RD Sharma Class 12 Exercise 3.1 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.1 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Updated on Jan 21, 2022 11:30 AM IST

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RD Sharma Class 12 Solutions Inverse Trigonometric Functions Exercise 3.1 has 14 questions, including subparts of Level 1 difficulty. However, these questions are fundamental concept-based and can be completed relatively quickly if you have gone through the theorems.

Inverse Trigonometric Functions Excercise: 3.1

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Function Exercise 3.1 Question 1 (i)

Answer:
$\frac{- π}{3}$
Hint:
$f (x) = y$
$f^{- 1} (y) = x$
Given:
$F i n d p r i n c i p a l v a l u e o f \sin^{- 1} (\frac{- \sqrt{3}}{2})$
Solution:
$L e t, \sin^{- 1} (\frac{- \sqrt{3}}{2}) = y$
$\Rightarrow \sin y = \frac{- \sqrt{3}}{2}$
$\Rightarrow \sin y = - \sin (\frac{π}{3}) = \sin (\frac{- π}{3})$
$R a n g e o f p r i n c i p a l v a l u e o f \sin^{- 1} i s (\frac{- π}{2}, \frac{π}{2}) a n d \sin (\frac{- π}{3}) = \frac{- \sqrt{3}}{2}$
$T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} (\frac{- \sqrt{3}}{2}) i s \frac{- π}{3}$

Inverse Trigonometric Function Exercise 3.1 Question 1 (ii)

Answer:

\frac{- π}{6}

Hint:

\cos (90^{\circ} + x) = - \sin x

Given:

F i n d p r i n c i p a l v a l u e o f \sin^{- 1} (\cos \frac{2 π}{3})

Solution:

L e t, \sin^{- 1} (\cos \frac{2 π}{3}) = y

\Rightarrow \sin y = (\cos \frac{2 π}{3})

\Rightarrow \sin y = \cos \frac{2 π}{3} = \cos (\frac{π}{2} + \frac{π}{6}) = - \sin (\frac{π}{6})

W e k n o w p r i n c i p a l v a l u e o f \sin^{- 1} i s [\frac{- π}{2}, \frac{π}{2}] a n d - \sin \sin (\frac{π}{6}) = \cos \cos (\frac{2 π}{3})

T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} [\cos (\frac{2 π}{3})] i s \frac{- π}{6} .

Inverse Trigonometric Function Exercise 3.1 Question 1 (iii)

Answer:
$\frac{π}{12}$
Hint:
$S e p a r a t e a n d r e o r g a n i z e t h e v a l u e s$
Given:
$F i n d p r i n c i p a l v a l u e o f \sin^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}})$
Solution:
$\sin^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}})$ $= \sin^{- 1} (\frac{\sqrt{3}}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}})$
$= \sin^{- 1} (\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}})$
$= \sin^{- 1} (\frac{\sqrt{3}}{2} \times \sqrt{1 - {(\frac{1}{\sqrt{2}})}^{2}} - \frac{1}{\sqrt{2}} \times \sqrt{1 - {(\frac{\sqrt{3}}{2})}^{2}})$
$= \sin^{- 1} (\frac{\sqrt{3}}{2}) - \sin^{- 1} (\frac{1}{\sqrt{2}})$
$= \frac{π}{3} - \frac{π}{4}$
$= \frac{π}{12}$
$T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) i s \frac{π}{2} .$

Inverse Trigonometric Function Exercise 3.1 Question 1 (iv)

Answer:

\frac{7 π}{12}

Hint:

S e p a r a t e a n d r e o r g a n i z e t h e v a l u e s

Given:

F i n d p r i n c i p a l v a l u e o f \sin^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})

Solution:

\sin^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})

= \sin^{- 1} (\frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}})

= \sin^{- 1} (\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}})

= \sin^{- 1} (\frac{\sqrt{3}}{2} \times \sqrt{1 - {(\frac{1}{\sqrt{2}})}^{2}} + \frac{1}{\sqrt{2}} \times \sqrt{1 - {(\frac{\sqrt{3}}{2})}^{2}})

= \sin^{- 1} (\frac{\sqrt{3}}{2}) + \sin^{- 1} (\frac{1}{\sqrt{2}})

= \frac{π}{3} + \frac{π}{4}

= \frac{7 π}{2}

T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}}) i s = \frac{7 π}{2} .

Inverse Trigonometric Function Exercise 3.1 Question 1 (v)

Answer:

\frac{- π}{4}

Hint:

\cos x = - \cos (180^{\circ} - x)

Given:

F i n d p r i n c i p a l v a l u e o f \sin^{- 1} (\cos (\frac{3 π}{4}))

Solution:

L e t, \sin^{- 1} (\cos (\frac{3 π}{4})) = y

\Rightarrow \sin y = \cos (\frac{3 π}{4})

\Rightarrow \sin y = \cos \frac{3 π}{4} = \cos (π - \frac{π}{4}) = - \cos (\frac{π}{4})

P r i n c i p a l v a l u e o f \cos^{- 1} i s [\frac{- π}{2}, \frac{π}{2}] a n d - \cos \cos (\frac{π}{4}) = \cos \cos (\frac{3 π}{4})

T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} (\cos (\frac{3 π}{4})) i s \frac{- π}{4} .

Inverse Trigonometric Function Exercise 3.1 Question 1 (vi)

Answer:
_{$\frac{π}{2}$}
Hint:
_{$\tan (π + x) = \tan x$}
Given:
$F i n d p r i n c i p a l v a l u e o f \sin^{- 1} (\tan (\frac{5 π}{4}))$
Solution:
$L e t, y = \sin^{- 1} (\tan (\frac{5 π}{4}))$
_{$\sin y = \tan (\frac{5 π}{4})$}
_{$\sin y = \tan (π + \frac{π}{4})$}
_{$= \tan (\frac{π}{4})$}
$= 1$
$= \sin (\frac{π}{2})$
$R a n g e P r i n c i p a l v a l u e o f \sin^{- 1} i s [\frac{- π}{2}, \frac{π}{2}] a n d \sin (\frac{π}{2}) = \tan (\frac{5 π}{4})$
$T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} (\tan (\frac{5 π}{4})) i s \frac{π}{2} .$

Inverse Trigonometric Function Exercise 3.1 Question 2 (i)

Answer:

\frac{- π}{3}

Hint:

2 \sin^{- 1} x = \sin^{- 1} (2 x \sqrt{1 - x^{2}})

Given:

F i n d p r i n c i p a l v a l u e o f \sin^{- 1} \frac{1}{2} - 2 \sin^{- 1} \frac{1}{\sqrt{2}}

Solution:

\sin^{- 1} \frac{1}{2} - 2 \sin^{- 1} \frac{1}{\sqrt{2}}

= \sin^{- 1} \frac{1}{2} - \sin^{- 1} (2 \times \frac{1}{\sqrt{2}} \sqrt{1 - {(\frac{1}{\sqrt{2}})}^{2}})

= \sin^{- 1} \frac{1}{2} - \sin^{- 1} 1

= \frac{π}{6} - \frac{π}{2}

= \frac{- π}{3}

T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} \frac{1}{2} - 2 \sin^{- 1} \frac{1}{\sqrt{2}} i s = \frac{- π}{3} .

Inverse Trigonometric Function Exercise 3.1 Question 2 (ii)

Answer:

\frac{π}{6}

Hint:

\sin^{- 1} x = y

\sin y = x

Given:

F i n d p r i n c i p a l v a l u e o f \sin^{- 1} {\cos (\sin^{- 1} (\frac{\sqrt{3}}{2}))}

Solution:

\sin^{- 1} {\cos (\sin^{- 1} (\frac{\sqrt{3}}{2}))}

= \sin^{- 1} [\cos (\frac{π}{3})] [∵ (\frac{\sqrt{3}}{2}) = \frac{π}{3}]

= (\frac{1}{2}) [∵ \cos \frac{π}{3} = \frac{1}{2}]

L e t, (\frac{1}{2}) = y

\sin \sin y = \frac{1}{2}

R a n g e o f \sin^{- 1} i s [\frac{- π}{2}, \frac{π}{2}] a n d \sin \sin (\frac{π}{6}) = \frac{1}{2}

T h e r e f o r e, p r i n c i p a l v a l u e o f \sin^{- 1} {\cos (\sin^{- 1} (\frac{\sqrt{3}}{2}))} i s \frac{π}{6} .

Inverse Trigonometric Function Exercise 3.1 Question 3 (i)

Answer:

[- 1, 1]

Hint:

T h e d o m a i n o f \sin^{- 1} x i s [- 1, 1]

Given:

\sin^{- 1} x^{2}

Explanation:

L e t f (x) = (x^{2})

T h e d o m a i n o f \sin^{- 1} i s [- 1, 1] w h i c h i m p l i e s t h a t t h e v a l u e o f x l i e s b e t w e e n - 1 a n d 1

i . e; - 1 \leq x \leq 1

b u t h e r e w e h a v e t h e f u n c t i o n x^{2}

s o, - 1 \leq x^{2} \leq 1 [A s x^{2} i s a l w a y s p o s i t i v e]

\Rightarrow 0 \leq x^{2} \leq 1

\Rightarrow | x | \leq 1

\Rightarrow x \leq 1 a n d x \geq - 1 [∵ a^{2} - b^{2} = (a + b) (a - b)]

H e n c e, t h e d o m a i n o f \sin^{- 1} x^{2} i s [- 1, 1] .

Inverse Trigonometric Function Exercise 3.1 Question 3 (ii)

Answer:

[- 1, 1]

Hint:

T h e d o m a i n o f \sin^{- 1} x i s [- 1, 1] a n d t h e d o m a i n o f \sin x i s R .

Given:

f (x) = \sin^{- 1} x + \sin x

Explanation:

W h e n w e f i n d t h e d o m a i n o f t h e f u n c t i o n w h i c h i s f u r t h e r t h e s u m o f t w o d i f f e r e n t f u n c t i o n s t h e n w e t a k e t h e i n t e r s e c t i o n o f t h e i r r e s p e c t i v e d o m a i n s .

[- 1, 1] \cap R

= [- 1, 1] \cap (- \infty, \infty)

= [- 1, 1]

Hence, the domain of

\sin^{- 1} x + \sin x

[- 1, 1]

Inverse Trigonometric Function Exercise 3.1 Question 3 (iii)

Answer:
$[- \sqrt{2}, 1] \cup [1, \sqrt{2}]$
Hint:
$T h e d o m a i n o f \sin^{- 1} x i s [- 1, 1] .$
Given:
$f (x) = \sin^{- 1} \sqrt{x^{2} - 1}$
Explanation:
$- 1 \leq \sqrt{x^{2} - 1} \leq 1$
$x^{2} - 1 \leq 1 [S q u a r i n g b o t h s i d e s]$
$x^{2} - 1 + 1 \leq 1 + 1$
$x^{2} \leq 2$
$- \sqrt{2} \leq x \leq \sqrt{2} \cdot \cdot \cdot \cdot (i)$
$A l s o, x^{2} - 1 \geq 0$
$x^{2} \geq 1$
$\Rightarrow x \leq - 1 a n d x \geq 1 \cdot \cdot \cdot \cdot (i i)$
$F r o m (i) a n d (i i), t h e v a l u e o f x l i e s b e t w e e n$
$1 \leq x \leq \sqrt{2} a n d - \sqrt{2} \leq x \leq - 1$
$T h e d o m a i n o f \sin^{- 1} \sqrt{x^{2} - 1} i s [- \sqrt{2,} - 1] \cup [1, \sqrt{2}]$

Inverse Trigonometric Function Exercise 3.1 Question 3 (iv

Answer:

[\frac{- 1}{2}, \frac{1}{2}]

Hint:

T h e d o m a i n o f \sin^{- 1} x i s [- 1, 1] .

Given:

f (x) = \sin^{- 1} x + \sin^{- 1} 2 x

Explanation:

N o w, w e f i n d t h e d o m a i n o f \sin^{- 1} 2 x

- 1 \leq 2 x \leq 1

\frac{- 1}{2} \leq x \leq \frac{1}{2}

T h e d o m a i n o f f (x) i s t h e i n t e r s e c t i o n o f d o m a i n s o f \sin^{- 1} x a n d \sin^{- 1} 2 x .

[\frac{- 1}{2}, \frac{1}{2}] \cap [- 1, 1]

= [\frac{- 1}{2}, \frac{1}{2}]

Inverse Trigonometric Function Exercise 3.1 Question 4

Answer:

4

Given:

\sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z + \sin^{- 1} t = 2 π t h e n f i n d x^{2} + y^{2} + z^{2} + t^{2}

Hint:

W e k n o w t h a t t h e r a n g e o f \sin^{- 1} i s [\frac{- π}{2}, \frac{π}{2}]

\sin^{- 1} x \leq \frac{π}{2}, \sin^{- 1} y \leq \frac{π}{2}, \sin^{- 1} z \leq \frac{π}{2}, \sin^{- 1} t \leq \frac{π}{2}

Solution:

\sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z + \sin^{- 1} t \leq 2 π a n d i t i s g i v e n t h a t L . H . S i s 2 π .

S o i t i s p o s s i b l e o n l y w h e n

\sin^{- 1} x = \frac{π}{2}, \sin^{- 1} y = \frac{π}{2}, \sin^{- 1} z = \frac{π}{2}, \sin^{- 1} t = \frac{π}{2}

x = 1, y = 1, z = 1, t = 1

∴ x^{2} + y^{2} + z^{2} + t^{2} = 4

Inverse Trigonometric Function Exercise 3.1 Question 5

Answer:

3

Given:

{(\sin^{- 1} x)}^{2} + {(\sin^{- 1} y)}^{2} + {(\sin^{- 1} z)}^{2} = \frac{3}{4} π^{2} t h e n f i n d x^{2} + y^{2} + z^{2} .

Hint:

W e k n o w t h a t t h e r a n g e o f \sin^{- 1} i s [\frac{- π}{2}, \frac{π}{2}]

\sin^{- 1} x \leq \frac{π}{2}, \sin^{- 1} y \leq \frac{π}{2}, \sin^{- 1} z \leq \frac{π}{2}

{(\sin^{- 1} x)}^{2} \leq \frac{π^{2}}{4}, {(\sin^{- 1} y)}^{2} \leq \frac{π^{2}}{4}, {(\sin^{- 1} z)}^{2} \leq \frac{π^{2}}{4}

{(\sin^{- 1} x)}^{2} + {(\sin^{- 1} y)}^{2} + {(\sin^{- 1} z)}^{2} \leq \frac{3}{4} π^{2} a n d i t i s g i v e n t h a t L . H . S i s \frac{3}{4} π^{2} .

S o i t i s p o s s i b l e o n l y w h e n

\sin^{- 1} x = \frac{π}{2}, \sin^{- 1} y = \frac{π}{2}, \sin^{- 1} z = \frac{π}{2}

x = 1, y = 1, z = 1

∴ x^{2} + y^{2} + z^{2} = 3

The concepts used in this exercise are:

Evaluating the Inverse of Sin and Cos functions
Finding the domain of inverse functions
Finding values of variables assigned to inverse equations

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