Proving Trig Identities

**orochimaru700** · March 10th 2010, 05:27 PM

Help proving these identities

1.) tanX-tanY / 1+tanX = cotY-cotX / cotXcotY+1

2.) sin5X + Sin3X / sin5X-sin3X = tan4X / tanX

**Grandad** · March 11th 2010, 12:15 AM

Hello orochimaru700

Originally Posted by orochimaru700

Help proving these identities

1.) tanX-tanY / 1+tanX = cotY-cotX / cotXcotY+1

2.) sin5X + Sin3X / sin5X-sin3X = tan4X / tanX

Can I suggest first that:

you should copy the question down carefully; and:
either learn how to write these things using LaTeX (lots of information available on this web-site); or:
use brackets correctly.

I'm sure you mean for number (1):

(tanX-tanY) / (1+tanXtanY) = (cotY-cotX) / (cotXcotY+1)

Anyway, enough of my ranting. Divide top-and-bottom of the fraction on the LHS by $\tan X\tan Y$ :

$\frac{\tan X - \tan Y}{1+ \tan X\tan Y}$

$=\frac{\dfrac{\tan X}{\tan X\tan Y} - \dfrac{\tan Y}{\tan X\tan Y}}{\dfrac{1}{\tan X\tan Y}+ \dfrac{\tan X\tan Y}{\tan X\tan Y}}$

$=\frac{\dfrac{1}{\tan Y} - \dfrac{1}{\tan X}}{\dfrac{1}{\tan X}\dfrac{1}{\tan Y}+ \dfrac{1}{\tan X\tan Y}}$

$=\frac{\cot Y - \cot X}{\cot X \cot Y +1}$

For (2) use the sum-to-product formulae. For example, the numerator becomes $2\sin 4x\cos x$ . Use a similar formula on the denominator, simplify and you're there.

Can you complete this now?

Grandad

**orochimaru700** · March 12th 2010, 04:57 PM

thnks for the advice and solution grandad, i'll try solve for # 2

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