Trig

**ruthvik** · December 17th 2008, 02:22 PM

tana^2 = sin(2a)

tried everything... -_-

**Mush** · December 17th 2008, 03:26 PM

Originally Posted by ruthvik

tana^2 = sin(2a)

tried everything... -_-

Do you mean:

$tan(a^2) = sin(2a)$

or

$tan^2(a) = sin(2a)$

?????

Doesn't really matter, neither of them are true.

**ruthvik** · December 17th 2008, 03:31 PM

Originally Posted by Mush

Do you mean:

$tan(a^2) = sin(2a)$

or

$tan^2(a) = sin(2a)$

?????

Doesn't really matter, neither of them are true.

this happened in different topic
im trying to solve for solutions
not proving it
and i meant this (tana) ^ 2 = sin(2a)

**o_O** · December 17th 2008, 03:43 PM

$\tan^2 a = \sin (2a)$

We'll make use of the identity: $\sin (2a) = \frac{2\tan a}{1 + \tan^2 a}$

So:
$\begin{aligned} \tan^2 a & = \frac{2\tan a}{1 + \tan^2 a} \\ \tan^2 a (1 + \tan^2 a) & = 2\tan a \\ & \ \vdots \\ \tan^4 a + \tan^2 a - 2\tan a & = 0 \\ \tan a \left(\tan^3 a + \tan a - 2\right) & = 0 \\ \tan a (\tan a - 1) (\tan^2 a + \tan a + 2) & = 0 \end{aligned}$

Set each factor equal to 0 and solve for $a$ . Note that the third factor (essentially a quadratic) does not have real solutions for $\tan a$ so don't worry about it.

**ruthvik** · December 17th 2008, 03:47 PM

Originally Posted by o_O

$\tan^2 a = \sin (2a)$

We'll make use of the identity: $\sin (2a) = \frac{2\tan a}{1 + \tan^2 a}$

So:
$\begin{aligned} \tan^2 a & = \frac{2\tan a}{1 + \tan^2 a} \\ \tan^2 a (1 + \tan^2 a) & = 2\tan a \\ & \ \vdots \\ \tan^4 a + \tan^2 a - 2\tan a & = 0 \\ \tan a \left(\tan^3 a + \tan a - 2\right) & = 0 \\ \tan a (\tan a - 1) (\tan^2 a + \tan a + 2) & = 0 \end{aligned}$

Set each factor equal to 0 and solve for $a$ . Note that the third factor (essentially a quadratic) does not have real solutions for $\tan a$ so don't worry about it.

this was perfect but the thing was this is bonus
which means he probably expects to us to do out whole thing
should i just do the last one too? by using quadratic forumula?

**o_O** · December 17th 2008, 03:48 PM

Yes . Or simply look at the discriminant and it'll show you that there are no real solutions to the quadratic.

**ruthvik** · December 17th 2008, 03:50 PM

Originally Posted by o_O

Yes . Or simply look at the discriminant and it'll show you that there are no real solutions to the quadratic.

i dont know how to do trig with imaginary numbers
i dont think he will except no real solutions for that one
ill try
thanks a lot
btw is there any way to prove sin2a equals the thing u gave me?
aight i got tana = o, pi, pi/4, 5pi/4

**o_O** · December 17th 2008, 04:01 PM

Exactly. That's why there are no solutions that are real numbers. There is no angle such that when you plug it into $\tan^2 a + \tan a + 2$ you will get 0.

To prove the identity: $\sin (2a) = \frac{2\tan a}{1+\tan^2a}$

Why don't you try

? It isn't too difficult really. Start off with using: $\sec^2 x = 1 + \tan^2 x$

**ruthvik** · December 17th 2008, 04:02 PM

Originally Posted by o_O

Exactly. That's why there are no solutions that are real numbers. There is no angle such that when you plug it into $\tan^2 a + \tan a + 2$ you will get 0.

To prove the identity: $\sin (2a) = \frac{2\tan a}{1+\tan^2a}$

Why don't you try

? It isn't too difficult really. Start off with using: $\sec^2 x = 1 + \tan^2 x$

nah i just said i changed the right side to tan by multiplying top and bottom by cosa^2

so 0,pi/4, pi, 5pi/4 r my solutions since
tana = 0 , a can only equal = {0, pi} in the interval i was given
tana = 1, a can only equal = (5pi/4, pi/4} in the interval between 0 and 2pi noninclusive

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