help me with this trig problem

[ryan882]

By writing cos x in terms of 1/2x, find an alternative expression for,
1-cosx
1+cosx

 

[MarkFL]

Use double-angle identities for cosine, since:

$\displaystyle \cos(x)=\cos\left(2\left(\dfrac{1}{2}x \right) \right)$

 

[soroban]

Hello, ryan882!

\(\displaystyle \text{By writing }\,\cos x\text{ in terms of }\frac{x}{2},\,\text{ find an alternative expression for: }\,
\dfrac{1-\cos x}{1+\cos x}\)

Use these two identities:

. . $\displaystyle \sin^2\!\frac{x}{2} \:=\:\dfrac{1-\cos x}{2} \quad\Rightarrow\quad 1 - \cos x \:=\:2\sin^2\!\frac{x}{2}$
. . $\displaystyle \cos^2\!\frac{x}{2} \:=\: \dfrac{1 + \cos x}{2} \quad\Rightarrow\quad 1 + \cos x \:=\:2\cos^2\!\frac{x}{2}$


$\displaystyle \text{Then: }\:\dfrac{1-\cos x}{1+\cos x} \:=\:\dfrac{2\sin^2\!\frac{x}{2}}{2\cos^2\!\frac{x}{2}} \:=\:\dfrac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} \:=\:\left(\dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\right)^2 \:=\:\tan^2\!\frac{x}{2}$

 

[ryan882]

Hello, ryan882!


Use these two identities:

. . $\displaystyle \sin^2\!\frac{x}{2} \:=\:\dfrac{1-\cos x}{2} \quad\Rightarrow\quad 1 - \cos x \:=\:2\sin^2\!\frac{x}{2}$
. . $\displaystyle \cos^2\!\frac{x}{2} \:=\: \dfrac{1 + \cos x}{2} \quad\Rightarrow\quad 1 + \cos x \:=\:2\cos^2\!\frac{x}{2}$


$\displaystyle \text{Then: }\:\dfrac{1-\cos x}{1+\cos x} \:=\:\dfrac{2\sin^2\!\frac{x}{2}}{2\cos^2\!\frac{x}{2}} \:=\:\dfrac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} \:=\:\left(\dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\right)^2 \:=\:\tan^2\!\frac{x}{2}$

Thanks alot