Question

How do you find the exact value of $\cos \left( \arcsin \left( \dfrac{5}{13} \right) \right)$?

Answer 1

In the given, we have been asked to find the exact value of a given expression. In order to find the value first we need to substitute $t=\arcsin \left( \dfrac{5}{13} \right)$, then simply after that we need to find the value of cos (t). By using the trigonometric identity i.e. ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$ and simplifying the expression further. In this way we will get the exact value of $\cos \left( \arcsin \left( \dfrac{5}{13} \right) \right)$.

Complete step by step solution:
We have given that,
$\Rightarrow \cos \left( \arcsin \left( \dfrac{5}{13} \right) \right)$
Substitute $t=\arcsin \left( \dfrac{5}{13} \right)$ in the above expression, we obtain
$\Rightarrow \cos t$
Now solving,
$t=\arcsin \left( \dfrac{5}{13} \right)$
Thus,
$\sin t=\left( \dfrac{5}{13} \right)$
Using the trigonometric identity i.e. ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$.
Applying this identity, we obtained
${{\cos }^{2}}t=1-{{\sin }^{2}}t$
Putting $\sin t=\left( \dfrac{5}{13} \right)$in the above expression, we get
${{\cos }^{2}}t=1-{{\left( \dfrac{5}{13} \right)}^{2}}$
Transposing the power of 2 to RHS of the above expression, we obtained
$\cos t=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}}$
Solving the above expression, we get
$\cos t=\sqrt{1-\left( \dfrac{25}{169} \right)}$
Solving the RHS of the above expression by taking the LCM, we get
$\cos t=\sqrt{\dfrac{169-25}{169}}=\sqrt{\dfrac{144}{169}}$
As we know that the value of square root of 144 is equal to 12 and the square root of 169 is equal to 13 i.e.
$\sqrt{144}=12$ And $\sqrt{169}=13$
Thus,
$\cos t=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}$
Therefore,
$\cos t=\dfrac{12}{13}$
Undo the substitution i.e.$t=\arcsin \left( \dfrac{5}{13} \right)$, we get
$\Rightarrow \cos \left( \arcsin \left( \dfrac{5}{13} \right) \right)=\dfrac{12}{13}$
Hence, this is the required answer.

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.