Ok so I've been given sin(2a) = (5/13) 0 deg< a < 45 deg.
How would I go about solving for sin.. I've been trying to use the double angel formula's and not much quite works.
There's another identity involving sin 2θ:
$\displaystyle sin 2\theta = \frac{2 tan \theta}{1 + tan^2 \theta}$.
Set sin 2θ equal to 5/13:
$\displaystyle \frac{5}{13} = \frac{2 tan \theta}{1 + tan^2 \theta}$
Cross multiply:
$\displaystyle 5(1 + tan^2 \theta) = 13(2 tan \theta)$
$\displaystyle 5 + 5 tan^2 \theta = 26 tan \theta$
You have a quadratic with tan θ:
$\displaystyle 5 tan^2 \theta - 26 tan \theta + 5 = 0$
Thank god it's factorable:
$\displaystyle (5 tan \theta - 1)(tan \theta - 5) = 0$
There are two possibilities for tan θ:
$\displaystyle 5 tan \theta - 1 = 0$
$\displaystyle tan \theta = 1/5$
$\displaystyle tan \theta - 5 = 0$
$\displaystyle tan \theta = 5$
Reject tan θ = 5 because you'll get an angle greater than 45°. So
$\displaystyle tan \theta = \frac{1}{5}$
Use the Pythagorean Theorem to get the hypotenuse (opposite = 1, adjacent = 5), and you'll get
$\displaystyle hyp = \sqrt{26}$.
Sine is defined as opposite over hypotenuse, so
$\displaystyle sin \theta = \frac{1}{\sqrt{26}}$.
I wish I would have known that definition of sin2a sometime yesterday...
If that particular definition derived from any others? the only formula for sin(2a) listed in my book is sin(2a) = 2cos(x)sin(x)...
This particular chapter seems to be horribly written.. many of the questions we are asked don't seemed to be covered in the book.. or require different methods than what we have been given..
I greatly appreciate the help.
I wasn't taught the other formula for sin 2θ either with I was in school. But it's not hard to derive it from the one we're familiar with:
$\displaystyle \begin{aligned}
sin 2\theta &= 2 sin \theta cos \theta \\
&= \frac{2 sin \theta}{sec \theta} \\
&= \frac{2 sin \theta}{sec \theta} \cdot \frac{sec \theta}{sec \theta} \\
&= \frac{2 sin \theta \cdot \frac{1}{cos \theta}}{sec^2 \theta} \\
&= \frac{2 tan \theta}{1 + tan^2 \theta}
\end{aligned}$
My suggestion would still work (great deal easier); take the arcsin of both sides and then the sin of both sides:
If $\displaystyle \sin(2a) = \frac{5}{13}$, then $\displaystyle 2a = \arcsin\frac{5}{13} \Rightarrow a = \frac{\arcsin\left(\frac{5}{13}\right)}{2} \Rightarrow \sin{a} = \sin\left({\frac{\arcsin\left(\frac{5}{13}\right)} {2}\right) =\frac{1}{\sqrt{26}}}.$
This works as well, but it requires you to find $\displaystyle \cos{a}$ so as to find $\displaystyle \sin{a}$. Now, $\displaystyle \sin{a}$ can be found in exactly
the same as $\displaystyle \cos{a}$, so why not just find $\displaystyle \sin{a}$ instead of finding $\displaystyle \cos{a}$ so as to find $\displaystyle \sin{a}$? In fact, finding
$\displaystyle \cos{a}$ involves bit more of work than finding $\displaystyle \sin{a}$, as you will need to find the missing side of the triangle.