Trigonometric Identity? - Deriving the Double Angle Formulas for Sine and Cosine

[jamesbob]

Trigonometric Identity??

I just can't figure this out. I don't think I've covered enough material to do this. Can anyone help? I've put the entire question but I am sure all i need is a little explanation and maybe the first answer and i could do the rest.

$$Let z = tan(\theta/2).// Show that cos\theta = \frac{1-z^2}{1+z^2} sin\theta = \frac{2z}{1+z^2} \frac{d\theta}{dz} = \frac{2}{1+z^2}$$

Thanks. Sorry i tried to use latex but can't figure out spaces.

 

[TD]

Start off with plugging it in and use that $\sec ^2 a = 1 + \tan ^2 a$.

For example, to start the first one:

$$\frac{{1 - z^2 }}{{1 + z^2 }} = \frac{{1 - \tan ^2 \left( {x/2} \right)}}{{1 + \tan ^2 \left( {x/2} \right)}} = \frac{{2 - \sec ^2 \left( {x/2} \right)}}{{\sec ^2 \left( {x/2} \right)}} = \frac{2}{{\sec ^2 \left( {x/2} \right)}} - 1$$

You're almost there, switch from sec to cos and use the double angle formula. Can you try the other ones now?

 

[VietDao29]

You should look up some double-angle formulae in your textbook (note that: $$x = 2 \times \frac{x}{2}$$), and some identities like:
$$1 + \tan ^ 2 x = \frac{1}{\cos ^ 2 x} = \sec ^ 2 x$$
I'll do the sin(x) for you:
$$\frac{2 \tan \left( \frac{x}{2} \right)}{1 + \left( \tan ^ 2 \frac{x}{2} \right)} = \frac{2 \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)}}{\frac{1}{\cos ^ 2 \left( \frac{x}{2} \right)}} = 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) = \sin (x)$$.
You can do the rest, right?
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Ooops, I didn't know you have posted, TD. Sorry to say nearly what you have said...

 

[TD]

No problem, just a little less work for jamesbob but I'll doubt he'll mind

 

[jamesbob]

Thanks so much for your help so far. I really appreciate the time you've spent to help me.

For the first one my complete working was (including every line, needed or not) :

$$\frac{{1 - z^2 }}{{1 + z^2 }} = \frac{{1 - \tan ^2 \left( {\theta/2} \right)}}{{1 + \tan ^2 \left( {\theta/2} \right)}}$$

here i said that

$$\tan^2\theta = \sec^2\theta - 1$$

so

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

so continuing we have:

$$\frac{2 - \sec^2(\frac{\theta}{2})}{\sec^2(\frac{\theta}{2})} = /frac{2}{\sec^2(\frac{\theta}{2})} = /frac{2}{\sec^2(\frac{\theta}{2})} - 1$$

here i said that $$\sec(x) = \frac{1}{\cos(x)}$$

so we have, switching to cos
$$\frac{2}{\frac{1}{\cos^2(\frac{\theta}{2})}} = 2\cos^2(\frac{\theta}{2}) - 1$$

and using $$2\cos^2x-1 = \cos2x$$
we have
$$\cos(2 x (\frac{\theta}{2}) = cos(\theta)$$

I hope this LaTeX coding works. I'll do another post for the next two answers.

 

[TD]

Apart from a few LaTeX-troubles, this seems alright

 

[jamesbob]

Thanks, yeah for some reason i just couldn't get those fractions to work when they seem identical to others. I just started using LaTeX today so i aint that experienced. For $$\sin(\theta) = \frac{2z}{1 + z^2}$$ i said that this is equal to

$$\frac{2\tan(\frac{\theta}{2})}{1 = \tan^2(\frac{\theta}{2})}$$ and I'm now realising its just the same as what's already been posted so I'm not going to do it again.

I'm having difficulty with the last one tho. Is it anything to do with

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

 

[TD]

You want to find $d\theta /dz$ but we know that $z = \tan \left( {\theta /2} \right)$. Try to rewrite this last expression so you have $\theta$ as a function of x instead of the other way arround it is now. Then you can find its derivative with respect to z which is just what you want.

 

[jamesbob]

Yeah that's what i had tried to do originally but got totally lost. best i could do was

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

and i reckon this is totally the wrong direction.

 

[VietDao29]

Thanks so much for your help so far. I really appreciate the time you've spent to help me.

For the first one my complete working was (including every line, needed or not) :

$$\frac{{1 - z^2 }}{{1 + z^2 }} = \frac{{1 - \tan ^2 \left( {\theta/2} \right)}}{{1 + \tan ^2 \left( {\theta/2} \right)}}$$

here i said that

$$\tan^2\theta = \sec^2\theta - 1$$

so

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

so continuing we have:

$$\frac{2 - \sec^2(\frac{\theta}{2})}{\sec^2(\frac{\theta}{2})} = /frac{2}{\sec^2(\frac{\theta}{2})} = /frac{2}{\sec^2(\frac{\theta}{2})} - 1$$

here i said that $$\sec(x) = \frac{1}{\cos(x)}$$

so we have, switching to cos
$$\frac{2}{\frac{1}{\cos^2(\frac{\theta}{2})}} = 2\cos^2(\frac{\theta}{2}) - 1$$

and using $$2\cos^2x-1 = \cos2x$$
we have
$$\cos(2 x (\frac{\theta}{2}) = cos(\theta)$$

I hope this LaTeX coding works. I'll do another post for the next two answers.

This looks fine (as TD has already confirmed you). But your 4th LaTeX part is wrong, it reads:
$$1 + \tan ^ 2 \left( \frac{\theta}{2} \right) = 1 + \sec ^ 2 \left( \frac{\theta}{2} \right)$$, when it should actually reads:
$$1 + \tan ^ 2 \left( \frac{\theta}{2} \right) = \sec ^ 2 \left( \frac{\theta}{2} \right)$$.
Other than that, everything looks good.
-------------------
Since you've posted your answer, I'd like to show you a little faster way:
Multiply both numerator, and denominator by cos²(theta / 2)
$$\frac{1 - \tan ^ 2 \left( \frac{\theta}{2} \right)}{1 + \tan ^ 2 \left( \frac{\theta}{2} \right)} = \frac{\cos ^ 2 \left( \frac{\theta}{2} \right)}{\cos ^ 2 \left( \frac{\theta}{2} \right)} \times \frac{1 - \tan ^ 2 \left( \frac{\theta}{2} \right)}{\frac{1}{\cos ^ 2 \left( \frac{\theta}{2} \right)}} = \frac{\cos ^ 2 \left( \frac{\theta}{2} \right) - \sin ^ 2 \left( \frac{\theta}{2} \right)}{1} = \cos (\theta)$$.
-------------------
$$z = \tan \left( \frac{\theta}{2} \right) \Rightarrow \frac{\theta}{2} = \arctan z \Rightarrow \theta = 2 \arctan z$$
Can you differentiate arctan(x) with respect to x?
ie, what's:
$$\frac{d(\arctan(x))}{dx} = ?$$
Have you covered it yet?
You can go from here, right?

 

[jamesbob]

umm not done arctan yet, any otherr way to do it?

 

[jamesbob]

This leads onto a question saying: a standard substitution to convert integrals of trigonometric functions into integrals of rational functions is the substitution

$$z = \tan(\frac{\theta}{2})$$​

Use this substitution to show that

$$\int\sec\thetad\theta = ln(\sec\theta + \tan\theta) + C$$

I don't know if this helps in the direction we need to take it. I can't see how to do this question either.

 

[VietDao29]

umm not done arctan yet, any otherr way to do it?

Okay, 1 other way is to note that:
arctan(tan(x)) = x (this is true for all $x \in ]- \pi / 2 ; \ \pi / 2[$, right?)
Let u = tan(x). We have:
arctan(u) = x
Differentiate both sides with respect to x, we have:
arctan'(u) u'(x) = 1 (We use the chain rule for the LHS)
So:
$$\arctan ' (u) = \frac{1}{u'(x)} = \frac{1}{\tan '(x)} = \frac{1}{\frac{1}{\cos ^ 2 x}} = \frac{1}{1 + \tan ^ 2 x} = \frac{1}{1 + u ^ 2}$$. So that means:
$$\arctan ' (x) = \frac{1}{1 + x ^ 2}$$. Okay, I think I've told you the answer. Do you get this?
---------------
For the next problem. Let $$t = \tan \left( \frac{\theta}{2} \right)$$
Before you integrate that expression. First, answer these question:
1. What's $\cos \theta$ in terms of t?
2. What's dx in terms of t, and dt?
3. What does the integral finally become?