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Question: How do you evaluate \(\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)\)?...

How do you evaluate cos⁡(arcsin(35))\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)?

Explanation

Solution

In this problem we need to calculate the value of cos⁡(arcsin(35))\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right). In the given function we can observe the arcsin\text{arcsin} function which is nothing but the inverse trigonometric function of sin⁡\sin , mathematically it is sin⁡−1{{\sin }^{-1}} function. So, we will assume the sin⁡−1{{\sin }^{-1}} function to be a variable, say yy. Now we will apply the trigonometric function sin⁡\sin on both sides of the equation and simplifies the equation to get the value of sin⁡y\sin y. After calculating the value of sin⁡y\sin y. We will use the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 to find the value of cos⁡y\cos y which is nothing but our required value.

Complete step by step answer:
Given that, cos⁡(arcsin(35))\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right).
In the above equation arcsin\text{arcsin} function is nothing but the sin⁡−1{{\sin }^{-1}} function. So, we are replacing it in the above equation, then we will get
cos⁡(arcsin(35))=cos⁡(sin⁡−1(35))\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)=\cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)
Considering the function sin⁡−1{{\sin }^{-1}} individually, and assuming it to a variable say yy, then we will get
y=sin⁡−1(35).....(i)y={{\sin }^{-1}}\left( \dfrac{3}{5} \right).....\left( \text{i} \right)
Applying the trigonometric function sin⁡\sin on both sides of the above equation, then we will get
⇒sin⁡(y)=sin⁡(sin⁡−1(35))\Rightarrow \sin \left( y \right)=\sin \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)
We know that the value of sin⁡(sin⁡−1(x))=sin⁡−1(sin⁡(x))=x\sin \left( {{\sin }^{-1}}\left( x \right) \right)={{\sin }^{-1}}\left( \sin \left( x \right) \right)=x, then we will get
⇒sin⁡y=35\Rightarrow \sin y=\dfrac{3}{5}
From the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, we can write the value of cos⁡y\cos y as
cos⁡y=1−sin⁡2y\cos y=\sqrt{1-{{\sin }^{2}}y}
Substituting the value of sin⁡y=35\sin y=\dfrac{3}{5} in the above equation, then we will get
cos⁡y=1−(35)2 ⇒cos⁡y=1−925 \begin{aligned} & \cos y=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \\\ & \Rightarrow \cos y=\sqrt{1-\dfrac{9}{25}} \\\ \end{aligned}
Simplifying the above equation by taking LCM, then we will have
⇒cos⁡y=25×1−925 ⇒cos⁡y=1625 \begin{aligned} & \Rightarrow \cos y=\sqrt{\dfrac{25\times 1-9}{25}} \\\ & \Rightarrow \cos y=\sqrt{\dfrac{16}{25}} \\\ \end{aligned}
We can write 16=4216={{4}^{2}}, 25=5225={{5}^{2}}. Now the above equation is modified as
⇒cos⁡y=(45)2 ⇒cos⁡y=45 \begin{aligned} & \Rightarrow \cos y=\sqrt{{{\left( \dfrac{4}{5} \right)}^{2}}} \\\ & \Rightarrow \cos y=\dfrac{4}{5} \\\ \end{aligned}
From equation (i)\left( \text{i} \right), substituting the value of yy in the above equation, then we will get

∴cos⁡(sin⁡−1(35))=45\therefore \cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)=\dfrac{4}{5}

Note: In this problem we have the used the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 after calculating the value of sin⁡y\sin y. We can also follow another method for finding the solution after calculating sin⁡y\sin y, which is by using the basic definition of trigonometric ratio sin⁡x\sin x we will compare it with the value of sin⁡y\sin y. Now we will construct a right-angled triangle with the data we have from the above comparison and try to calculate the remaining data by using the Pythagoras theorem. After having all the data of the triangle, we can use definitions of basic trigonometric ratios and calculate whatever ratio we want.