Question

How do you evaluate $\sin \left( {2\arccos \left( {\dfrac{3}{5}} \right)} \right)?$

Answer 1

Hint : As we know that above question contains a trigonometric function. In this question there is a term called “ arcos”. We should know that it is the inverse of the cosine function. It means that the $\arccos$ function returns the angle whose cosine value is give. We will apply the trigonometric identities to solve this question.

Complete step by step solution:
As per the question we have $\sin \left( {2\arccos \left( {\dfrac{3}{5}} \right)} \right)$ .
We know the property which says that $\cos (\arccos (x)) = x$ . We will first square the given expression and take the square root :
$\sqrt {{{\sin }^2}\left( {2\arccos \left( {\dfrac{3}{5}} \right)} \right)}$ .
We know the property which is
${\sin ^2}x + {\cos ^2}x = 1$ , which means that ${\sin ^2}x$ can be written as $1 - {\cos ^2}$ .
By applying this in the above expression we have
$\sqrt {1 - {{\cos }^2}\left( {2\arccos \left( {\dfrac{3}{5}} \right)} \right)}$ .
There is a double angle formula of cosine with many form i.e.
$\cos 2a = {\cos ^2}a - {\sin ^2}a = 2{\cos ^2}a - 1 = 1 - 2{\sin ^2}a$ .
Here we have $a = \arccos \dfrac{3}{5}$ , so we apply the double angle formula.
By applying the second formula, since we need it in terms of cosine we have $\sqrt {1 - {{(2{{\cos }^2}\left( {\arccos \left( {\dfrac{3}{5}} \right)} \right) - 1)}^2}}$ .
By squaring and on further simplifying we have $\sqrt {1 - 4 \cdot {{\cos }^4}\left( {\arccos \left( {\dfrac{3}{5}} \right)} \right) + 4{{\cos }^2}\left( {\arccos \left( {\dfrac{3}{5}} \right)} \right) - 1}$ . Now we can replace $\cos (\arccos (x)) = x$ .
It gives us $\sqrt {4 \times {{\left( {\dfrac{3}{5}} \right)}^2} - 4 \times {{\left( {\dfrac{3}{5}} \right)}^4}} = \sqrt {4 \times \dfrac{9}{{25}} - 4 \times \dfrac{{81}}{{625}}}$ ,
We will now solve it $\sqrt {\dfrac{{36}}{{25}} - \dfrac{{324}}{{625}}} = \sqrt {\dfrac{{576}}{{625}}}$ .
Hence the required value is $\dfrac{{24}}{{25}}$ .
So, the correct answer is “ $\dfrac{{24}}{{25}}$ ”.

Note : Before solving this kind of question we should be fully aware of the trigonometric ratios, their functions, their identities and formulas. Here we have been asked to find the value of the function with the given sine and $\arccos$ , which is the inverse function of cosine.