Answer:
Step-by-step explanation:
First, we know that the sin function is odd which means:
sin(-x) = -sin(x).
Secondly evaluating an inverse trigonometric function with a normal trigonometric function as the argument can be rewritten as an algebraic expression.
Let [tex]t = \sin(-\frac{11\pi}{4}) = - \sin(\frac{11\pi}{4})[/tex]
We know the certain identity.
[tex]\sin(\theta) = \sin(2\pi + \theta)[/tex]
We use it to evaluate sin(11 pi / 4).
[tex]\sin(\frac{11 \pi}{4}) = \sin({\frac{8\pi}{4} + \frac{3 \pi}{4}}) = \sin(2\pi + \frac{3 \pi}{4}) = \sin(\frac{3\pi}{4})[/tex]
Another helping identity is the following:
[tex]\sin(\theta) = \sin(\pi - \theta)[/tex]
[tex]\sin(\frac{3\pi}{4}) = \sin(\pi - \frac{3\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}[/tex]
But let's not forget that t = -sin(11 pi/4) = - sqrt(2) / 2
Now we end up with the following equation.
[tex]\cos^{-1}(-\frac{\sqrt{2}}{2}) = x\\\cos(x) = -\frac{\sqrt{2}}{2} => x = \frac{3\pi}{4}[/tex]
