# Find a in terms of b

• Apr 30th 2011, 09:51 AM
Joker37
Find a in terms of b
For $\displaystyle \pi<a<\frac{3\pi}{2}$ with $\displaystyle cos(a) = -sin(b)$ where $\displaystyle 0<b<\frac{\pi}{2}$, find $\displaystyle a$ in terms of $\displaystyle b$.

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!
• Apr 30th 2011, 11:16 AM
HappyJoe
Quote:

Originally Posted by Joker37
For $\displaystyle \pi<a<\frac{3\pi}{2}$ with $\displaystyle cos(a) = -sin(b)$ where $\displaystyle 0<b<\frac{\pi}{2}$, find $\displaystyle a$ in terms of $\displaystyle b$.

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 $\displaystyle x$ is arbitrary:

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

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

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

whence

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

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

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

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

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

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

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

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

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

or

$\displaystyle a = -b+\frac{3\pi}{2}.$
• Apr 30th 2011, 06:25 PM
Joker37
I don't understand all of the above.

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

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

Where did I go wrong?
• Apr 30th 2011, 06:29 PM
topsquark
Quote:

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

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

How is
$\displaystyle cos(a) = cos( \pi + a)$

-Dan
• Apr 30th 2011, 06:31 PM
Joker37
Quote:

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

-Dan

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

?

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

?

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

Is this the right methodology?
• Apr 30th 2011, 07:01 PM
topsquark
Quote:

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

?

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

?

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

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
• Apr 30th 2011, 07:13 PM
Joker37
Quote:

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.
• Apr 30th 2011, 07:38 PM
topsquark
Quote:

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