# Thread: [SOLVED] How can i show this...trig identities??

1. ## [SOLVED] How can i show this...trig identities??

How can i show that

$2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)$

2. Originally Posted by maths900
How can i show that

$2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)$

Formula

$
\sqrt{\frac{1+cos(\theta)}{2}} = cos(\frac{\theta}{2})
$

$

\sqrt{\frac{1-cos(\theta)}{2}} = sin(\frac{\theta}{2})
$

Now its better if you try it and ask incase of further troubles

3. Originally Posted by maths900

How can i show that $2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)$
this is not an identity! for example it's not true if $\theta = 2\pi.$ in order to make it an identity we need $\frac{\theta}{2}$ to be in the quadrant 1 or 4, where cosine is positive. the proof then is very easy.

just use the identities: $1+\cos(2x) = 2 \cos^2 x$ and $1-\cos(2x)=2\sin^2x.$

4. Hello,
Originally Posted by maths900
How can i show that

$2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)$
$\cos(2x)=1-2 \sin^2(x)$
Hence $2 \sin^2(\theta/4)=1-\cos(\theta/2)$

So $4 \sin^2(\theta/4)=2-2\cos(\theta/2)$

Then $\cos(2x)=2 \cos^2(x)-1 \Rightarrow 2\cos(\theta)=(2 \cos^2(\theta/2))^2-2 \Rightarrow (2\cos^2(\theta/2))^2=2+2 \cos(\theta)$
---> $2 \cos(\theta/2)=\pm \sqrt{2+2 \cos(\theta)}$

Now you have to see in which quadrant $\theta/2$ is...

Originally Posted by ADARSH
Formula

$
\sqrt{\frac{1+cos(\theta)}{2}} = cos(\frac{\theta}{2})
$

$

\sqrt{\frac{1-cos(\theta)}{2}} = sin(\frac{\theta}{2})
$

Now its better if you try it and ask incase of further troubles
A cosine or a sine can be negative...

5. Hello, maths900!

We need the double-angle identities:

. . $\sin^2\!A \:=\:\frac{1-\cos2A}{2}$

. . $\cos^2\!A \:=\:\frac{1+\cos2A}{2}$

Show that: . $2-\sqrt{2+2\cos\theta}\;=\;4\sin^2\!\left(\tfrac{\th eta}{4}\right)$

The left side is:

. . $2 - \sqrt{2(1 + \cos\theta)}$

. . $=\;2-\sqrt{4\left(\frac{1+\cos\theta}{2}\right)}$

. . $=\;2 - \sqrt{4\cos^2\!\tfrac{\theta}{2}}$

. . $=\;2-2\cos\tfrac{\theta}{2}$

. . $=\;2\left(1 - \cos\tfrac{\theta}{2}\right)
$

. . $= \;4\left(\frac{1-\cos\frac{\theta}{2}}{2}\right)$

. . $=\;4\sin^2\!\tfrac{\theta}{4}$

6. Thanks everyone, thought i'd have to use the double angle identities but i'm not as good with them as you guys are, thank you