1. Prove(9)

Prove that $\frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=tan\fr ac{\theta}{2}$.

Hence or otherwise, find the values of $\theta$ between $0$ and $\pi$ which satisfies the equation $1+sin\theta=2cos\theta$. where $cos
\theta$
is not equals to zero

2. Originally Posted by Punch
Prove that $\frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=tan\fr ac{\theta}{2}$.

Hence or otherwise, find the values of $\theta$ between $0$ and $\pi$ which satisfies the equation $1+sin\theta=2cos\theta$. where $cos
\theta$
is not equals to zero
insert all the half angles stuff inside ..

knowing these identities ,

$\sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$

$\cos \theta=1-2\sin^2 \frac{\theta}{2}$

so

$\frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

bring out the common factor and do some necessary cancellings , then you will have your proof .

insert all the half angles stuff inside ..

knowing these identities ,

$\sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$

$\cos \theta=1-2\sin^2 \frac{\theta}{2}$

so

$\frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

bring out the common factor and do some necessary cancellings , then you will have your proof .
I don't see how the formulas used are related, what is the key idea to it?

To convert all to sin or cos?

4. Originally Posted by Punch

Prove that $\frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=tan\fr ac{\theta}{2}$.
Hi Punch,

you can also start with $Tan\frac{\theta}{2}=\frac{Sin(\frac{\theta}{2})}{C os(\frac{\theta}{2})}$

then...

$\frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{ Sin(\frac{\theta}{2})}{Cos(\frac{\theta}{2})}$ ?

giving...

$Cos\frac{\theta}{2}-Cos\theta\ Cos\frac{\theta}{2}+Sin\theta\ Cos\frac{\theta}{2}=Sin\frac{\theta}{2}+Cos\theta\ Sin\frac{\theta}{2}+Sin\theta\ Sin\frac{\theta}{2}$ ?

Now we can use the CosACosB, SinASinB and CosASinB identities to get

$\ Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta-\frac{1}{2}Cos\frac{\theta}{2}+\frac{1}{2}Sin\frac {3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}$

$=Sin\frac{\theta}{2}+\frac{1}{2}Sin\frac{3}{2}\the ta-\frac{1}{2}Sin\frac{\theta}{2}+\frac{1}{2}Cos\frac {\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta$ ?

$\frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}$

$=\frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}$ ? yes, it's true

5. Originally Posted by Archie Meade
Hi Punch,

you can also start with $Tan\frac{\theta}{2}=\frac{Sin(\frac{\theta}{2})}{C os(\frac{\theta}{2})}$

then...

$\frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{ Sin(\frac{\theta}{2})}{Cos(\frac{\theta}{2})}$ ?

giving...

$Cos\frac{\theta}{2}-Cos\theta\ Cos\frac{\theta}{2}+Sin\theta\ Cos\frac{\theta}{2}=Sin\frac{\theta}{2}+Cos\theta\ Sin\frac{\theta}{2}+Sin\theta\ Sin\frac{\theta}{2}$ ?

Now we can use the CosACosB, SinASinB and CosASinB identities to get

$\ Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta-\frac{1}{2}Cos\frac{\theta}{2}+\frac{1}{2}Sin\frac {3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}$

$=Sin\frac{\theta}{2}+\frac{1}{2}Sin\frac{3}{2}\the ta-\frac{1}{2}Sin\frac{\theta}{2}+\frac{1}{2}Cos\frac {\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta$ ?

$\frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}$

$=\frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}$ ? yes, it's true
My teacher once said that to solve a proving question, we have to prove from left to right or from right to left. And that we can't solve it by manipulating both sides and eventually getting $A=A$

insert all the half angles stuff inside ..

knowing these identities ,

$\sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$

$\cos \theta=1-2\sin^2 \frac{\theta}{2}$

so

$\frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$

bring out the common factor and do some necessary cancellings , then you will have your proof .
I think I need more help, I didn't solved it using this..

7. The calculations I showed prove that both sides are equal, Punch.

If $\frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{ Sin\frac{\theta}{2}}{Cos\frac{\theta}{2}}$

Then $Cos\frac{\theta}{2}(1-Cos\theta+Sin\theta)$ must equal $Sin\frac{\theta}{2}(1+Cos\theta+Sin\theta)$

The remaining calculations examine whether or not this is true,
but i know what you mean.

Using the double-angle formulae for CosA and SinA

$\frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{ 1-\frac{1-tan^2\frac{\theta}{2}}{1+tan^2{\frac{\theta}{2}}}+ \frac{2tan\frac{\theta}{2}}{1+tan^2{\theta}{2}}}{1 +\frac{1-tan^2\frac{\theta}{2}}{1+tan^2\frac{\theta}{2}}+\f rac{2tan\frac{\theta}{2}}{1+tan^2\frac{\theta}{2}} }$

$=\frac{\frac{1+tan^2\frac{\theta}{2}-(1-tan^2\frac{\theta}{2})+2tan\frac{\theta}{2}}{1+tan ^2\frac{\theta}{2}}}{\frac{1+tan^2\frac{\theta}{2} +1-tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}}{1+tan^ 2\frac{\theta}{2}}}$

$=\frac{\frac{1+tan^2\frac{\theta}{2}-1+tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}}{1+ta n^2\frac{\theta}{2}}}{\frac{1+tan^2\frac{\theta}{2 }+1-tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}}{1+tan^ 2\frac{\theta}{2}}}$

$=\frac{2tan^2\frac{\theta}{2}+2tan\frac{\theta}{2} }{2+2tan\frac{\theta}{2}}$

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

insert all the half angles stuff inside ..

knowing these identities ,

$\sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$

$\cos \theta=1-2\sin^2 \frac{\theta}{2}$

so

$\frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$ -----1

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$ ---------2

$\frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$ --------3

bring out the common factor and do some necessary cancellings , then you will have your proof .
Which step don u understand ? 1,2 or 3 , or the steps after that ?

9. u can also use the following identities:
1)1-cos(theta)=2sin^2(theta/2)
2)1+cos(theta)=2cos^2(theta/2)

10. After looking at this now, I still do not understand the workings provided.

Sorry but I guess I need a complete and detailed working.

11. Hi Punch,

if you get stuck at any of the steps 1 to 7, let me know.

$\frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=Tan\fr ac{\theta}{2}$

The evaluation of the left fraction is $Tan\frac{\theta}{2}$

This is "tangent of the half-angle",
therefore we may express the $Cos\theta$ and $Sin\theta$ from the left using "half-angle formulae".

The "double angle formulae" for $Cos\theta$ and $Sin\theta$ are

$Cos2\theta=\frac{1-Tan^2\theta}{1+Tan^2\theta},\ Sin2\theta=\frac{2Tan\theta}{1+Tan^2\theta}$ (1)

If we replace $\theta$ with $\frac{\theta}{2}$, we will have the "half-angle" formulae.

Therefore the half-angle formulae for $Cos\theta$ and $Sin\theta$ are

$Cos\theta=\frac{1-Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\ frac{\theta}{2}\right)}$

$Sin\theta=\frac{2Tan\left(\frac{\theta}{2}\right)} {1+Tan^2\left(\frac{\theta}{2}\right)}$ (2)

Then, we can write the ratio on the left in terms of $Tan\frac{\theta}{2}$

$\left[\frac{1-\frac{1-Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\ frac{\theta}{2}\right)}+\frac{2Tan\left(\frac{\the ta}{2}\right)}{1+Tan^2\left(\frac{\theta}{2}\right )}}{1+\frac{1-Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\ frac{\theta}{2}\right)}+\frac{2Tan\left(\frac{\the ta}{2}\right)}{1+Tan^2\left(\frac{\theta}{2}\right )}}\right]$ (3)

Writing 1 as $\frac{1+Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^ 2\left(\frac{\theta}{2}\right)}$

$\frac{\frac{1+Tan^2\left(\frac{\theta}{2}\right)-\left[1-Tan^2\left(\frac{\theta}{2}\right)\right]+2Tan\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\ frac{\theta}{2}\right)}}{\frac{1+Tan^2\left(\frac{ \theta}{2}\right)+1-Tan^2\left(\frac{\theta}{2}\right)+2Tan\left(\frac {\theta}{2}\right)}{1+Tan^2\left(\frac{\theta}{2}\ right)}}$ (4)

gives $\frac{1+Tan^2\left(\frac{\theta}{2}\right)-\left[1-Tan^2\left(\frac{\theta}{2}\right)\right]+2Tan\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\ frac{\theta}{2}\right)+1-Tan^2\left(\frac{\theta}{2}\right)+2Tan\left(\frac {\theta}{2}\right)}$ (5)

$=\frac{2Tan^2\left(\frac{\theta}{2}\right)+2Tan\le ft(\frac{\theta}{2}\right)}{2+2Tan\left(\frac{\the ta}{2}\right)}$ (6)

$\frac{2\left[Tan\left(\frac{\theta}{2}\right)\right]\left[Tan\left(\frac{\theta}{2}\right)+1\right]}{2\left[Tan\left(\frac{\theta}{2}\right)+1\right]}=Tan\left(\frac{\theta}{2}\right)$ (7)

12. Okay. I haven't seen what you have mentioned in (1) which refers to the formulas used. The other part are pretty much about simplifying which are rather okay with me.

13. Hi Punch,

Those identities are listed as standard, though they can be developed from

$Cos(A+B)=CosACosB-SinASinB\ \Rightarrow\ Cos(2\theta)=Cos(\theta+\theta)=Cos^2\theta-Sin^2\theta$

$=\frac{Cos^2\theta-Sin^2\theta}{1}$

and since $Cos^\theta+Sin^2\theta=1$

this is $\frac{Cos^2\theta-Sin^2\theta}{Cos^2\theta+Sin^2\theta}=\frac{\frac{ 1}{Cos^2\theta}}{\frac{1}{Cos^2\theta}}\frac{Cos^2 \theta-Sin^2\theta}{Cos^2\theta+Sin^2\theta}$ $=\frac{\frac{Cos^2\theta}{Cos^2\theta}-\frac{Sin^2\theta}{Cos^2\theta}}{\frac{Cos^2\theta }{Cos^2\theta}+\frac{Sin^2\theta}{Cos^2\theta}}$

$=\frac{1-Tan^2\theta}{1+Tan^2\theta}$

$Sin(A+B)=SinACosB+CosASinB$

$Sin(\theta+\theta)=Sin(2\theta)=2Sin\theta\ Cos\theta$

$=\frac{\frac{1}{Cos^2\theta}}{\frac{1}{Cos^2\theta }}2Sin\theta\ Cos\theta=\frac{2\frac{Sin\theta}{Cos\theta}}{\fra c{Cos^2\theta+Sin^2\theta}{Cos^2\theta}}$

$=\frac{2Tan\theta}{1+Tan^2\theta}$