Understanding 2t , 1-t^2 etc

**200001** · April 30th 2010, 09:09 AM

Hi
Can someone explain the following diagram I have drawn.
For example where the t and 2 came from and their use in identities as I cannot make the relationship between them in trig identities involving tan
Thanks

**Rpellar** · April 30th 2010, 11:07 AM

Using pythagorean theorem you can see the relationship between the sides.

$A^2 = B^2 + C^2$
Where A is the hypotenuse (across from the 90 degree angle) and B and C are the vertical and horizontal sides.
So,
$A^2 = (2t)^2 + (1-t^2)^2$
$A^2 = 4t^2 + (1-2t^2+t^4) = t^4 +2t^2 + 1$
The right side is a perfect square and thus simplifies to:
$A^2 = (t^2 + 1)^2$
Rooting both side we see A, the hypotenuse, is $t^2+1$
Does this clarify things for you?

**200001** · April 30th 2010, 11:16 AM

Hi
Thanks, I get that bit but Im not sure on its use in double angle idnetities and when and where to use it

**running-gag** · April 30th 2010, 11:29 AM

Hi

$\sin 2t = 2 \sin t \cos t$

$\cos 2t = \cos^2 t - \sin^2 t$

Therefore $\tan 2t = \frac{2 \sin t \cos t}{\cos^2 t - \sin^2 t}$

Dividing numerator and denominator by $\cos^2 t$

$\tan 2t = \frac{2 \tan t}{1 - \tan^2 t}$

Let T = tan(t)

$\tan 2t = \frac{2T}{1 - T^2}$

**200001** · April 30th 2010, 10:20 PM

beautiful
thanks very much

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