Double angle trigonometry....NEED HELP PLEASE!!

Hi guys, I am new to this forum and really need some help.

I have been trying to complete a maths assignment for my college work but have come across a question that I am totally stuck on and my lecturer wont help me!!

Can someone help me please?!?!

The question is:

Give the exact value for the following:

2tan75°

______

1-tanē75°

If someone could please please help me..

I need to know the working and each step that was taken to get to the final answer....

Please explain :)

Thanks :)

Re: Double angle trigonometry....NEED HELP PLEASE!!

If you write the formula out using sine and cosine equivalents you get:

$\displaystyle \frac {2 \tan (75)}{1 - \tan ^2 (75)} = \frac { \frac {2\sin (75)}{\cos (75)}}{1 - \frac {sin^2 (75)}{\cos^2 (75)}}$

Now simplify - multiply numerator and denominator by $\displaystyle \cos^2 (75)$:

$\displaystyle \frac {2 \sin (75) \cos (75)}{\cos^2 (75) - \sin ^2 (75)}$

Now the numerator is in the form of the identity $\displaystyle \sin (2x) = 2 \sin x \cos x$, and the denominator is like $\displaystyle \cos (2x) = \cos^2x - \sin^2x$. Here x = 75, so the original equation is equivalent to:

$\displaystyle \frac {\sin (150)}{\cos (150)} $

You can know determine an exact value for this, since you know the values for both $\displaystyle \sin(150)$ and $\displaystyle \cos (150)$.

Re: Double angle trigonometry

Re: Double angle trigonometry

An alternative method is to use a compound angle formula, since 75 = 30 + 45.

Re: Double angle trigonometry

I really have no idea how to use compound formulas....could you run through it please?

Re: Double angle trigonometry

Re: Double angle trigonometry

Quote:

Originally Posted by

**maximus** I really have no idea how to use compound formulas....could you run through it please?

You should memorize the compound angle formulas for sin(a+b) and cosine(a+b). The others can be derived from these:

$\displaystyle \sin(a+b) = \sin(a) \cps(b)+ \cos(a) \sin(b)$

$\displaystyle \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b) $

So for your problem you have

$\displaystyle \tan (75) = \frac {\sin(30+45)}{\cos(30+45)} = \frac {\sin(30)\cos(45) + \cos(30) \sin(45)}{\cos(30)\cos(45) - \sin(30| \sin(45)}$

Substitute the values for $\displaystyle \sin(30), \ \cos(30), \ \sin45), \ \cos(45)$ and then it's just a matter of working the algebra.

Alternativey - if you can remember the compound angle for tangents:

$\displaystyle \tan(a+b) = \frac {\tan(a) \pm \tab (b)}{1 \mp \tan \\(a) \tab (b)} $

then by inspection you can see that for your problem if you make $\displaystyle a = b = 75$, the answer is simply $\displaystyle \tan(75+75) = \tan (150)$.

Re: Double angle trigonometry

I think you mean $\displaystyle \tan{(a + b)} \equiv \frac{\tan{(a)} \pm \tan{(b)}}{1 \mp \tan{(a)}\tan{(b)}}$.

Re: Double angle trigonometry

Quote:

Originally Posted by

**Prove It** I think you mean $\displaystyle \tan{(a + b)} \equiv \frac{\tan{(a)} \pm \tan{(b)}}{1 \mp \tan{(a)}\tan{(b)}}$.

Yes - thanks for catching my error!