## sabika13 3 years ago Write an equation this is equivalent to f(x)=-5sin(x-pi/2) - 8 using trig.function of x

1. richyw

ok so you have $f(x)=-5\sin\left(x-\frac{\pi}{2}\right)-8$ For now just focus on $g(x)=\sin\left(x-\frac{\pi}{2}\right)$ what does this do to the graph?

2. richyw

where $$x=\pi/2$$, what is $\sin\left(x-\frac{\pi}{2}\right)$also brb. just gotta hand in something quick...

3. sabika13

okay.. will it be sin(x-pi/2)=0??

4. sabika13

i dont really understand the question itself

5. richyw

sorry open study died on me.

6. richyw

You are correct that $$g(\pi/2)=0$$ You can repeat this with some other key x-values $$x=\pi/2,\quad x=\pi ,\quad x=0$$ and find that what you get is $\sin\left(x-\frac{\pi}{2}\right)=-\cos\pi$

7. richyw

you will eventually just memorize these phase shifts, but if you ever forget on a test, you can always make a quick sketch using the method I explained. So now remember that we called $g(x)=\sin\left(x-\frac{\pi}{2}\right)=-\cos x$ Going back to the original function you have $f(x)=-5\sin\left(x-\frac{\pi}{2}\right)-8$$f(x)=-5g(x)-8$So the hard part is already done. The 8 is all by itself so it just shifts the up and down, and the -5 just makes the amplitude of the wave greater. All you have to do now is plug in the expression you worked out for g(x). $f(x)=-5(-\cos x)-8$$f(x)=5\cos x -8$

8. richyw

Oops made a mistake, in the second last post. It should say : You are correct that $$g(π/2)=0$$ You can repeat this with some other key x-values $$x=π/2,\quad x=π,\quad x=0$$ and find that what you get is $\sin\left( x−\frac{\pi}{2}\right)=−\cos x$

9. sabika13

sorry i fell alseep -_- .. accoriding the the cofunction identities isnt it : cosx=sin(pi/2-x)? so using that you can get -cosx=sin(x-pi/2) (sorrydont understand)