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Question: The value of \({\sin ^{ - 1}}\left( {\sin 12} \right) + {\cos ^{ - 1}}\left( {\cos 12} \right)\) is ...

The value of sin⁡−1(sin⁡12)+cos⁡−1(cos⁡12){\sin ^{ - 1}}\left( {\sin 12} \right) + {\cos ^{ - 1}}\left( {\cos 12} \right) is equal to:
A.Zero
B.24−2π24 - 2\pi
C.4π−244\pi - 24
D.None of these

Explanation

Solution

Hint: We will try to calculate the value of sin⁡−1(sin⁡12){\sin ^{ - 1}}\left( {\sin 12} \right) and cos⁡−1(cos⁡12){\cos ^{ - 1}}\left( {\cos 12} \right) separately. Here, the range of sin⁡−1(sin⁡12){\sin ^{ - 1}}\left( {\sin 12} \right) must lie between −π2 - \dfrac{\pi }{2} and π2\dfrac{\pi }{2}. The range of cos⁡−1(cos⁡12){\cos ^{ - 1}}\left( {\cos 12} \right) must lie between 0 and π\pi .

Complete step-by-step answer:
Here the required expression can be broken into two parts. We will calculate the value separately for sin⁡−1(sin⁡12){\sin ^{ - 1}}\left( {\sin 12} \right)and cos⁡−1(cos⁡12){\cos ^{ - 1}}\left( {\cos 12} \right).
Let us first calculate the value of sin⁡−1(sin⁡12){\sin ^{ - 1}}\left( {\sin 12} \right).
Let us consider the value of sin⁡−1(sin⁡12){\sin ^{ - 1}}\left( {\sin 12} \right) to be equal to θ\theta .
Then sin⁡θ=sin⁡12\sin \theta = \sin 12
The general solution for the above equation will be given as θ=2πk+12\theta = 2\pi k + 12, where kk is an integer. But the range of θ\theta must lie between −π2 - \dfrac{\pi }{2} to π2\dfrac{\pi }{2}, as the principle range of the sin⁡−1θ{\sin ^{ - 1}}\theta lies between −π2 - \dfrac{\pi }{2} to π2\dfrac{\pi }{2}. Therefore the value of kk must be so chosen that the value of θ\theta lies between −π2 - \dfrac{\pi }{2} to π2\dfrac{\pi }{2}.
−π2<2πk+12- \dfrac{\pi }{2} < 2\pi k + 12 and
2πk+12<π22\pi k + 12 < \dfrac{\pi }{2}
and kk is an integer.
−14−6π<k- \dfrac{1}{4} - \dfrac{6}{\pi } < k
k⩾−2k \geqslant - 2 as kk is an integer.
Similarly,
k<\-14−6πk < \- \dfrac{1}{4} - \dfrac{6}{\pi }
k⩽−2k \leqslant - 2
Therefore the value of kk is −2 - 2.
And thus θ=2π(−2)+12\theta = 2\pi \left( { - 2} \right) + 12
θ=−4π+12\theta = - 4\pi + 12
We will do the same steps to find the value of cos⁡−1(cos⁡12){\cos ^{ - 1}}\left( {\cos 12} \right).
Let cos⁡−1(cos⁡12)=φ{\cos ^{ - 1}}\left( {\cos 12} \right) = \varphi
Then cos⁡φ=cos⁡12\cos \varphi = \cos 12
The general solution for the above equation will be given as φ=2πk±12\varphi = 2\pi k \pm 12, where kk is an integer. But the range of φ\varphi must lie between 0 to π\pi , as the principle range of the cos⁡−1φ{\cos ^{ - 1}}\varphi lies between 0 to π\pi . Therefore the value of kk must be so chosen that the value of φ\varphi lies between 0 to π\pi .
0<2πk±120 < 2\pi k \pm 12 and
2πk±12<π2\pi k \pm 12 < \pi
and kk is an integer.
±6π<k\pm \dfrac{6}{\pi } < k
k⩾−2k \geqslant - 2or k⩾2k \geqslant 2,as kk is an integer.
Similarly,
k<12±6πk < \dfrac{1}{2} \pm \dfrac{6}{\pi }
k⩽2k \leqslant 2 or k⩽−2k \leqslant - 2
Therefore the value of kk satisfying the equations is 2, along with taking a subtraction between 12 and 2πk2\pi k.
And thus φ=2π(2)−12\varphi = 2\pi \left( 2 \right) - 12
φ=4π−12\varphi = 4\pi - 12
Adding both values will give us the value for the required expression.
−4π+4π+12−12=0- 4\pi + 4\pi + 12 - 12 = 0

Note: The general solution for sin⁡x=sin⁡θ\sin x = \sin \theta is given by x=2πk+θx = 2\pi k + \theta , where kk is an integer and the range of xx lies between −π2 - \dfrac{\pi }{2} and π2\dfrac{\pi }{2}. Similarly, the general solution for cos⁡x=cos⁡θ\cos x = \cos \theta is given by x=2πk±θx = 2\pi k \pm \theta , where kk is an integer and the range of xx lies between 0 and π\pi .