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Question: If we have an expression \({\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z + {\sin ^{ - 1}}t = 2...

If we have an expression sin⁡−1x+sin⁡−1y+sin⁡−1z+sin⁡−1t=2π{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z + {\sin ^{ - 1}}t = 2\pi then, find the value of x2+y2+z2+t2{x^2} + {y^2} + {z^2} + {t^2} .

Explanation

Solution

The above inverse-trigonometry equation can be solved as the basic idea that can be drawn from it that the domain of the sin⁡−1x{\sin ^{ - 1}}x lies between [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] . The maximum value of the sin⁡−1x{\sin ^{ - 1}}x is π2\dfrac{\pi }{2} and the value of the given expression can also be found.

Complete step-by-step solution:
The given equation is sin⁡−1x+sin⁡−1y+sin⁡−1z+sin⁡−1t=2π{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z + {\sin ^{ - 1}}t = 2\pi
As it is known that the value of sin⁡−1x∈[−π2,π2]{\sin ^{ - 1}}x \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]
∵\because It can be seen that the max value of sin⁡−1x{\sin ^{ - 1}}x is π2\dfrac{\pi }{2} .
Also the value of RHS of the above equation is 2π2\pi .
∴\therefore Each of sin⁡−1x, sin⁡−1y, sin⁡−1z and sin⁡−1t{\sin ^{ - 1}}x,{\text{ }}{\sin ^{ - 1}}y,{\text{ }}{\sin ^{ - 1}}z{\text{ and }}{\sin ^{ - 1}}t must be equal to π2\dfrac{\pi }{2} ∴ sin⁡−1x=sin⁡−1y=sin⁡−1z=sin⁡−1t=π2\therefore {\text{ }}{\sin ^{ - 1}}x = {\sin ^{ - 1}}y = {\sin ^{ - 1}}z = {\sin ^{ - 1}}t = \dfrac{\pi }{2}
⇒sin⁡−1x=π2\Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{2}
⇒x=sin⁡π2\Rightarrow x = \sin \dfrac{\pi }{2}
⇒x=1\Rightarrow x = 1
Again,
y=z=t=sin⁡π2y = z = t = \sin \dfrac{\pi }{2}
∴y=1\therefore y = 1
z=1z = 1
And,
t=1t = 1
On putting the values of x, y, z and t in x2+y2+z2+t2{x^2} + {y^2} + {z^2} + {t^2}
⇒x2+y2+z2+t2=12+12+12+12\Rightarrow {x^2} + {y^2} + {z^2} + {t^2} = {1^2} + {1^2} + {1^2} + {1^2}
On simplifying further,
⇒x2+y2+z2+t2=1+1+1+1\Rightarrow {x^2} + {y^2} + {z^2} + {t^2} = 1 + 1 + 1 + 1
On adding,
⇒x2+y2+z2+t2=4\Rightarrow {x^2} + {y^2} + {z^2} + {t^2} = 4
∴\therefore The value of x2+y2+z2+t2{x^2} + {y^2} + {z^2} + {t^2} is 4.

Note: This is trigonometry-based question and in order to solve this the identities related to it must be known. The equation should be solved carefully. The quadrant in which it lies must be seen carefully and according to that the sign of the trigonometric-function is determined. Remember that in the first quadrant all trigonometry functions are positive, in the second quadrant only sin⁡θ\sin \theta is positive, in the third quadrant only tan⁡θ\tan \theta is positive and in the fourth quadrant only cos⁡θ\cos \theta is positive. Calculations should be done carefully to avoid any mistake. After the final answer is found out it can be checked that whether it satisfies the original equation given in the question by simply substituting its value in the equation and if it does not satisfy the equation then the solution must be rechecked. The equation should be solved in accordance with the identities which would result in the correct solution. Always try to solve the question step by step so that the wrong step can be determined and changed.