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# Class 12 RD Sharma Solutions- Chapter 4 Inverse Trigonometric Functions – Exercise 4.1

• Difficulty Level : Easy
• Last Updated : 03 Jan, 2021

### Question 1. Find the principal value of each of the following :

(i) Sin-1(- âˆš3/2)

(ii) Sin-1(cos 2Ï€/3)

(iii)Sin-1(-âˆš3 – 1/2âˆš2)

(iv) Sin-1(âˆš3 + 1/2âˆš2)

(v) Sin-1(cos 3Ï€/4)

(vi) Sin-1(tan 5Ï€/4)

Solution:

(i) Sin-1 (-âˆš3/ 2)

= Sin-1 [sin (- Ï€ / 3)]

= – Ï€ / 3     Ans.

(ii) Sin-1 (cos 2Ï€ / 3)

= Sin-1 (- 1 / 2)

= Sin-1 (- Ï€ / 6)

= – Ï€ / 6   Ans.

(iii) Sin-1 (âˆš3 – 1/2âˆš2)

= Sin-1 ( sin Ï€ / 12 )

= Ï€ / 12  Ans.

(iv) Sin-1 (âˆš3 + 1/2 âˆš2)

= Sin-1 (5Ï€ / 12)

= 5Ï€ / 12   Ans.

(v) Sin-1 (cos 3Ï€ / 4)

= Sin-1 (- âˆš2 / 2)

= [Sin-1 (- Ï€ / 4)]

= – Ï€ / 4   Ans.

(vi) Sin-1 (tan 5Ï€ / 4)

= Sin-1 (1)

= Sin-1 [sin (Ï€ / 2)]

= Ï€ / 2    Ans.

### Question 2.

(i) Sin-1 1/2  -2 Sin-1 1/âˆš2

(ii) Sin-1 {cos (Sin-1 âˆš3 / 2)}

Solution:

(i) Sin-1 1/2  -2  Sin-1 1/ âˆš2

= Sin-1 1/2 – Sin-1 [ 2 x 1/ âˆš2 âˆš1-  ]

=   Sin-1 1/2 – Sin-1 (1)

=  Ï€/6 –  Ï€ /2

=  Ï€ / 3   Ans.

(ii) Sin-1 { cos ( Sin-1 sin  Ï€ / 3 )}

=   Sin-1 { cos ( Ï€ / 3 ) }

=    Sin-1 { 1/2 }

=   Sin-1 { sin  Ï€ / 6 }

=  Ï€ / 6   Ans.

### Question 3.  Find the domain of each of the following functions :

(i) f(x) = Sin-1 x2

(ii) f(x) = Sin-1x + sinx

(iii) f(x) = Sin-1âˆšx2 – 1

(iv) f(x) = Sin-1x + Sin-1 2x

Solution:

(i) Domain of  Sin-1 lies between the interval [ -1 , 1 ]

and x2 âˆˆ [ 0 , 1 ] as x2 can not be negative .

So, x âˆˆ [ -1 , 1 ]

Hence, the domain of the function f(x) = [ -1 , 1 ]   Ans.

(ii) Let f(x) = g(x) + h(x) , where g(x) =  Sin-1x  and h(x) = sinx respectively.

Therefore , the domain of f(x) is given by the intersection of the domain g(x) & h(x) .

The domain of g(x) = [ -1 , 1 ]

The domain of h(x) = [ – âˆž , âˆž ]

Thus, the interaction of g(x) and h(x) is [ -1 , 1 ]

Hence , the domain of f(x) is [ -1 , 1 ]    Ans.

(iii) As we know , the domain of Sin-1 x is [ -1 , 1 ]

Therefore , domain of Sin-1  âˆšx2 – 1 will also lies in the interval [ -1 , 1 ]

:. x2 – 1 âˆˆ [ 0 , 1 ] as square root cannot be negative .

=> x2 âˆˆ [ 1 , 2 ]

=> x âˆˆ [ – âˆš2 , -1 ] U [ 1 , âˆš2 ]

Hence, the domain of function f(x) = [ – âˆš2 , -1 ] U [ 1 , âˆš2 ]   Ans.

(iv) Let f(x) = g(x) + h(x), where g(x) = Sin-1 x  x and h(x) = Sin-1 2x

Therefore, the domain of f(x) will be given by the intersection of g(x) and h(x) .

the domain of g(x) = [ -1 , 1 ]

lly , the domain of h(x) = [ -1/2 , 1/2 ]

g(x) âˆ© h(x) = [ -1 , 1 ]  âˆ© [ -1/2 , 1/2 ]

Hence, the domain of the function f(x) =  [- 1/ 2 , 1/2 ]    Ans.

### Question 4. If sin-1x + sin-1y + sin-1z + sin-1t = 2Ï€, then find the value of x2 + y2 + z2 + t2 .

Solution:

As we already know, Range of sin-1 is [ – Ï€ / 2 , Ï€ / 2 ]

Given: (sin-1x) + (sin-1y) +(sin-1y)+(sin-1t) = 2 Ï€

So, each takes the value of Ï€ / 2

:. x = 1, y = 1, z = 1 & t = 1

Hence, x2 + y2+ z2 + t2 = 1+ 1 + 1 + 1 = 4    Ans.

### Question 5. If (sin-1x)2 + ( sin-1y )2 + ( sin-1y )2  = 3Ï€2/4, find the value of  x2 + y2 + z2             .

Solution:

As we already know , Range of   sin-1 is [ – Ï€ / 2 , Ï€ / 2 ]

Given:     ( sin-1x )2 + ( sin-1y )2 + ( sin-1y )2 = 3Ï€2/4

:. each takes the value of  Ï€ / 2

x = 1, y = 1 & z = 1 .

Hence ,  x2 + y2 + z2 = 1 + 1 + 1 = 3    Ans.

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