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Question: Write the principal value of \(\left( {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}+{{\cos }^{-1}}\dfrac{-1}{2} ...

Write the principal value of (cos⁡−132+cos⁡−1−12)\left( {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}+{{\cos }^{-1}}\dfrac{-1}{2} \right)

Explanation

Solution

Hint: We know that the principal value of cos⁡−1{{\cos }^{-1}} lies from 0 to π0\text{ to }\pi . Now we will use this fact to find the value of cos⁡−132{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} and cos⁡−1−12{{\cos }^{-1}}\dfrac{-1}{2} separately and then we will add these two values to get our final answer and we will see that if it lies between [0,π]\left[ 0,\pi \right] or not.

Complete step-by-step answer:
Let’s start solving this question.
We know that the principal value of cos⁡−1{{\cos }^{-1}} lies from [0,π]\left[ 0,\pi \right] .
First we will find the value of cos⁡−132{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} .
We know that cos⁡π6=32\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}
Therefore we get,
cos⁡−132=π6{{\cos }^{-1}}\dfrac{\sqrt{3}}{2}=\dfrac{\pi }{6} which lies between [0,π]\left[ 0,\pi \right].
Now we will find the value of cos⁡−1−12{{\cos }^{-1}}\dfrac{-1}{2}.
We know that cos is negative in the second quadrant.
We know thatcos⁡(π−π3)=−12\cos \left( \pi -\dfrac{\pi }{3} \right)=\dfrac{-1}{2} .
Therefore we get,
cos⁡−1−12=2π3{{\cos }^{-1}}\dfrac{-1}{2}=\dfrac{2\pi }{3} which lies between [0,π]\left[ 0,\pi \right].
Now substituting the value of cos⁡−132=π6{{\cos }^{-1}}\dfrac{\sqrt{3}}{2}=\dfrac{\pi }{6} and cos⁡−1−12=2π3{{\cos }^{-1}}\dfrac{-1}{2}=\dfrac{2\pi }{3} we get,
(cos⁡−132+cos⁡−1−12)=2π3+π6=5π6\left( {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}+{{\cos }^{-1}}\dfrac{-1}{2} \right)=\dfrac{2\pi }{3}+\dfrac{\pi }{6}=\dfrac{5\pi }{6} which lies between [0,π]\left[ 0,\pi \right].
Hence, the answer will be 5π6\dfrac{5\pi }{6} .

Note: We have used some known results like cos⁡π6=32\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2} and cos⁡(π−π3)=−12\cos \left( \pi -\dfrac{\pi }{3} \right)=\dfrac{-1}{2}, one must remembered these results so that there is no problem while understanding the solution. One can also use the formula for cos⁡−1a+cos⁡−1b=cos⁡−1(ab+(1−a2)(1−b2)){{\cos }^{-1}}a+{{\cos }^{-1}}b={{\cos }^{-1}}\left( ab+\sqrt{\left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right)} \right) and then substitute the value of ‘a’ and ‘b’ to get the final answer, the answer that we get from this method must be same as the answer that we get from previous method.