Solveeit Logo
Login

Question

Question: Write the value of given inverse trigonometric function \(\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}...

Write the value of given inverse trigonometric function sin⁡[π3−sin⁡−1(−12)]\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right] .

Explanation

Solution

Hint: For solving this question we will use two important trigonometric results. First, we will use the formula of the inverse trigonometric functions to write sin⁡−1(−12)=−π6{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=-\dfrac{\pi }{6} in the given term. After that, we will use one of the basic formulas of trigonometric ratio, i.e. sin⁡π2=1\sin \dfrac{\pi }{2}=1 for giving the final answer for the question correctly.

Complete step-by-step solution -
Given:
We have to find the value of the following:
sin⁡[π3−sin⁡−1(−12)]\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right]
Now, we will simplify the term sin⁡[π3−sin⁡−1(−12)]\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right] and try to find its correct value.
Now, before we proceed we should know the following formulas:
sin⁡−1(12)=π6....................(1) sin⁡π2=1.................................(2) \begin{aligned} & {{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}....................\left( 1 \right) \\\ & \sin \dfrac{\pi }{2}=1.................................\left( 2 \right) \\\ \end{aligned}
Now, we will use the above two formulas to solve this question.
We have, sin⁡[π3−sin⁡−1(−12)]\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right] .
Now, as we know that, sin⁡−1(−x)=−sin⁡−1(x){{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}\left( x \right) for x∈[−1,1]x\in \left[ -1,1 \right] and −1<−12<1-1<-\dfrac{1}{2}<1 so, we can write sin⁡−1(−12)=−sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{-1}{2} \right)=-{{\sin }^{-1}}\left( \dfrac{1}{2} \right) . Then,
sin⁡[π3−sin⁡−1(−12)] ⇒sin⁡[π3−(−sin⁡−1(12))] ⇒sin⁡[π3+sin⁡−1(12)] \begin{aligned} & \sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right] \\\ & \Rightarrow \sin \left[ \dfrac{\pi }{3}-\left( -{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right] \\\ & \Rightarrow \sin \left[ \dfrac{\pi }{3}+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right] \\\ \end{aligned}
Now, we will use the formula from the equation (1) to write sin⁡−1(12)=π6{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6} in the term sin⁡[π3+sin⁡−1(12)]\sin \left[ \dfrac{\pi }{3}+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right] . Then,
sin⁡[π3+sin⁡−1(12)] ⇒sin⁡[π3+π6] ⇒sin⁡π2 \begin{aligned} & \sin \left[ \dfrac{\pi }{3}+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right] \\\ & \Rightarrow \sin \left[ \dfrac{\pi }{3}+\dfrac{\pi }{6} \right] \\\ & \Rightarrow \sin \dfrac{\pi }{2} \\\ \end{aligned}
Now, we will use the formula from the equation (2) to write sin⁡π2=1\sin \dfrac{\pi }{2}=1 in the above line. Then,
sin⁡π2 ⇒1 \begin{aligned} & \sin \dfrac{\pi }{2} \\\ & \Rightarrow 1 \\\ \end{aligned}
Now, from the above result, we conclude that the value of the expression sin⁡[π3−sin⁡−1(−12)]\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right] will be equal to one. Then,
sin⁡[π3−sin⁡−1(−12)]=1\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right]=1
Thus, sin⁡[π3−sin⁡−1(−12)]=1\sin \left[ \dfrac{\pi }{3}-{{\sin }^{-1}}\left( -\dfrac{1}{2} \right) \right]=1.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, though the question is easy, but we should apply the formulas of inverse trigonometric functions, for example: function y=sin⁡−1xy={{\sin }^{-1}}x is defined for x∈[−1,1]x\in \left[ -1,1 \right] and its range is y∈[−π2,π2]y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] then, yy is the principal value of sin⁡−1x{{\sin }^{-1}}x . Then, we should solve without any mathematical error and avoid making calculation mistakes while solving to get the correct answer.