Thread: Double angle trigonometry

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

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 $\displaystyle \cos (150)$.

Thankyou so much

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

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

List of trigonometric identities - Wikipedia, the free encyclopedia

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 $\displaystyle \tan(75+75) = \tan (150)$.

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

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!