Solveeit Logo
Login

Question

Question: Evaluate: \({\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) + {\sin ^{ - 1}}\left( {\sin \...

Evaluate: cos⁡−1(cos⁡2π3)+sin⁡−1(sin⁡2π3){\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right)
(A) 4π3\dfrac{{4\pi }} {3}
(B) π2\dfrac{\pi } {2}
(C) 3π4\dfrac{{3\pi }} {4}
(D) π\pi

Explanation

Solution

The given expressions can be evaluated by using the properties of inverse trigonometric functions.
According to property of inverse trigonometric functions, if x∈[0,π]x \in \left[ {0,\pi } \right]; the value of sin⁡−1(sin⁡x){\sin ^{ - 1}}\left( {\sin x} \right) is xx.
Also, if x∈[−π2,π2]x \in \left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right];
the value of cos⁡−1(cos⁡x){\cos ^{ - 1}}\left( {\cos x} \right) is xx.

Complete step by step solution:
The given expression is cos⁡−1(cos⁡2π3)+sin⁡−1(sin⁡2π3){\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right).
In order to evaluate the above expression, we need to solve each of the terms by using the properties of inverse trigonometric functions.
It is known that if x∈[0,π]x \in \left[ {0,\pi } \right];
the value of sin⁡−1(sin⁡x){\sin ^{ - 1}}\left( {\sin x} \right) is xx.
Also, if x∈[−π2,π2]x \in \left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right];
the value of cos⁡−1(cos⁡x){\cos ^{ - 1}}\left( {\cos x} \right)is xx.
So, sin⁡−1(sin⁡x)=x,∀x∈[−π2,π2]{\sin ^{ - 1}}\left( {\sin x} \right) = x,\forall x \in \left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right] and cos⁡−1(cos⁡x)=x;∀x∈[0,π]{\cos ^{ - 1}}\left( {\cos x} \right) = x;\forall x \in \left[ {0,\pi } \right].
Now, simplify the term cos⁡−1(cos⁡2π3){\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right).
In the above expression, the angle is 2π3\dfrac{{2\pi }} {3}
which belongs to the interval [−π2,π2]\left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right].
So, the value of the expression cos⁡−1(cos⁡2π3){\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) is 2π3\dfrac{{2\pi }} {3} that is, cos⁡−1(cos⁡2π3)=2π3{\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) = \dfrac{{2\pi }} {3}.
Also, simplify the term sin⁡−1(sin⁡2π3){\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right).
In the above expression, the angle is 2π3\dfrac{{2\pi }} {3} which does not belong to the interval [−π2,π2]\left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right].
So, rewrite the angle 2π3\dfrac{{2\pi }} {3} as,
2π3=3π−π3 =3π3−π3 =π−π3  \dfrac{{2\pi }} {3} = \dfrac{{3\pi - \pi }} {3} \\\ = \dfrac{{3\pi }} {3} - \dfrac{\pi } {3} \\\ = \pi - \dfrac{\pi } {3} \\\
Substitute π−π3\pi - \dfrac{\pi } {3}
for 2π3\dfrac{{2\pi }} {3}
in the expression sin⁡−1(sin⁡2π3){\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right)
sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡(π−π3)){\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi } {3}} \right)} \right)
Also, it is known that sin⁡(π−π3)=sin⁡π3\sin \left( {\pi - \dfrac{\pi } {3}} \right) = \sin \dfrac{\pi } {3}
Substitute sin⁡π3\sin \dfrac{\pi } {3} for sin⁡(π−π3)\sin \left( {\pi - \dfrac{\pi } {3}} \right) in the sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡(π−π3)){\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi } {3}} \right)} \right)
sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡(π−π3)) sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡π3)  {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi } {3}} \right)} \right) \\\ {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi } {3}} \right) \\\
In the above expression, it is seen that the angle is π3\dfrac{\pi } {3} which belongs to the interval [−π2,π2]\left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right]
Rewrite the expression sin⁡−1(sin⁡π3){\sin ^{ - 1}}\left( {\sin \dfrac{\pi } {3}} \right) by using the property sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x
sin⁡−1(sin⁡π3)=π3{\sin ^{ - 1}}\left( {\sin \dfrac{\pi } {3}} \right) = \dfrac{\pi } {3}
Substitute the values 2π3\dfrac{{2\pi }} {3}
for cos⁡−1(cos⁡2π3){\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right)
and π3\dfrac{\pi } {3}
for sin⁡−1(sin⁡π3){\sin ^{ - 1}}\left( {\sin \dfrac{\pi } {3}} \right)
in the expression cos⁡−1(cos⁡2π3)+sin⁡−1(sin⁡2π3){\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right)
and simplify.
cos⁡−1(cos⁡2π3)+sin⁡−1(sin⁡2π3)=2π3+π3 =2π+π3 =3π3 =π  {\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right) = \dfrac{{2\pi }} {3} + \dfrac{\pi } {3} \\\ = \dfrac{{2\pi + \pi }} {3} \\\ = \dfrac{{3\pi }} {3} \\\ = \pi \\\
Thus, the required value of the given expression is π\pi .
Hence, the required correct answer is (D).

Note: To evaluate the given expression it is necessary to use the properties of inverse trigonometric functions for cos⁡−1{\cos ^{ - 1}} and sin⁡−1{\sin ^{ - 1}} . We must know that the inverse trigonometric function cos⁡−1{\cos ^{ - 1}} lies in the interval [0,π]\left[ {0,\pi } \right] . Also, the inverse trigonometric function sin⁡−1{\sin ^{ - 1}} lies in the interval [−π2,π2]\left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right]