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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  $$\dfrac {-1}{2}=\dfrac {-\pi}{6}$$

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

#### 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 $$\dfrac {-1}{2}=\dfrac {2\pi}{3}$$.

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

#### 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 $$-1=\dfrac {-\pi}{4}$$.

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

#### 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 $$x$$.

#### 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 $$x$$.

#### 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

#### 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