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Question: How do you find the exact values of \[\sin \theta \] and \[\tan \theta \] when \[\cos \theta =\dfrac...

How do you find the exact values of sin⁡θ\sin \theta and tan⁡θ\tan \theta when cos⁡θ=12\cos \theta =\dfrac{1}{2} ?

Explanation

Solution

Hint : In the given question, we have been asked to find the exact values of trigonometric functions and it is given that cos⁡θ=12\cos \theta =\dfrac{1}{2} . In order to find the exact values, we need to use the definition of cosine function, we will get the value of base and hypotenuse of a unit circle right triangle and later using Pythagoras theorem we will find the value of the perpendicular side. Then using the definition of sine and tangent function of trigonometry, we will get to know the exact values of these two functions.

Complete step-by-step answer :
We have given that,
cos⁡θ=12\cos \theta =\dfrac{1}{2}
Using the definition of cos function to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.
cos⁡θ=basehypotenuse=12\cos \theta =\dfrac{base}{hypotenuse}=\dfrac{1}{2}
Finding the perpendicular side of the unit circle triangle. Since we know the values of base and the hypotenuse.
Using the Pythagoras theorem, i.e.
(hypotenuse)2=(base)2+(perpendicular)2{{\left( hypotenuse \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}
perpendicular=(base)2+(hypotenuse)2perpendicular=\sqrt{{{\left( base \right)}^{2}}+{{\left( hypotenuse \right)}^{2}}}
Putting the values, we get
perpendicular=(1)2+(2)2=1+2=±3perpendicular=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}=\sqrt{1+2}=\pm \sqrt{3}
Using the definition of sine function,
Taking, perpendicular=3perpendicular=\sqrt{3}
sin⁡θ=perpendicularhypotenuse=32\sin \theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{\sqrt{3}}{2}
Taking, perpendicular=−3perpendicular=-\sqrt{3}
sin⁡θ=perpendicularhypotenuse=−32\sin \theta =\dfrac{perpendicular}{hypotenuse}=-\dfrac{\sqrt{3}}{2}
Now if we divide sin⁡θ\sin \theta by cos⁡θ\cos \theta ,, we will get
sin⁡θcos⁡θ=perpendicularhypotenusebasehypotenuse=perndicularbase\dfrac{\sin \theta }{\cos \theta }=\dfrac{\dfrac{perpendicular}{hypotenuse}}{\dfrac{base}{hypotenuse}}=\dfrac{perndicular}{base} $$$$
tan⁡θ=sin⁡θcos⁡θ=perpendicularbase\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{perpendicular}{base}
Therefore,
Taking, perpendicular=3perpendicular=\sqrt{3}
tan⁡θ=perpendicularbase=31=3\tan \theta =\dfrac{perpendicular}{base}=\dfrac{\sqrt{3}}{1}=\sqrt{3}
Taking, perpendicular=−3perpendicular=-\sqrt{3}
tan⁡θ=perpendicularbase=−31=−3\tan \theta =\dfrac{perpendicular}{base}=-\dfrac{\sqrt{3}}{1}=-\sqrt{3}

Note : One must be careful while noted down the values to avoid any error in the answer. We must know the basic concept of trigonometry and also the definition of the trigonometric function. The sine, cosine and the tangent are the three basic functions in introduction to trigonometry which shows the relation between all the sides of the triangles.