Trigonometry question.

• February 6th 2011, 06:43 AM
Googl
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.
• February 6th 2011, 07:02 AM
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.
• February 6th 2011, 09:10 AM
Googl
Quote:

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))
Attachment 20699
• February 6th 2011, 09:28 AM
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
• February 6th 2011, 09:55 AM
Googl
Quote:

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.
Attachment 20700

So this is correct?
Attachment 20701

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.
• February 6th 2011, 10:06 AM
Soroban
Hello, Googl!

Quote:

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

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

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

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

• February 6th 2011, 11:20 AM
Quote:

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.
Attachment 20700

So this is correct?
Attachment 20701

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

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