Question

With reference to the principal values, if sin-1x + sin-1y + sin-1z = $\frac {3π}{2}$, then x100 + y100 + z100 =?

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (C)

3

Answer 2

sin-1x + sin-1y + sin-1z = $\frac {3π}{2}$
We know that the principal values of sin-1θ lie between -$\frac {π}{2}$ and $\frac {π}{2}$. Since the sum of the three angles is equal to $\frac {3π}{2}$, it means that each angle must be equal to $\frac {π}{2}$. Therefore, we have:
sin-1x = $\frac {π}{2}$
sin-1y = $\frac {π}{2}$
sin-1z = $\frac {π}{2}$
Taking the sine of both sides of these equations:
sin(sin-1x) = sin $\frac {π}{2}$
sin(sin-1y) = sin $\frac {π}{2}$
sin(sin-1z) = sin $\frac {π}{2}$
Using the inverse sine function's property sin(sin-1θ) = θ, we simplify:
x = 1
y = 1
z = 1
Now, we can calculate the sum of their 100th powers:
x100 + y100 + z100 = 1100 + 1100 + 1100 = 1 + 1 + 1 = 3
Therefore, the value of x100 + y100 + z100 is 3.
Among the given options, (C) 3 is the correct answer.