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Question: If, \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \dfrac{{3\pi }}{2}\], then the value of ...

If, sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \dfrac{{3\pi }}{2}, then the value of x9+y9+z9−1x9y9z9{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} is
A. 0
B. 1
C. 2
D. 3

Explanation

Solution

we have to find the value of x, y and z from the given equation using the values of trigonometric standard angles. The domain and range of sine inverse function is (-1,1) and (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) respectively. This will be used in finding the values of x, y and z. Then, we have to put those values in the equation whose answer we have to find and by solving step by step we will get our answer.

Complete step by step answer:
Given:
sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \dfrac{{3\pi }}{2}
We can also rewrite this equation as
sin⁡−1x+sin⁡−1y+sin⁡−1z=π2+π2+π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \dfrac{\pi }{2} + \dfrac{\pi }{2} + \dfrac{\pi }{2}
So, this means the maximum value in the range of sine inverse x is π2\dfrac{\pi }{2} .
Sum of three inverse of sine is 3×π23 \times \dfrac{\pi }{2}
This states that every sine inverse function is equal to π2\dfrac{\pi }{2}. Then,
sin⁡−1x=π2{\sin ^{ - 1}}x = \dfrac{\pi }{2}, sin⁡−1y=π2{\sin ^{ - 1}}y = \dfrac{\pi }{2} and sin⁡−1z=π2{\sin ^{ - 1}}z = \dfrac{\pi }{2}
So, x=sin⁡π2x = \sin \dfrac{\pi }{2}, y=sin⁡π2y = \sin \dfrac{\pi }{2}and z=sin⁡π2z = \sin \dfrac{\pi }{2}
We know that the value of sin⁡π2\sin \dfrac{\pi }{2}is 1. So,
X = 1, y = 1 and z = 1.
Now we will put value of x, y and z in the equation x9+y9+z9−1x9y9z9{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}}and we will get,
x9+y9+z9−1x9y9z9=19+19+19−119×19×19{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = {1^9} + {1^9} + {1^9} - \dfrac{1}{{{1^9} \times {1^9} \times {1^9}}}
x9+y9+z9−1x9y9z9=1+1+1−11×1×1{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 1 + 1 + 1 - \dfrac{1}{{1 \times 1 \times 1}}
x9+y9+z9−1x9y9z9=3−11{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 3 - \dfrac{1}{1}
x9+y9+z9−1x9y9z9=3−1{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 3 - 1
x9+y9+z9−1x9y9z9=2{x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 2

So, the correct answer is “Option C”.

Note: The range of sine inverse function is (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). So, the maximum value that sine inverse function can take is π2\dfrac{\pi }{2}. Because of this we have split 3π2\dfrac{{3\pi }}{2}into three π2\dfrac{\pi }{2}. Sum of sin⁡−1x+sin⁡−1y+sin⁡−1z{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z can be 3π2\dfrac{{3\pi }}{2}only if value of each function is equal or less than π2\dfrac{\pi }{2}. This thing is very important to solve the question. And same is with other trigonometric functions.