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Question: How do you find the remaining trigonometric functions of \[\theta \] given \( \cos \theta = \dfrac{{...

How do you find the remaining trigonometric functions of θ\theta given cos⁡θ=32\cos \theta = \dfrac{{\sqrt 3 }}{2} and θ\theta terminates in the fourth quadrant?

Explanation

Solution

Hint : Trigonometric ratios are the ratios of any two sides of a right-angled triangle, they establish a relationship between any two sides of the right-angled triangle and one of its angles other than the right angle. There are six types of trigonometric functions – sine, cosecant, cosine, secant, tangent and cotangent. The sine, cosine and tangent functions are the main functions while cosecant, secant and cotangent are their reciprocals respectively (the tangent function is equal to the ratio of the sine function and the cosine function). All the trigonometric ratios are linked with each other by various identities and other relations too. Using those relations and identities, we will find the remaining trigonometric ratios when one of them is known.

Complete step-by-step answer :
We are given cos⁡θ=32\cos \theta = \dfrac{{\sqrt 3 }}{2}
We know that –
sec⁡θ=1cos⁡θ ⇒sec⁡θ=132 ⇒sec⁡θ=23   \sec \theta = \dfrac{1}{{\cos \theta }} \\\ \Rightarrow \sec \theta = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} \\\ \Rightarrow \sec \theta = \dfrac{2}{{\sqrt 3 }} \;
We also know that –
cos⁡2θ+sin⁡2θ=1 ⇒sin⁡2θ=1−34 ⇒sin⁡2θ=14 ⇒sin⁡θ=±12   {\cos ^2}\theta + {\sin ^2}\theta = 1 \\\ \Rightarrow {\sin ^2}\theta = 1 - \dfrac{3}{4} \\\ \Rightarrow {\sin ^2}\theta = \dfrac{1}{4} \\\ \Rightarrow \sin \theta = \pm \dfrac{1}{2} \;
But we know that the sine function is negative in the fourth quadrant, so the positive value is rejected.
⇒sin⁡θ=−12\Rightarrow \sin \theta = - \dfrac{1}{2}
cos⁡ecθ=1sin⁡θ=1−12=−2\cos ec\theta = \dfrac{1}{{\sin \theta }} = \dfrac{1}{{ - \dfrac{1}{2}}} = - 2
⇒cos⁡ecθ=−2\Rightarrow \cos ec\theta = - 2
Now, tan⁡θ=sin⁡θcos⁡θ=−1232=−12×23=−13\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\dfrac{{ - 1}}{2}}}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{ - 1}}{2} \times \dfrac{2}{{\sqrt 3 }} = - \dfrac{1}{{\sqrt 3 }}
⇒tan⁡θ=−13\Rightarrow \tan \theta = - \dfrac{1}{{\sqrt 3 }}
We know cot⁡θ=1tan⁡θ=1−13=−3\cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{1}{{ - \dfrac{1}{{\sqrt 3 }}}} = - \sqrt 3
⇒cot⁡θ=−3\Rightarrow \cot \theta = - \sqrt 3
Hence, when cos⁡θ=32\cos \theta = \dfrac{{\sqrt 3 }}{2} , we get sec⁡θ=23, sin⁡θ=−12, csc⁡θ=−2, tan⁡θ=−13 and cot⁡θ=−3\sec \theta = \dfrac{2}{{\sqrt 3 }},\,\sin \theta = - \dfrac{1}{2},\,\csc \theta = - 2,\,\tan \theta = - \dfrac{1}{{\sqrt 3 }}\,and\,\cot \theta = - \sqrt 3 .

Note : A graph is divided into four quadrants, trigonometric functions have different signs in different quadrants. All the trigonometric ratios are positive in the first quadrant, sine and cosecant are positive in the first quadrant while others are negative, tangent and cotangent are positive in the third quadrant while the remaining are negative, and cosine and secant functions are positive in the fourth quadrant while all the other trigonometric functions are negative in the fourth quadrant.