Double angle trigonometry

**maximus** · March 21st 2013, 12:39 PM

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-tan²75°

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

**ebaines** · March 21st 2013, 02:09 PM

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

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

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

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

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

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

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

**maximus** · March 21st 2013, 03:28 PM

Thankyou so much

**Prove It** · March 21st 2013, 03:38 PM

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

**maximus** · March 21st 2013, 03:44 PM

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

**Prove It** · March 21st 2013, 04:04 PM

List of trigonometric identities - Wikipedia, the free encyclopedia

**ebaines** · March 22nd 2013, 05:46 AM

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:

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

So for your problem you have

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

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

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

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

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

**Prove It** · March 22nd 2013, 05:53 AM

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

**ebaines** · March 22nd 2013, 07:03 AM

Originally Posted by Prove It

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

Yes - thanks for catching my error!

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