Need help in solving trigo an

**ZGOON** · September 22nd 2012, 10:37 PM

Given that tan(2A) = (2tanA)/(1-tan^2 A)

show that tan(22.5) = sqrt(2) - 1

**MarkFL** · September 22nd 2012, 11:03 PM

Define $A\equiv22.5^{\circ}$ and write the given double angle identity for tangent as:

$\tan(2(22.5^{\circ}))=\frac{2\tan(22.5^{\circ})}{1-\tan^2(22.5^{\circ})}$

$\tan(45^{\circ})=\frac{2\tan(22.5^{\circ})}{1-\tan^2(22.5^{\circ})}$

$1=\frac{2\tan(22.5^{\circ})}{1-\tan^2(22.5^{\circ})}$

Can you finish?

**ZGOON** · September 22nd 2012, 11:09 PM

Originally Posted by MarkFL2

Define $A\equiv22.5^{\circ}$ and write the given double angle identity for tangent as:

$\tan(2(22.5^{\circ}))=\frac{2\tan(22.5^{\circ})}{1-\tan^2(22.5^{\circ})}$

$\tan(45^{\circ})=\frac{2\tan(22.5^{\circ})}{1-\tan^2(22.5^{\circ})}$

$1=\frac{2\tan(22.5^{\circ})}{1-\tan^2(22.5^{\circ})}$

Can you finish?

I managed to get that part...
and i let tan(22.5) be Y
it becomes eqn [y^2 + 2y - 1 = 0]
but somehow i cant get the sqrt(2)

*im using the cross multiply method to solve the eqn

**MarkFL** · September 22nd 2012, 11:14 PM

You have the correct quadratic...now either complete the square or use the quadratic formula.

You will get 2 roots. You will need to decide which is extraneous, either by substituting the roots into the original equation, or some knowledge of which quadrant the angle is in and what sign tangent has there...

**ZGOON** · September 23rd 2012, 12:11 AM

Ohhhh! I'm im able to solve it now!!! THANKS THANKS THANKS!!!

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