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Question: How to simplify \[\sin \left( {{{\sec }^{ - 1}}\left( x \right)} \right)\]?...

How to simplify sin⁡(sec⁡−1(x))\sin \left( {{{\sec }^{ - 1}}\left( x \right)} \right)?

Explanation

Solution

Here, we will use the Inverse Trigonometric Identity and assume the inverse trigonometric function to be some variable. We will then simplify the equation further using the trigonometric identities and basic mathematical operation to get the required answer. Trigonometric identity is an equation which is always true for all the variables.

Formula Used:
We will use the following formula:

  1. Inverse Trigonometric Identity: sec⁡−1(x)=cos⁡−1(1x){\sec ^{ - 1}}\left( x \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right)
  2. Trigonometric Identity: sin⁡2y+cos⁡2y=1{\sin ^2}y + {\cos ^2}y = 1

Complete step by step solution:
We are given a Trigonometric equation sin⁡(sec⁡−1(x))\sin \left( {{{\sec }^{ - 1}}\left( x \right)} \right).
Inverse Trigonometric Identity: sec⁡−1(x)=cos⁡−1(1x){\sec ^{ - 1}}\left( x \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right) . But, this trigonometric identity is true only when ∣x∣≥1\left| x \right| \ge 1 since the equation is valid for values in the Domain of sec⁡−1(x){\sec ^{ - 1}}\left( x \right).
Now, by using the Trigonometric Identity, we get
⇒sin⁡(sec⁡−1(x))=sin⁡(cos⁡−1(1x))\Rightarrow \sin \left( {{{\sec }^{ - 1}}\left( x \right)} \right) = \sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right).
Let us consider the trigonometric function
y=cos⁡−1(1x)y = {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right) ………………………………………………..(1)\left( 1 \right)
Rewriting the equation, we get
⇒(1x)=cos⁡y\Rightarrow \left( {\dfrac{1}{x}} \right) = \cos y
Squaring on both the sides, we get
⇒(1x)2=cos⁡2y\Rightarrow {\left( {\dfrac{1}{x}} \right)^2} = {\cos ^2}y
⇒1x2=cos⁡2y\Rightarrow \dfrac{1}{{{x^2}}} = {\cos ^2}y
Now we can write the identity sin⁡2y+cos⁡2y=1{\sin ^2}y + {\cos ^2}y = 1 as cos⁡2y=1−sin⁡2y{\cos ^2}y = 1 - {\sin ^2}y.
Using this trigonometric identity in above equation, we get
⇒1x2=1−sin⁡2y\Rightarrow \dfrac{1}{{{x^2}}} = 1 - {\sin ^2}y
Rewriting the equation, we get
⇒sin⁡2y=1−1x2\Rightarrow {\sin ^2}y = 1 - \dfrac{1}{{{x^2}}}
By taking LCM on the RHS, we get
⇒sin⁡2y=x2−1x2\Rightarrow {\sin ^2}y = \dfrac{{{x^2} - 1}}{{{x^2}}}
Taking square root on both the sides, we get
⇒sin⁡y=x2−1x\Rightarrow \sin y = \dfrac{{\sqrt {{x^2} - 1} }}{x}
By substituting equation (1)\left( 1 \right) in above equation, we get
⇒sin⁡(cos⁡−1(1x))=x2−1x\Rightarrow \sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right) = \dfrac{{\sqrt {{x^2} - 1} }}{x}
Substituting sec⁡−1(x)=cos⁡−1(1x){\sec ^{ - 1}}\left( x \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{x}} \right) in the above equation, we get
⇒sin⁡(sec⁡−1(x))=x2−1x\Rightarrow \sin \left( {{{\sec }^{ - 1}}\left( x \right)} \right) = \dfrac{{\sqrt {{x^2} - 1} }}{x}

Therefore, the value of sin⁡(sec⁡−1(x))\sin \left( {{{\sec }^{ - 1}}\left( x \right)} \right)isx2−1x\dfrac{{\sqrt {{x^2} - 1} }}{x}.

Note:
We know that Trigonometric Equation is defined as an equation involving the trigonometric ratios. We should know that we have many trigonometric identities which are related to all the other trigonometric equations. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle. Trigonometric Ratios are used to find the relationships between the sides of a right angle triangle. We should know that the inverse trigonometric function is used to find the missing angles in a right angled triangle whereas the trigonometric function is used to find the missing sides in a right angled triangle.