Find the values of $a$ and $b$ ~ Trigonometry

Question

The function $f$, where $f(x) = a \sin x+b$, is defined for the domain $0 \leq x \leq 2\pi$. Given that $f(\frac{1}{2}\pi)=2$ and that $f(\frac{3}{2}\pi)=-8$,

find the values of $a$ and $b$.

I know we have to solve it simultaneously,

$$a \sin (\frac{1}{2}\pi) + b = 2) ---- (1)$$

$$a\sin(\frac{3}{2}\pi)+b=-8----(2)$$

From $(1)$: $$b=2-a\sin(\frac{1}{2}\pi)----(3)$$

And we substitute equation $3$ in $2$:

$$a\sin(\frac{3}{2}\pi) + 2 - a\sin(\frac{1}{2}\pi)+8=0$$

What I did is: $$a(\sin(\frac{3}{2}\pi) - \sin(\frac{1}{2}\pi)) = -10$$.

$$0.055a=10$$

Then solved $a$.

My value of $a$ is $-182$ but the real answer is $5$.

Help? Urgent?

Answer 1

Since $\sin (\frac{3\pi}{2}) = -1$ and $\sin (\frac{\pi}{2}) = 1$ then

$a \sin (\frac{3\pi}{2}) + 2 - a \sin (\frac{\pi}{2}) + 8 = -a + 2 - a + 8 = -2a + 10$ and this is equal to zero. You should be able to get $a=5$ from here. It's possible you were using a calculator set in degrees instead of radians.

Answer 2

we have $f(\pi/2)=a\sin(\pi/2)+b=2$ therefore we get $a+b=2$ since $\sin(\pi/2)=1$ and $f(3/2\pi)=a\sin(3/2\pi)+b=-8$ therefore we have $-a+b=-8$ since $\sin(3/2\pi)=-1$ now you have two equations $a+b=2$ and $-a+b=-8$ we get $a=5$ and $b=-3$