1. ## half angle formula

Express cosA in terms of t when tan(A/2)=t

i checked the solution but i don't understand it.

solution:

using the half-angle formula for tan, tan^2(A/2)=(1-cosA)/(1+cosA)

therefore cosA=(1-tan^2(A/2))/(1+tan^2(A/2))=(1-t^2)/(1+t^2)

what i don't understand is how they get cosA in terms of tan^2(A/2).
how i would solve it is to multiply both sides of tan^2(A/2)=(1-cosA)/(1+cosA) with (1+cosA), but that won't get me cosA=(1-tan^2(A/2))/(1+tan^2(A/2))

2. ## Re: half angle formula

Originally Posted by muddywaters
Express cosA in terms of t when tan(A/2)=t

i checked the solution but i don't understand it.

solution:

using the half-angle formula for tan, tan^2(A/2)=(1-cosA)/(1+cosA)

therefore cosA=(1-tan^2(A/2))/(1+tan^2(A/2))=(1-t^2)/(1+t^2)

what i don't understand is how they get cosA in terms of tan^2(A/2).
how i would solve it is to multiply both sides of tan^2(A/2)=(1-cosA)/(1+cosA) with (1+cosA), but that won't get me cosA=(1-tan^2(A/2))/(1+tan^2(A/2))
Note that :

$\cos^2 \left(\frac{A}{2}\right) = \frac{1}{1+t^2} ~\text{and}~\sin^2 \left(\frac{A}{2}\right) = \frac{t^2}{1+t^2}$

hence :

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

3. ## Re: half angle formula

sorry, i don't get it >.<

the question is this:

what i dont understand is this:
question 2.bmp

4. ## Re: half angle formula

$\tan^2\left(\frac{a}{2}\right)=\frac{1-\cos(a)}{1+\cos(a)}$
$\Leftrightarrow \tan^2\left(\frac{a}{2}\right)[1+\cos(a)]=1-\cos(a)$
$\Leftrightarrow \tan^2\left(\frac{a}{2}\right)+\tan^2\left(\frac{a }{2}\right)\cos(a)=1-\cos(a)$
$\Leftrightarrow \tan^2\left(\frac{a}{2}\right)\cos(a)+\cos(a)=1-\tan^2\left(\frac{a}{2}\right)$
$\Leftrightarrow \cos(a)\left[\tan^2\left(\frac{a}{2}\right)+1\right]=1-\tan^2\left(\frac{a}{2}\right)$
$\Leftrightarrow \cos(a)=\frac{1-\tan^2\left(\frac{a}{2}\right)}{1+\tan\left(\frac{ a}{2}\right)}$
$\Rightarrow \cos(a)=\frac{1-t^2}{1+t^2}$

5. ## Re: half angle formula

wow awesome. i just didn't see that! thanks so much! super fast reply there's no one i could ask, thanks ^^