Verifying a half angle formula

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July 1st 2012, 07:50 AM
dudewithmathproblems

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Verifying a half angle formula

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I have tried this one several times and I feel like I am going in circles. Probably has an easy solution and I will be embarrassed. Any help or guidance will be appreciated. Thank You

I used this to write the equation: Online LaTeX Equation Editor - create, integrate and download
July 1st 2012, 08:31 AM
BobP

Re: Verifying a half angle formula

On the RHS, write $\tan x$ as $\sin x/\cos x$ and $\sec x$ as $1/\cos x,$ multiply top and bottom by $\cos x$ and then, (assuming that you are allowed to), make use of the identities

$\sin 2A=\2\sin A\cos A$

and

$\cos 2A=2\cos^{2}A-1.$
July 1st 2012, 12:33 PM
Soroban

Re: Verifying a half angle formula

Hello, dudewithmathproblems!

Quote:

$\text{Prove: }\:\tan\frac{x}{2} \;=\; \frac{\tan x}{\sec x + 1}$

We can work from the LHS . . .

$\tan\frac{x}{2} \;=\;\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \;=\; \frac{\sqrt{\frac{1-\cos x}{2}}}{\sqrt{\frac{1+\cos x}{2}}} \;=\;\sqrt{\frac{1-\cos x}{1+\cos x}} \;=\;\sqrt{\frac{1-\cos x}{1+\cos x}\cdot\frac{1+\cos x}{1+\cos x}}$

. . . . . $=\;\sqrt{\frac{1-\cos^2x}{(1+\cos x)^2}} \;=\; \sqrt{\frac{\sin^2x}{(1+\cos x)^2}} \;=\;\frac{\sin x}{1+\cos x}$

Divide numerator and denominator by $\cos x\!:$

. . . . . $\frac{\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x} + \dfrac{\cos x}{\cos x}} \;=\;\frac{\tan x}{\sec x + 1}$

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