Find a in terms of b

**Joker37** · April 30th 2011, 09:51 AM

For $\pi<a<\frac{3\pi}{2}$ with $cos(a) = -sin(b)$ where $0<b<\frac{\pi}{2}$ , 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!

**HappyJoe** · April 30th 2011, 11:16 AM

Originally Posted by Joker37

For $\pi<a<\frac{3\pi}{2}$ with $cos(a) = -sin(b)$ where $0<b<\frac{\pi}{2}$ , 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 $\pi<a<\frac{3\pi}{2}$ , 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 $0<b<\frac{\pi}{2}$ , 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

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

**Joker37** · April 30th 2011, 06:25 PM

I don't understand all of the above.

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

However the answer is:
$\frac{3\pi}{2}-b$

Where did I go wrong?

**topsquark** · April 30th 2011, 06:29 PM

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

**Joker37** · April 30th 2011, 06:31 PM

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?

**topsquark** · April 30th 2011, 07:01 PM

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

**Joker37** · April 30th 2011, 07:13 PM

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.

**topsquark** · April 30th 2011, 07:38 PM

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

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