Question

How do you simplify $\sin \left( 2{{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)$ ?

Answer 1

Hint: Here we have to simplify the value given. Firstly by using the double angle formula $\sin 2A=2\sin A\cos A$ we will expand the value given and as we know $\cos {{\cos }^{-1}}A=A$ we will get value of one of the term. Then we will use the square relation ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and simplify our terms further. Finally we will get the square root value which is our desired answer.

Complete step by step answer:
We have to simplify the value given as follows:
$\sin \left( 2{{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)$……$\left( 1 \right)$
Now as know the double angle formula given as:
$\sin 2A=2\sin A\cos A$
Using the above formula in equation (1) where $A={{\cos }^{-1}}\left( \dfrac{4}{5} \right)$ we get,
$\Rightarrow 2\sin \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)\cos \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)$
As by inverse trigonometric function we know that $\cos {{\cos }^{-1}}A=A$ where in this case $A=\dfrac{4}{5}$ we get,
$\Rightarrow 2\sin \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)\times \dfrac{4}{5}$
$\Rightarrow \dfrac{8}{5}\sin \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)$….$\left( 2 \right)$
Now as we can see that we have sine and inverse cosine which don’t have any formula for simplification so we will use the square relation given as follows to simplify our value:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$
So,
$\Rightarrow \sin x=\pm \sqrt{1-{{\cos }^{2}}x}$
Using the above relation in equation (2) where $x={{\cos }^{-1}}\left( \dfrac{4}{5} \right)$ we get,
$\Rightarrow \pm \dfrac{8}{5}\sqrt{1-{{\cos }^{2}}\left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)}$
As ${{\cos }^{2}}x={{\left( \cos x \right)}^{2}}$
$\Rightarrow \pm \dfrac{8}{5}\sqrt{1-{{\left( \cos \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right) \right)}^{2}}}$
Now using $\cos {{\cos }^{-1}}A=A$ where in this case $A=\dfrac{4}{5}$ we get,
$\Rightarrow \pm \dfrac{8}{5}\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}$
$\Rightarrow \pm \dfrac{8}{5}\sqrt{1-\dfrac{16}{25}}$
Simplifying it further we get,
$\Rightarrow \pm \dfrac{8}{5}\sqrt{\dfrac{25-16}{25}}$
$\Rightarrow \pm \dfrac{8}{5}\sqrt{\dfrac{9}{25}}$
As $\sqrt{9}=3$ and $\sqrt{25}=5$ using it above we get,
$\Rightarrow \pm \dfrac{8}{5}\times \dfrac{3}{5}$
$\Rightarrow \pm \dfrac{24}{25}$
We got our answer as $\pm \dfrac{24}{25}$
Hence on simplify $\sin \left( 2{{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)$ we get the answer as $\pm \dfrac{24}{25}$ .

Note:
 The key point to note in this problem is that we should always try to convert all trigonometric and inverse trigonometric ratios into the same trigonometric ratio as it simplifies the calculation. Here we converted sin in terms of cosine because we had the inverse of cosine as angle. So we can get a simplified form.