Proving a Trigonometric Identity

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Dec 20th 2009, 06:51 PM
KelvinScale

Proving a Trigonometric Identity

Hi everyone:

Is there any way to prove the following identity:

tan(A/2)=(sinA)/(1+cosA)

without drawing a diagram? I know that it is a basic trig identity, but I don't know how to prove it by manipulating hte formulas.

Thanks!
Dec 20th 2009, 07:30 PM
bigwave

use sin(x/2) / cos(x/2)

$\tan{\frac{A}{2}}=\frac{\sin{A}}{1+\cos{A}}$

this derived from by using

$\tan{\frac{A}{2}}=\frac{\sin\frac{A}{2}}{\cos\frac {A}{2}}$

$ \frac {\pm\sqrt{\frac{1-\cos{A}}{2}}} {\pm\sqrt{\frac{1+\cos{A}}{2}}} $

$=\pm\sqrt\frac{\left(1-\cos{A}\right)\times\left(1+cos{A}\right)} {\left(1+\cos{A}\right)\times\left(1+\cos{A}\right )} $

$=\pm\sqrt\frac{1-\cos^2{A}} {\left(1+\cos{A}\right)^2} $

$=\pm\vert\frac{\sin{A}}{1-\cos{A}}\vert$

$1-\cos{A}$ in never negative, so the sign of the fractional expression depends only on the sign of $\sin{A}$
Dec 20th 2009, 07:36 PM
Soroban

Hello, Kelvin!
We need these two identities: . $\begin{array}{ccc}\sin\frac{\theta}{2} &=& \sqrt{\dfrac{1-\cos\theta}{2}} \\ \\[-4mm]\cos\frac{\theta}{2} &=& \sqrt{\dfrac{1+\cos\theta}{2}} \end{array}$

Quote:

Prove: . $\tan\frac{A}{2} \:=\:\frac{\sin A}{1+\cos A}$

$\tan\frac{A}{2} \;=\;\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \;=\;\frac{\sqrt{\dfrac{1-\cos A}{2}}} {\sqrt{\dfrac{1 + \cos A}{2}}} \;=\; \sqrt{\frac{1-\cos A}{1+\cos A}}$

Multiply by $\frac{1+\cos A}{1 + \cos A}$

. . $\sqrt{\frac{1-\cos A}{1+\cos A}\cdot\frac{1+\cos A}{1 + \cos A}} \;=\;\sqrt{\frac{1-\cos^2\!A}{(1+\cos A)^2}} \;=\;\sqrt{\frac{\sin^2\!A}{(1+\cos A)^2}} \;=\; \frac{\sin A}{1 + \cos A}$
Dec 20th 2009, 07:38 PM
I-Think

$tan(\frac{A}{2})=\frac{sinA}{1+cosA}$

Manipulating R.H.S.
$\frac{2sin(\frac{A}{2}){cos(\frac{A}{2})}}{1+cos^2 (\frac{A}{2})-sin^2(\frac{A}{2})}$

Note that
$1=sin^2(\frac{A}{2})+cos^2(\frac{A}{2})$

Introduce into the equation and you should solve your problem

Edit
In the time I took to write this I was beaten by 2 other forum users
I really need to take typing classes.
On the other hand, at least I provided a different method
Dec 20th 2009, 07:41 PM
KelvinScale

Thank you so much!
Dec 20th 2009, 09:11 PM
KelvinScale

Okay, Similar Question.

Prove that:

tan(A/2)=(1+sinA-cosA)/(1+sinA+cosA)

I tried replacing tan(A/2) with sinA/(1+cosA), but I could not find a way to add the extra components to both the numerator and denominator. Gosh this is frustrating.
Dec 21st 2009, 05:55 AM
Soroban

Helolo, Kelvin!

This one is tricky . . .

Quote:

Prove: . $\tan\frac{A}{2} \;=\;\frac{1+\sin A-\cos A}{1+\sin A+\cos A}$

Start with the right side . . .

Multiply by $\frac{1+\sin A - \cos A}{1 + \sin A - \cos A}$

. . $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}\cdot{\color{blue}\frac{1 + \sin A - \cos A}{1 + \sin A - \cos A}} \;=\;\frac{(1+\sin A - \cos A)^2}{(1+\sin A)^2 - \cos^2\!A}$

. . $=\; \frac{1 + 2\sin A - 2\cos A + \sin^2\!A - 2\sin A\cos A + \cos^2\!A}{1 + 2\sin A + \sin^2\!A - \cos^2\!A}$

. . $=\; \frac{1 + \overbrace{\sin^2\!A + \cos^2\!A}^{\text{This is 1}} + 2\sin A - 2\cos A - 2\sin A\cos A}{2\sin A + \sin^2\!A + \underbrace{1 - \cos^2\!A}_{\text{This is }\sin^2\!A}}$

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

. . $=\; \frac{1 + \sin A - \cos A - \sin A\cos A}{\sin A(1 + \sin A)} \;=\;\frac{(1 + \sin A) - \cos A(1 + \sin A)}{\sin A(1 + \sin A)}$

. . $=\; \frac{(1+\sin A)(1 - \cos A)}{\sin A(1 + \sin A)} \;=\;\frac{1-\cos A}{\sin A}$

Multiply by $\frac{1+\cos A}{1 + \cos A}\!:\quad \frac{1-\cos A}{\sin A}\cdot{\color{blue}\frac{1+\cos A}{1+\cos A}} \;=\; \frac{1-\cos^2\!A}{\sin A(1 + \cos A)}$

. . . . . . . . . . . . . $=\;\frac{\sin^2\!A}{\sin A(1 + \cos A)} \;=\;\frac{\sin A}{1 + \cos A}$

Finally: . $\frac{\sin A}{1 + \cos A} \;=\;\tan\frac{A}{2} \quad \hdots\;There!$
Dec 21st 2009, 07:00 AM
Krizalid

be careful with what angle you're working on, those identities are not true for all angles.