Question

Evaluate the following:
(i) $\sin ({{\cot }^{-1}}x)$
(ii) $\sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right)$

Answer 1

Hint: In the first part we will assume ${{\cot }^{-1}}x=\theta$ and then we will convert it in terms of sin and cos and use the formula ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to get the answer. Then in the second part of the question we will use ${{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x$ and then substitute the angle for which the value is given and get our answer.
Complete step-by-step answer:
(i) The expression mentioned in the first part of the question is $\sin ({{\cot }^{-1}}x).........(1)$
Now let ${{\cot }^{-1}}x=\theta ........(2)$
Now rearranging equation (2) we get,
$\cot \theta =x........(3)$
Now converting cot in terms of cos and sin in equation (3) we get,
$\dfrac{\cos \theta }{\sin \theta }=x........(4)$
Now squaring both sides in equation (4) we get,
$\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }={{x}^{2}}........(5)$
Now cross multiplying the terms in equation (5) we get,
${{\cos }^{2}}\theta ={{x}^{2}}{{\sin }^{2}}\theta ........(6)$
Now we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and hence using this in equation (6) we get,
$1-{{\sin }^{2}}\theta ={{x}^{2}}{{\sin }^{2}}\theta ........(7)$
Now simplifying and rearranging the terms in equation (7) and solving we get,

& \Rightarrow {{x}^{2}}{{\sin }^{2}}\theta +{{\sin }^{2}}\theta =1 \\\ & \Rightarrow {{\sin }^{2}}\theta (1+{{x}^{2}})=1 \\\ & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{1+{{x}^{2}}}........(8) \\\ \end{aligned}$$ Taking square root of both sides of equation (8) we get, $$\Rightarrow \sin \theta =\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}}........(9)$$ Now substituting from equation (2) in equation (9) we get, $$\Rightarrow \sin ({{\cot }^{-1}}x)=\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}}$$ Hence $$\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}}$$ is the answer. (ii) The expression mentioned in the second part of the question is $$\sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right)$$. $$\Rightarrow \sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right).........(1)$$ We know that the domain of $${{\sin }^{-1}}x$$ is between -1 and +1. Also the range is between $$-\dfrac{\pi }{2}$$ and $$\dfrac{\pi }{2}$$. Also we know that $${{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x$$ and using this in equation (1) we get, $$\Rightarrow \sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right).........(2)$$ We know that $$\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$$ and substituting this in equation (2) we get, $$\Rightarrow \sin \left( \dfrac{\pi }{2}+{{\sin }^{-1}}\left( \sin \dfrac{\pi }{3} \right) \right).........(3)$$ Now simplifying and rearranging in equation (3) we get, $$\Rightarrow \sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{3} \right).........(4)$$ Now taking the LCM and solving in equation (4) we get, $$\Rightarrow \sin \left( \dfrac{5\pi }{6} \right)=\sin \left( \pi -\dfrac{\pi }{6} \right)=\sin \dfrac{\pi }{6}=\dfrac{1}{2}$$ Hence $$\dfrac{1}{2}$$ is the answer. Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identity $$\sin \left( \pi -x \right)=\sin x$$ as we can write cos in place of sin in equation (4) in part (ii).