 x
So, we get Base=22xBase = 2\sqrt 2 x
Hence, cos⁡θ=22x3x=223\cos \theta = \dfrac{{2\sqrt 2 x}}{{3x}} = \dfrac{{2\sqrt 2 }}{3}
So, the value of cos⁡θ\cos \theta given that sin⁡θ=13\sin \theta = \dfrac{1}{3} is cos⁡θ=223\cos \theta = \dfrac{{2\sqrt 2 }}{3}
So, the correct answer is “ cos⁡θ=223\cos \theta = \dfrac{{2\sqrt 2 }}{3} ”.

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.