Double and Half Angle Trouble...

**topdnbass** · November 10th 2009, 04:35 PM

There are two equations in my HW that im having trouble with. They happen to be next to each other so I took a picture.

They are separated by a white line, the answer I got is in blue and the answer the book gives is in red.

Really appreciate the assistance, as I have a test tomorrow and this stuff will be included.

First one is half angle, second is double..

What am I missing here? The formulas are correct, and I believe I'm doing them right.

**adkinsjr** · November 10th 2009, 04:49 PM

Your answer to the second problem is correct, you just have to simplify it further. $\sqrt{168}=2\sqrt{42}$ .

I don't understand what you did on the first problem. You need to find $tan(\frac{1}{2}\theta)$ right?

**topdnbass** · November 10th 2009, 04:58 PM

Thanks for the response, I feel stupid now

For the first one, yes we're given $tan(theta) = sqrt(7)/3$
Then you need to find $tan(theta/2)$ or as you said $tan(\frac{1}{2}\theta)$ .

I made the assumption that since $sinx/cosx = tanx$ that $sinx=sqrt(7)$ and $cosx=3$ .

So i substituted those into a tangent formula and ended up with the answer which I circled in blue.

**adkinsjr** · November 10th 2009, 05:30 PM

Originally Posted by topdnbass

Thanks for the response, I feel stupid now

For the first one, yes we're given $tan(theta) = sqrt(7)/3$
Then you need to find $tan(theta/2)$ or as you said $tan(\frac{1}{2}\theta)$ .

I made the assumption that since $sinx/cosx = tanx$ that $sinx=sqrt(7)$ and $cosx=3$ .

So i substituted those into a tangent formula and ended up with the circled blue one.

Ok, I see where you went wrong. You can't make that assumption. Sine can't equal the $\sqrt{7}$ because sine is always less than 1. To solve this one, I would just use the formula for the tangent of twice an angle:

$tan(2A)=\frac{2tan(A)}{1-tan^2(A)}$

If we let $\theta=2A$ then $A=\frac{1}{2}\theta$ . Therefore:

$tan(\theta)=\frac{2tan(\frac{1}{2}\theta)}{1-tan^2(\frac{1}{2}\theta)}$

$\frac{\sqrt{7}}{3}=\frac{2tan(\frac{1}{2}\theta)}{ 1-tan^2(\frac{1}{2}\theta)}$

If you do the algebra you should obtain a quadratic equation where you can solve for $tan(\frac{1}{2}\theta)$

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