# Need help with prooving

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• May 12th 2006, 03:27 AM
xanith_47
Need help with prooving
Im not a very good mathematician, and im only in grade 10..and need some help

Question is: Show
sin(180-sinB)
____________ = -tanB
cos(180-B)

Sorry if this question is really easy, i must really suck.
Ive tried doing it:

sinB
____ = tan B
cosB

.: sin B- sinx
___________ = -tanx
cos B- cosx

Is this method right, and in an exam will i get full marks for it?

Another question:
Simplify cos(90-b)
_________
sin(180-b)

I havent the slightest idea what to do.
I understand that cosx over sinx= 1/2 x tanx..but the values of cos and sin are different in this question... so is it implying that the answer is:

cos90- cos b
_____________ = 1/4 x tan (since cos over sin is 1/2tan, this is half)
sin 180- sin b

And i need to solve for b....

Thanks in advance.
• May 12th 2006, 04:09 AM
CaptainBlack
Quote:

Originally Posted by xanith_47
Im not a very good mathematician, and im only in grade 10..and need some help

Question is: Show
sin(180-B)
____________ = -tanB
cos(180-B)

Use the difference formulas for $\displaystyle \cos$ and $\displaystyle \sin$:

$\displaystyle \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$

and:

$\displaystyle \cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$

So if $\displaystyle A=180^\circ$:

$\displaystyle \sin(180-B)=\sin(180)\cos(B)-\cos(180)\sin(B)$$\displaystyle =\sin(B) \displaystyle \cos(180-B)=\cos(180)\cos(B)+\sin(180)\sin(B)$$\displaystyle =-\cos(B)$

Hence:

$\displaystyle \frac{\sin(180-B)}{\cos(180-B)}=\frac{\sin(B)}{-\cos(B)}=-tan(B)$

RonL