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Inverse Trigonometric Functions And Their Values

Learn inverse trigonometric functions & sine, cosine, tangent functions in calculus. Find simplification of expression containing a trigonometric and an inverse trigonometric function.

Inverse Sine Functions

  • All trigonometric functions are many one or not one to one, therefore, their inverse cannot be found. Consider \(f(x)=sin\,x\).
  • If we apply the horizontal line test we see that \(f\) is not one to one.

     

But if we restrict the domain of \(sin\,x\) to \(\left [ \dfrac {-\pi}{2},\dfrac {\pi}{2} \right]\)then graph is 

 

Now the function becomes one-to-one.

We say that the inverse function of the restricted sine function is \(sin^{-1}x\) or  arc \((sin\,x)\)

  • If \(f(x)=sin\,x\), then \(f^{-1}(x)=sin^{-1}\,x\) = arc \((sin\,x)\)
  • \(y=sin^{-1}\,x\Rightarrow sin\,y=x.\;\;\left ( -\dfrac {\pi}{2}\leq y \leq \dfrac {\pi}{2} \right)\)
  • If \(-1\leq x\leq 1\) , then \(sin^{-1}\,x\) is the angle between \(-\dfrac {\pi}{2}\) and \(\dfrac {\pi}{2}\) whose \(sin\) is \(x\).
  • \(sin^{-1}\left ( \dfrac {-1}{2}\right)=\) angle between \(-\dfrac {\pi}{2}\) and \(\dfrac {\pi}{2}\) whose sine is  \(\dfrac {-1}{2}=\dfrac {-\pi}{6}\)

\(\Rightarrow sin^{-1}\left (-\dfrac {1}{2}\right)=\dfrac {-\pi}{6}\)

Illustration Questions

The value of \(sin^{-1} \left (\dfrac {-\sqrt 3}{2}\right)\) is

A \( \dfrac {\pi}{6}\)

B \( \dfrac {-\pi}{3}\)

C \( \dfrac {-\pi}{6}\)

D \( \dfrac {\pi}{3}\)

×

\(sin^{-1} \left (\dfrac {-\sqrt 3}{2}\right)=\) Angle between \(-\dfrac {\pi}{2}\) and \(\dfrac {\pi}{2}\) whose sine  is \( \dfrac {-\sqrt 3}{2}= \dfrac {-\pi }{3}\)

\(\therefore \;sin^{-1}\left ( \dfrac {-\sqrt 3}{2}\right)= \dfrac {-\pi }{3}\)

The value of \(sin^{-1} \left (\dfrac {-\sqrt 3}{2}\right)\) is

A

\( \dfrac {\pi}{6}\)

.

B

\( \dfrac {-\pi}{3}\)

C

\( \dfrac {-\pi}{6}\)

D

\( \dfrac {\pi}{3}\)

Option B is Correct

Inverse Cosine Functions

  • Consider, \(f(x)=cos\,x\)
  • If we apply the horizontal line test, we see that \(f\) is not one to one and its inverse is not possible.

  • Now we restrict the domain of \(f\) to \((0, \pi)\) or \(0\leq x\leq\pi\) then we see that \(f\)is one to one.

  • We say that the inverse of this restricted cosine function is inverse cosine function or \(cos^{-1}\,x\) or arc \((cos\,x)\).
  • If  \(f(x)=cos\,x\), then \(f^{-1}(x)=cos^{-1}\,x\) or arc \((cos\,x)\)
  • If \(y=cos^{-1}\,x\Rightarrow cos\,y=x\;\;\;(0\leq y \leq \pi)\)
  • If  \(-1\leq x \leq 1\), then \(cos^{-1}\,x\) is the angle between 0 and \(\pi\), whose cosine is \(x\).
  • \(cos^{-1}\,\Big(\dfrac {1}{2}\Big)\) is the angle between 0 and \(\pi\), whose cosine is \(\dfrac {-1}{2}=\dfrac {2\pi}{3}\).

\(\therefore\;cos^{-1}\Big(-\dfrac {1}{2}\Big)=\dfrac {2\pi}{3}\)

Illustration Questions

The value of \(cos^{-1}\Big(\dfrac {\sqrt 3}{2}\Big)\) is 

A \(\dfrac {\pi}{6}\)

B \(\dfrac {-\pi}{6}\)

C \(\dfrac {\pi}{3}\)

D \(\dfrac {-\pi}{3}\)

×

\(cos^{-1}\,\Big(\dfrac {\sqrt 3}{2}\Big)\)= Angle between 0 and \(\pi\), whose cosine is \(\dfrac {\sqrt 3}{2}=\dfrac {\pi}{6}\).

\(\therefore\;cos^{-1}\Big(\dfrac {\sqrt 3}{2}\Big)=\dfrac {\pi}{6}\)

The value of \(cos^{-1}\Big(\dfrac {\sqrt 3}{2}\Big)\) is 

A

\(\dfrac {\pi}{6}\)

.

B

\(\dfrac {-\pi}{6}\)

C

\(\dfrac {\pi}{3}\)

D

\(\dfrac {-\pi}{3}\)

Option A is Correct

Inverse of Tangent Function

  • Consider \(f(x)=tan\,x\)
  • If we apply the horizontal line test we see that \(f\) is not one-to-one.

  • Now if we restrict the domain to  \(\left (\dfrac {-\pi}{2}\,,\dfrac {\pi}{2} \right)\) i.e. \(\dfrac {-\pi}{2}\,<x<\dfrac {\pi}{2}\)then we see that \(tan\,x\) has become one-to-one.

The inverse function of this restricted tangent function is called inverse tangent function or \(tan^{-1}\,x\) or arc\((tan\,x)\).

  • If \(f(x)=tan\,x\) ,then  \(f^{-1}(x)=tan^{-1}\,x\) arc \((tan\,x)\)
  • If \(y=tan^{-1}\,x\;\Rightarrow x = tan\,y\)    \(\left (\dfrac {-\pi}{2}\,<y<\dfrac {\pi}{2}\right)\)
  • If \(x\in R\) then \(tan^{-1}\,x\) is the angle between \(\dfrac {-\pi}{2}\) and \(\dfrac {\pi}{2}\) whose tangent is \(x\).
  •  \(tan^{-1}\,(-1)\) is the angle between \(\dfrac {-\pi}{2}\) and \(\dfrac {\pi}{2}\) whose tangent is \(-1=\dfrac {-\pi}{4}\).

\(\therefore \;\; tan^{-1}(-1)=\dfrac {-\pi}{4}\)

Illustration Questions

Find the value of \(tan^{-1} \left ( \dfrac {-1}{\sqrt 3} \right)\).

A \(\dfrac {\pi}{ 3}\)

B \(\dfrac {-\pi}{ 6}\)

C \(\dfrac {-\pi}{ 3}\)

D \(\dfrac {-\pi}{ 2} \)

×

\(tan^{-1} \left ( \dfrac {-1}{\sqrt 3} \right)=\)Angle between \(\dfrac {-\pi}{2}\) and \(\dfrac {\pi}{2}\) whose tangent is \(\dfrac {-1}{\sqrt 3}=\dfrac {-\pi}{6}\).

\(\therefore \;\;tan^{-1} \left ( \dfrac {-1}{\sqrt 3} \right)=\dfrac {-\pi}{6}\)

Find the value of \(tan^{-1} \left ( \dfrac {-1}{\sqrt 3} \right)\).

A

\(\dfrac {\pi}{ 3}\)

.

B

\(\dfrac {-\pi}{ 6}\)

C

\(\dfrac {-\pi}{ 3}\)

D

\(\dfrac {-\pi}{ 2} \)

Option B is Correct

Cancelation Formulae for Sine Functions

  • \(sin^{-1}\,(sin\,x)=x\) if \(\dfrac {-\pi}{2}\leq x \leq \dfrac {\pi}{2}\)
  • \(sin\,(sin^{-1}\,x)=x\) if \(-1\leq x \leq 1\)

The above formulae are a direct result of the fact that \(fof^{-1}(x)=x\) for all valid \(x\).

Illustration Questions

The value of \(sin^{-1} \Big ( sin\,\dfrac {13\,\pi}{6}\Big)\) is 

A \(\dfrac {7\,\pi}{6}\)

B \(\dfrac {\pi}{6}\)

C \(\dfrac {13\,\pi}{6}\)

D \(\dfrac {\pi}{4}\)

×

\(sin^{-1}\,(sin\,x)=x\) only if \(\dfrac {-\pi}{2}\leq x \leq \dfrac {\pi}{2}\)

We reduce the angle \(\dfrac {13\,\pi}{6}\)  so that, it comes in the desired range  \(\left (\dfrac {-\pi}{2},\dfrac {\pi}{2}\right)\).

\(sin\,\dfrac {13\,\pi}{6}=sin\,\left (2\pi+\dfrac {\pi}{6}\right)=sin\,\dfrac {\pi}{6}\)

\(\therefore \;sin^{-1}\,\left (sin\,\dfrac {13\,\pi}{6}\right) =sin^{-1}\,\left (sin\,\dfrac {\pi}{6}\right)=\dfrac {\pi}{6}\)

The value of \(sin^{-1} \Big ( sin\,\dfrac {13\,\pi}{6}\Big)\) is 

A

\(\dfrac {7\,\pi}{6}\)

.

B

\(\dfrac {\pi}{6}\)

C

\(\dfrac {13\,\pi}{6}\)

D

\(\dfrac {\pi}{4}\)

Option B is Correct

Cancelation Formulae for Cosine Functions

  • \(cos^{-1}\,(cos\,x)=x\) if \(0\leq x \leq \pi\)
  • \(cos\,(cos^{-1}\,x)=x\) if \(-1\leq x \leq 1\)

The above formulae are a direct result of the fact that \(fof^{-1}(x)=x\) for all valid \(x\).

Illustration Questions

The value of \(cos^{-1} \Big ( cos\,\dfrac {13\,\pi}{6}\Big)\) is 

A \(\dfrac {\pi}{6}\)

B \(\dfrac {\pi}{3}\)

C \(\dfrac {-\pi}{3}\)

D \(\dfrac {\pi}{2}\)

×

\(cos^{-1}\,(cos\,x)=x\) only if \(0\leq x \leq \pi\) 

We reduce the angle \(\dfrac {13\,\pi}{6}\) , so that it comes in the desired range \((0, \pi)\) .

\(\Rightarrow cos\,\dfrac {13\,\pi}{6}\)

\(=cos\,\left (2\pi+\dfrac {\pi}{6}\right)\)

\(=cos\,\dfrac {\pi}{6}\)

\(\therefore \;\;cos^{-1}\,\left ( cos\,\dfrac {13\,\pi}{6}\right)=cos^{-1}\,\left (cos\,\dfrac {\pi}{6}\right)\)

\(=\dfrac {\pi}{6}\)

The value of \(cos^{-1} \Big ( cos\,\dfrac {13\,\pi}{6}\Big)\) is 

A

\(\dfrac {\pi}{6}\)

.

B

\(\dfrac {\pi}{3}\)

C

\(\dfrac {-\pi}{3}\)

D

\(\dfrac {\pi}{2}\)

Option A is Correct

Simplification of Expression Containing a Trigonometric and an Inverse Trigonometric Function 

  • Sometimes we come across certain expressions which contain a trigonometric and an inverse trigonometric expression .We can simplify these expressions by using basic trigonometric rules. 

          e.g Consider the expression  

          \(tan (sin^{-1}x)\)

Let  \(sin^{-1} x=y \Rightarrow x=sin \,y\)    where \(y\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) 

Now if   \(sin \,y = x \Rightarrow cos\,y = \sqrt{1-x^2}\)  

\(\Rightarrow sec\,y =\dfrac{1}{\sqrt{1-x^2}} \Rightarrow tan\,y \sqrt{\dfrac{1}{1-x^2}-1} = \sqrt{\dfrac{x^2}{1-x^2}}\)

\(\Rightarrow tan\,y = \dfrac{x}{\sqrt{1-x^2}} \to\)  which is the required expression

Illustration Questions

Simplify the expression  \(f(x)= sin (tan^{-1} x)\)

A \(\dfrac {x}{\sqrt{1-x^2}}\)

B \(\dfrac {x}{\sqrt{1+x^2}}\)

C \(\dfrac {\sqrt{1+x^2}}{x}\)

D \(\dfrac {\sqrt{1-x^2}}{x}\)

×

Use one of the basic identities 

1. \(sin^2\,x +cos^2\,x =1 \,\forall \,x\)

2. \(sec^2\,x - tan^2\,x =1 \,\forall \,x \in R - \left\{(2n+1)\dfrac{\pi}{2}\right\}\)

3. \(cosec^2\,x - cot^2\,x =1 \,\forall\,x \in R - \{n\,\pi\}\) 

\(\Rightarrow f(x) =sin(tan^{-1}x) =sin\,y\)  where  \(tan^{-1}x=y\)

\(\Rightarrow x=tan\,y \to \) now expres \(sin\,y\) in terms of  \(x \) 

\(tan\,y= x\Rightarrow sec\,y = \sqrt{1+x^2} \Rightarrow cos\,y =\dfrac{1}{\sqrt{1+x^2}}\)

\( \Rightarrow sin\,y= \sqrt{1-\dfrac{1}{1+x^2}} =\sqrt{\dfrac{x^2}{1+x^2}} = \dfrac{x}{\sqrt{1+x^2}} \)

\(\therefore\; sin(tan^{-1}x) = sin\,y = \dfrac{x}{\sqrt{1+x^2}}\)

Simplify the expression  \(f(x)= sin (tan^{-1} x)\)

A

\(\dfrac {x}{\sqrt{1-x^2}}\)

.

B

\(\dfrac {x}{\sqrt{1+x^2}}\)

C

\(\dfrac {\sqrt{1+x^2}}{x}\)

D

\(\dfrac {\sqrt{1-x^2}}{x}\)

Option B is Correct

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