Question

Pleaseeee help w these math questions! please show work!!! 50 points

Answer 1

9. You need an exact answer. So I will give you it but I'll also give you an approximation because it might help you understand it more.

Because we know the cosine fo theta, we can evaluate the inverse cosine of -2/3 first to find theta.

$sin^{-1}(-2/3)$ = 131.81° approx.

Because 131.81° is in the second quadrant its third quadrant partner is 228.19°. Which would make the sine 0.75 approximately.

But we need an exact answer I suppose so which is this disgusting mess...$sin(2\pi -cos^{-1}(\frac{-2}{3}))$. I know it looks scary but it is basically  all the steps we just did but without evaluating anything.

This can be simplified using: sin(x− y) = sinxcosy−cosxsiny

To... $-sin(cos^{-1}(\frac{-2}{3}))$

answer: $-sin(cos^{-1}(\frac{-2}{3}))$

10. Okay. So because we have a point we can say that...

θ = $tan^{-1}(-9/11)$

sin(θ) = +sin($tan^{-1}(-9/11)$)

11. arcsin(-0.37) = $sin^{-1}(-0.37)$ = -21.72°+2kπ or 201.72°+2kπ approx. where k has to be an integer

answer: idk if you want one or more solutions so I gave you them all.

12. arccos(-√3/2) = $cos^{-1}(\frac{-\sqrt{3} }{2}) = \frac{5\pi }{6}+2k\pi, \frac{7\pi }{6}+2k\pi$

Remember that "k" must be an integer.

answer: 5π/6