Question

Prove that ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{33}}{{65}}} \right)$

Answer 1

To prove the statement at first we have to assume each term in LHS as separate variables. For example consider ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)$ be equal to a variable x and ${\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)$ be equal to another variable y. Then By using the property of trigonometric inverse function we will find out $\cos x$ and $\cos y$. Then we will find corresponding values of $\sin x$ and $\sin y$. Finally we have to find out the value of $\cos (x + y)$ by substituting the values of $\cos x$,$\cos y$,$\sin x$ and $\sin y$ obtained earlier and by using the properties of trigonometric inverse function we can prove the given statement.

Complete step by step answer:
The LHS of the statement is given by
$LHS = {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)$ ………………………… (1)
Let ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = x$ ………………………… (2)
And ${\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = y$ ………………………… (3)
But we know the property of the trigonometric function that if ${\cos ^{ - 1}}\theta = A$ then $\theta = \cos A$. Hence using this property we can write eq. (2) and (3) as
$\cos x = \dfrac{4}{5}$ ………………………… (4)
And $\cos y = \dfrac{{12}}{{13}}$ ………………………… (5)
Again we know the sine and cosine functions are related by
$\sin \theta = \sqrt {1 - {{\cos }^2}\theta }$ ………………………… (6)
Applying these formulae to eq. (4) and (5), we will obtain,

$$\sin x = \sqrt {1 - {{\left( {\dfrac{4}{5}} \right)}^2}} \\\ = \sqrt {\dfrac{9}{{25}}} \\\ = \dfrac{3}{5} \\\$$

                            ………………………………………… (7)  

And

$$\sin y = \sqrt {1 - {{\left( {\dfrac{{12}}{{13}}} \right)}^2}} \\\ = \sqrt {\dfrac{{25}}{{169}}} \\\ = \dfrac{5}{{13}} \\\$$

                        ……………………………………………. (8)  

We know the formulae that
$\cos (x + y) = \cos x\cos y - \sin x\sin y$ ……………………………. (9)
Now substituting the values of eq. (2), (3), (4), (5), (6) and (7) in eq. (9) we will get,

$$\cos \left[ {{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)} \right] = \left( {\dfrac{4}{5}} \right)\left( {\dfrac{{12}}{{13}}} \right) - \left( {\dfrac{3}{5}} \right)\left( {\dfrac{5}{{13}}} \right) \\\ = \dfrac{{48}}{{65}} - \dfrac{3}{{13}} \\\ = \dfrac{{33}}{{65}} \\\$$

           ……………………………….. (10)  

Using the property of inverse trigonometric function we can write eq. (10) as
${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{33}}{{65}}} \right)$
Now the statement is proved.

Note: In alternative method we can apply direct formula given by, ${\cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}\left[ {AB - \sqrt {\left( {1 - {A^2}} \right)\left( {1 - {B^2}} \right)} } \right]$ to prove the statement.