TRIGONOMETRIC FUNCTIONS-Exercise 3.2
Maths - Exercise
Find the values of other five trigonometric functions in Exercises 1 to 5.
Q1. \(\cos x=-\dfrac{1}{2},\; x \) lies in third quadrant.
Solution
Given that \(\cos x = -\dfrac{1}{2}\) and the angle \(x\) lies in the third quadrant, both \(\sin x\) and \(\cos x\) are negative while \(\tan x\) is positive
\[ \begin{aligned} \cos x &= -\dfrac{1}{2} \\ \sin^{2} x &= 1 - \cos^{2} x \\ &= 1 - \left( \dfrac{-1}{2} \right)^{2} \\ &= 1 - \dfrac{1}{4} \\ &= \dfrac{3}{4} \\ \sin x &= -\dfrac{\sqrt{3}}{2} \end{aligned} \]
\[ \begin{aligned} \tan x &= \dfrac{\sin x}{\cos x} \\ &= \dfrac{-\sqrt{3}/2}{-1/2} \\ &= \sqrt{3} \end{aligned} \]
\[ \begin{aligned} \cot x &= \dfrac{1}{\tan x} \\ &= \dfrac{1}{\sqrt{3}} \end{aligned} \]
\[ \begin{aligned} \sec x &= \dfrac{1}{\cos x} \\ &= \dfrac{1}{-1/2} \\ &= -2 \end{aligned} \]
\[ \begin{aligned} \text{cosec}\, x &= \dfrac{1}{\sin x} \\ &= \dfrac{1}{-\sqrt{3}/2} \\ &= -\dfrac{2}{\sqrt{3}} \end{aligned} \]
Q2. \(\sin x=\dfrac{3}{5},\; x\) lies in second quadrant.
Solution
Given that \(\sin x = \dfrac{3}{5}\) and the angle \(x\) lies in the second quadrant, \(\sin x\) is positive while \(\cos x\) and \(\tan x\) are negative
\[ \begin{aligned} \sin x &= \dfrac{3}{5} \\ \cos^{2} x &= 1 - \sin^{2} x \\ &= 1 - \left( \dfrac{3}{5} \right)^{2} \\ &= 1 - \dfrac{9}{25} \\ &= \dfrac{25 - 9}{25} \\ &= \dfrac{16}{25} \\ \cos x &= \pm \sqrt{\dfrac{16}{25}} \\ &= \pm \dfrac{4}{5} \\ \cos x &= -\dfrac{4}{5} \end{aligned} \]
\[ \begin{aligned} \tan x &= \dfrac{\sin x}{\cos x} \\ &= \dfrac{3/5}{-4/5} \\ &= -\dfrac{3}{4} \end{aligned} \]
\[ \begin{aligned} \cot x &= \dfrac{1}{\tan x} \\ &= -\dfrac{4}{3} \end{aligned} \]
\[ \begin{aligned} \sec x &= \dfrac{1}{\cos x} \\ &= \dfrac{1}{-4/5} \\ &= -\dfrac{5}{4} \end{aligned} \]
\[ \begin{aligned} \text{cosec}\, x &= \dfrac{1}{\sin x} \\ &= \dfrac{1}{3/5} \\ &= \dfrac{5}{3} \end{aligned} \]
Q3. \(\cot x =\dfrac{3}{4},\; x\) lies in third quadrant.
Solution
Given that \(\cot x = \dfrac{3}{4}\) and the angle \(x\) lies in the third quadrant, both \(\sin x\) and \(\cos x\) are negative while \(\tan x\) is positive
\[ \begin{aligned} \cot x &= \dfrac{3}{4} \\ \tan x &= \dfrac{1}{\cot x} \\ &= \dfrac{4}{3} \end{aligned} \]
\[ \begin{aligned} \sec^{2} x &= 1 + \tan^{2} x \\ &= 1 + \left( \dfrac{4}{3} \right)^{2} \\ &= 1 + \dfrac{16}{9} \\ &= \dfrac{9 + 16}{9} \\ &= \dfrac{25}{9} \\ \sec x &= \pm \sqrt{\dfrac{25}{9}} \\ &= \pm \dfrac{5}{3} \\ \sec x &= -\dfrac{5}{3} \end{aligned} \]
\[ \begin{aligned} \cos x &= \dfrac{1}{\sec x} \\ &= \dfrac{1}{-5/3} \\ &= -\dfrac{3}{5} \end{aligned} \]
\[ \begin{aligned} \sin^{2} x &= 1 - \cos^{2} x \\ &= 1 - \left( \dfrac{-3}{5} \right)^{2} \\ &= 1 - \dfrac{9}{25} \\ &= \dfrac{25 - 9}{25} \\ &= \dfrac{16}{25} \\ \sin x &= \pm \sqrt{\dfrac{16}{25}} \\ &= \pm \dfrac{4}{5} \\ \sin x &= -\dfrac{4}{5} \end{aligned} \]
\[ \begin{aligned} \text{cosec}\, x &= \dfrac{1}{\sin x} \\ &= \dfrac{1}{-4/5} \\ &= -\dfrac{5}{4} \end{aligned} \]
Q4. \(\sec x=\dfrac{13}{5},\;x\) lies in fourth quadrant.
Solution
Given that \(\sec x = \dfrac{13}{5}\) and the angle \(x\) lies in the fourth quadrant, \(\cos x\) is positive while \(\sin x\) and \(\tan x\) are negative
\[ \begin{aligned} \sec x &= \dfrac{13}{5} \\ \cos x &= \dfrac{1}{\sec x} \\ &= \dfrac{1}{13/5} \\ &= \dfrac{5}{13} \end{aligned} \]
\[ \begin{aligned} \sin^{2} x &= 1 - \cos^{2} x \\ &= 1 - \left( \dfrac{5}{13} \right)^{2} \\ &= 1 - \dfrac{25}{169} \\ &= \dfrac{169 - 25}{169} \\ &= \dfrac{144}{169} \\ \sin x &= \pm \sqrt{\dfrac{144}{169}} \\ &= \pm \dfrac{12}{13} \\ \sin x &= -\dfrac{12}{13} \end{aligned} \]
\[ \begin{aligned} \text{cosec}\, x &= \dfrac{1}{\sin x} \\ &= \dfrac{1}{-12/13} \\ &= -\dfrac{13}{12} \end{aligned} \]
\[ \begin{aligned} \tan x &= \dfrac{\sin x}{\cos x} \\ &= \dfrac{-12/13}{5/13} \\ &= -\dfrac{12}{5} \end{aligned} \]
\[ \begin{aligned} \cot x &= \dfrac{1}{\tan x} \\ &= \dfrac{1}{-12/5} \\ &= -\dfrac{5}{12} \end{aligned} \]
Q5. \(\tan x = -\dfrac{5}{12},\;x\) lies in second quadrant.
Solution
Given that \(\tan x = -\dfrac{5}{12}\) and the angle \(x\) lies in the second quadrant, \(\sin x\) is positive while \(\cos x\) and \(\tan x\) are negative
\[ \begin{aligned} \tan x &= -\dfrac{5}{12} \\ \cot x &= \dfrac{1}{\tan x} \\ &= \dfrac{1}{-5/12} \\ &= -\dfrac{12}{5} \end{aligned} \]
\[ \begin{aligned} \sec^{2} x &= 1 + \tan^{2} x \\ &= 1 + \left( -\dfrac{5}{12} \right)^{2} \\ &= 1 + \dfrac{25}{144} \\ &= \dfrac{144 + 25}{144} \\ &= \dfrac{169}{144} \\ \sec x &= \pm \sqrt{\dfrac{169}{144}} \\ &= \pm \dfrac{13}{12} \\ \sec x &= -\dfrac{13}{12} \end{aligned} \]
\[ \begin{aligned} \cos x &= \dfrac{1}{\sec x} \\ &= \dfrac{1}{-13/12} \\ &= -\dfrac{12}{13} \end{aligned} \]
\[ \begin{aligned} \sin^{2} x &= 1 - \cos^{2} x \\ &= 1 - \left( -\dfrac{12}{13} \right)^{2} \\ &= 1 - \dfrac{144}{169} \\ &= \dfrac{169 - 144}{169} \\ &= \dfrac{25}{169} \\ \sin x &= \pm \sqrt{\dfrac{25}{169}} \\ &= \pm \dfrac{5}{13} \\ \sin x &= \dfrac{5}{13} \end{aligned} \]
\[ \begin{aligned} \text{cosec}\, x &= \dfrac{1}{\sin x} \\ &= \dfrac{1}{5/13} \\ &= \dfrac{13}{5} \end{aligned} \]
Find the values of the trigonometric functions in Exercises 6 to 10.
Q6. \(\sin 765°\)
Solution
Using the periodicity of the sine function, the given angle can be reduced to an equivalent acute angle
\[ \begin{aligned} \sin 765^\circ &= \sin \left( 720^\circ + 45^\circ \right) \\ &= \sin 45^\circ \\ &= \dfrac{1}{\sqrt{2}} \end{aligned} \]
Q7. \(\text{cosec}\, (– 1410°)\)
Solution
Using the periodicity and odd–even properties of trigonometric functions, the given angle is reduced to a standard angle
\[ \begin{aligned} \text{cosec}\,(-1410^\circ) &= \text{cosec}\,\left(-1410^\circ + 1440^\circ\right) \\ &= \text{cosec}\,(-30^\circ) \\ &= -\text{cosec}\,(30^\circ) \\ &= -\dfrac{1}{\sin 30^\circ} \\ &= -2 \end{aligned} \]
Q8. \(\tan \dfrac{19\pi}{3}\)
Solution
Using the periodicity of the tangent function, the given angle can be reduced to its principal equivalent
\[ \begin{aligned} \tan \dfrac{19\pi}{3} &= \tan \left( 6\pi + \dfrac{\pi}{3} \right) \\ &= \tan \dfrac{\pi}{3} \\ &= \sqrt{3} \end{aligned} \]
Q9. \(\\sin (-\dfrac{11\pi}{3})\)
Solution
Using the periodicity property of the sine function, the given negative angle is reduced to an equivalent acute angle
\[ \begin{aligned} \sin \left( -\dfrac{11\pi}{3} \right) &= \sin \left( -\dfrac{11\pi}{3} + 4\pi \right) \\ &= \sin \dfrac{\pi}{3} \\ &= \dfrac{\sqrt{3}}{2} \end{aligned} \]
Q10. \(\cot (-\dfrac{15\pi}{4})\)
Solution
Using the periodicity of the cotangent function, the given angle can be reduced to an equivalent standard angle
\[ \begin{aligned} \cot \left( -\dfrac{15\pi}{4} \right) &= \cot \left( -\dfrac{15\pi}{4} + 4\pi \right) \\ &= \cot \left( \dfrac{\pi}{4} \right) \\ &= 1 \end{aligned} \]
