How can $\sin(2*\arcsin(3/5))$ be evaluated by hand?

Question

Although it is dandy enough to simply go on Wolfram Alpha to see that the answer is 24/25, I would like to learn how to prove that by hand, if possible. Unfortunately, arcsin(3/5) is a transcendental number, and it seems to go have endless digits.

Is there a way to evaluate sin(2*arcsin(3/5)) to be 24/25 without jumping into computer functions?

Answer 1

Let's use a right triangle. The simple one, 3-4-5.

I can't draw a diagram since I suck at those, but let the angle $\theta$ be opposite of the side of length 3. So $\sin(\theta)=\frac{3}{5}$ and $\cos(\theta)=\frac{4}{5}.$ (This is actually derivable since $\cos^2\theta=1-\sin^2\theta=\frac{16}{25}.$ Note that this doesn't make $\cos\theta=-\frac{4}{5}$ thanks to arcsin's definition.)

Now, $\sin(2\theta)=2\sin(\theta)\cos(\theta)=2\cdot \frac{3}{5}\cdot \frac{4}{5}=\boxed{\frac{24}{25}}.$

Answer 2

You can use the double angle formula to write $$\begin{eqnarray}\sin(2\arcsin(3/5)) &=& 2\sin(\arcsin(3/5))\cos(\arcsin(3/5))\\ &=& 2\sin(\arcsin(3/5))\sqrt{1-\sin^2(\arcsin(3/5))}\end{eqnarray}$$

We have $\sin(\arcsin(3/5)) =3/5$, so can plug this in to get the answer.

Answer 3

Use $\sin(2x) =2\sin(x)\cos(x)$ and then $\cos(x)=\sqrt{1-\sin^2(x)}$.

Answer 4

for any $\theta$ we have $$ \sin 2\theta = 2 \sin \theta\cos \theta $$ if
$$ \theta = \arcsin \frac35 $$ then, trivially, $$ \sin \theta = \frac35 $$ using $$ \cos^2 \theta +\sin^2 \theta = 1 $$ can you compute $\cos \theta$ to finish off? (how do you interpret the fact that there are two possible values of $\cos \theta$?)

Answer 5

$\begin{array}\\ \sin(2\arcsin(x)) &=2\sin(\arcsin(x))\cos(\arcsin(x))\\ &=2x\sqrt{1-\sin^2(\arcsin(x))}\\ &=2x\sqrt{1-x^2}\\ \end{array} $

Putting $x = 3/5$, since $\sqrt{1-x^2} = 4/5$, I get $2\dfrac35 \dfrac45 =\dfrac{24}{25} $.

Answer 6

By formula $\sin x=2\sin x\cos x$ we write $$\color{red}{\sin(2\arcsin\dfrac35)}=2\sin(\arcsin\dfrac35)\cos(\arcsin\dfrac35)$$ By formula $\cos x=\sqrt{1-\sin^2x}$ $$\color{red}{\sin(2\arcsin\dfrac35)}=2\sin(\arcsin\dfrac35)\sqrt{1-\sin^2(\arcsin\dfrac35)}$$ But $\sin(\arcsin x)=x$ then $$\color{red}{\sin(2\arcsin\dfrac35)}=2\dfrac35\sqrt{1-(\dfrac35)^2}$$ $$\color{red}{\sin(2\arcsin\dfrac35)}=2\dfrac35\sqrt{\dfrac{16}{25}}=\color{blue}{\dfrac{24}{25}}$$