# exact values practice question

• Mar 12th 2009, 06:03 AM
jvignacio
exact values practice question
Suppose cos A = $\displaystyle -\frac{7}{25}$ and tan B = $\displaystyle \frac{15}{8}$
where A is in the second quadrant and B is in the third quadrant. Find the exact values of:

sin(A-B):

tan(A-B):

any help much appreciated.

jv
• Mar 12th 2009, 06:16 AM
red_dog
$\displaystyle \sin A=\sqrt{1-\cos^2A}=\frac{24}{25}$

$\displaystyle \sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}=-\frac{15}{17}$

$\displaystyle \cos B=-\frac{1}{\sqrt{1+\tan^2B}}=-\frac{8}{17}$

$\displaystyle \tan B=\frac{15}{8}$

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

$\displaystyle \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$
• Mar 14th 2009, 03:24 AM
jvignacio
Quote:

Originally Posted by red_dog
$\displaystyle \sin A=\sqrt{1-\cos^2A}=\frac{24}{25}$

$\displaystyle \sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}=-\frac{15}{17}$

$\displaystyle \cos B=-\frac{1}{\sqrt{1+\tan^2B}}=-\frac{8}{17}$

$\displaystyle \tan B=\frac{15}{8}$

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

$\displaystyle \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

ahh yeah question how did u get $\displaystyle \frac{24}{25}$ from $\displaystyle \sqrt{1-\cos^2A}$ ?

EDIT: sorry i understand now $\displaystyle \sqrt{1-(\frac{-7}{25})^2}]$
• Mar 14th 2009, 03:47 AM
jvignacio
Quote:

Originally Posted by red_dog
$\displaystyle \sin A=\sqrt{1-\cos^2A}=\frac{24}{25}$

$\displaystyle \sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}=-\frac{15}{17}$

$\displaystyle \cos B=-\frac{1}{\sqrt{1+\tan^2B}}=-\frac{8}{17}$

$\displaystyle \tan B=\frac{15}{8}$

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

$\displaystyle \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

sorry question how did u get $\displaystyle sin B=-\frac{\tan B}{\sqrt{1+\tan^2B}}$ 3rd quadrant is tan, yes but then u got another tan ontop where numerator is . why?