1. ## Trigonometry question.

Hi all,

I am new here by the way, I have a small problem for a question I can complete in Compound Angle identities

Prove that:
sinθ = tan (θ/2)(1 + cosθ)

Look forward to the answer. I can almost work it work it out but I get stuck somewhere near the end.

2. Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.

3. Originally Posted by Plato
Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
I understand that. This is not homework, it's for my own understanding of Trig (revision). Okay how I did it...

sinθ = tan (θ/2)(1 + cosθ)
tan (θ/2) (1 + cos (θ/2 + θ/2))
tan (θ/2)(cos^2(θ/2) - sin^2(θ/2))
tan (θ/2)(cos^2(θ/2) + sin^2(θ/2) + cos^2(θ/2) - sin^2(θ/2))

4. Let $A = \dfrac{\theta}{2}$ (purely to save me effort)

From line 2: $\tan(A)(1+\cos(2A)) = \dfrac{\sin(A)}{\cos(A)} \cdot 2\cos^2A = 2 \sin A \cos A = 2 \sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)$

There is a common identity you can use now to get the LHS

5. Originally Posted by e^(i*pi)
Let $A = \dfrac{\theta}{2}$ (purely to save me effort)

From line 2: $\tan(A)(1+\cos(2A)) = \dfrac{\sin(A)}{\cos(A)} \cdot 2\cos^2A = 2 \sin A \cos A = 2 \sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)$

There is a common identity you can use now to get the LHS
Hello e^(i*pi)

I actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sinθ.

So this is correct?

Thanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.

6. Hello, Googl!

$\text{Prove that: }\:\sin\theta \:=\:\tan \frac{\theta}{2}(1 + \cos\theta)$

Identities: . $\begin{array}{ccccccc}\cos^2\!\frac{\theta}{2} &=& \dfrac{1+\cos\theta}{2} &\Rightarrow& 1 + \cos\theta &=& 2\cos^2\!\frac{\theta}{2} \\ \\[-3mm]
\sin\theta &=& 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}$

$\text{On the right side we have:}$

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

7. Originally Posted by Googl
Hello e^(i*pi)

I actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sinθ.

So this is correct?

Thanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.
Yes, that's good when starting at the right.

To start at the left and end up at the right...

$\displaystyle\ sin\theta=sin\left(\frac{\theta}{2}+\frac{\theta}{ 2}\right)=2sin\frac{\theta}{2}cos\frac{\theta}{2}$

using $sin2A=2sinAcosA$

which gets us to the half angle.
To get tan, we need cos under the sine, so multiply by

$\displaystyle\ 1=\frac{cos\frac{\theta}{2}}{cos\frac{\theta}{2}}$

$\displaystyle\ 2sin\frac{\theta}{2}cos\frac{\theta}{2}\left[\frac{cos\frac{\theta}{2}}{cos\frac{\theta}{2}}\ri ght]=\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}\l eft[2cos^2\frac{\theta}{2}\right]$

Now use

$cos^2A=\frac{1}{2}(1+cos2A)\Rightarrow\ 2cos^2A=1+cos2A$

to get

$\displaystyle\ sin\theta=tan\frac{\theta}{2}\left[1+cos\left(\frac{\theta}{2}+\frac{\theta}{2}\right )\right]$