# Thread: Find a in terms of b

1. ## Find a in terms of b

For $\pi with $cos(a) = -sin(b)$ where $0, find $a$ in terms of $b$.

I am terrible at maths so please show any simple easy to understand working out so that I can understand better.

Any help would be appreciated!

2. Originally Posted by Joker37
For $\pi with $cos(a) = -sin(b)$ where $0, find $a$ in terms of $b$.

I am terrible at maths so please show any simple easy to understand working out so that I can understand better.

Any help would be appreciated!
Recall the following "transition formula" for going from cosine to sine, where $x$ is arbitrary:

$\cos(x) = \sin(x+\frac{\pi}{2}).$

Since the numbers $\frac{\pi}{2}$ and $-\frac{3\pi}{2}$ are different by $2\pi$, and since sine is periodic with period $2\pi$, we have also that

$\sin(x+\frac{\pi}{2})=\sin(x-\frac{3\pi}{2}),$

whence

$\cos(x) = \sin(x-\frac{3\pi}{2}).$

Since $\pi, we have that $a-\frac{3\pi}{2}$ lies in the interval $(-\frac{\pi}{2},0).$

Since you want to solve $\cos(a) = -\sin(b)$ for $a$, we have by the transition formula above that we might just as well solve

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

for $a$, where the last equality follows from sine being an odd function.

Since $0, we have that $-b$ lies in the interval $(-\frac{\pi}{2},0)$. Hence both $a-\frac{3\pi}{2}$ and $-b$ are numbers in the interval $(-\frac{\pi}{2},0)$, on which sine is injective, so in solving

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

for $a$, we can apply $\sin^{-1}$ to both sides to obtain

$a-\frac{3\pi}{2} = -b,$

or

$a = -b+\frac{3\pi}{2}.$

3. I don't understand all of the above.

Below was my attempt:
$cos(a)=-sin(b)$
$cos(\pi+a)=-cos(\frac{\pi}{2}-b)$
$\pi+a=\frac{\pi}{2}-b$
$a=\frac{-\pi}{2}-b$

$\frac{3\pi}{2}-b$

Where did I go wrong?

4. Originally Posted by Joker37
I don't understand all of the above.

Below was my attempt:
$cos(a)=-sin(b)$
$cos(\pi+a)=-cos(\frac{\pi}{2}-b)$
How is
$cos(a) = cos( \pi + a)$

-Dan

5. Originally Posted by topsquark
How is
$cos(a) = cos( \pi + a)$

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

?

But still, how would this change the result?
edit: $-\pi -a = \frac{\pi}{2}-b$

?

edit2: $a=\frac{3\pi}{2}-b$

Is this the right methodology?

6. Originally Posted by Joker37
$cos(a)=-sin(b)$
$-cos(\pi+a)=-cos(\frac{\pi}{2}-b)$

?

But still, how would this change the result?
edit: $-\pi -a = \frac{\pi}{2}-b$

?

edit2: $a=\frac{3\pi}{2}-b$

Is this the right methodology?
Looks good to me. But as HappyJoe's post indicates there are a few landmines we need to step over to say that we can do this.

-Dan

7. Originally Posted by topsquark
Looks good to me. But as HappyJoe's post indicates there are a few landmines we need to step over to say that we can do this.

-Dan
What sort of things?
I don't really understand HappyJoe's working out.
I'm terrible at maths.

8. Originally Posted by Joker37
What sort of things?
I don't really understand HappyJoe's working out.
I'm terrible at maths.
I think it's best for you to go through HappyJoe's post and let us know what line or lines you don't understand. We can help you better that way.

-Dan