Samacheer Kalvi Class 12 Maths Solutions Chapter 4 Inverse Trigonometric Functions Exercise 4.1

 

Question 1. Find all the values of x such that
(i) sin x = 0
(ii) sin x = -1
Answer:
(i) For \( \sin x = 0 \), the general solution for x is \( x = n\pi \), where n is an integer.
Given the values for n as \( n = 0, \pm 1, \pm 2, \pm 3, \ldots, \pm 10 \), the values of x are all integer multiples of \( \pi \) from \( -10\pi \) to \( 10\pi \). This represents points where the sine wave crosses the x-axis.
(ii) For \( \sin x = -1 \), the general solution for x is \( x = (4n-1)\frac{\pi}{2} \), where n is an integer.
Given the values for n as \( n = 0, \pm 1, \pm 2, \pm 3, 4 \), we can find the specific values of x:
For \( n = 0 \), \( x = (4(0)-1)\frac{\pi}{2} = -\frac{\pi}{2} \)
For \( n = 1 \), \( x = (4(1)-1)\frac{\pi}{2} = \frac{3\pi}{2} \)
For \( n = -1 \), \( x = (4(-1)-1)\frac{\pi}{2} = -\frac{5\pi}{2} \)
For \( n = 2 \), \( x = (4(2)-1)\frac{\pi}{2} = \frac{7\pi}{2} \)
For \( n = -2 \), \( x = (4(-2)-1)\frac{\pi}{2} = -\frac{9\pi}{2} \)
For \( n = 3 \), \( x = (4(3)-1)\frac{\pi}{2} = \frac{11\pi}{2} \)
For \( n = -3 \), \( x = (4(-3)-1)\frac{\pi}{2} = -\frac{13\pi}{2} \)
For \( n = 4 \), \( x = (4(4)-1)\frac{\pi}{2} = \frac{15\pi}{2} \)
In simple words: When sine of x is zero, x can be any whole number multiple of pi. When sine of x is minus one, x can be specific odd multiples of pi divided by two, like \( -\frac{\pi}{2} \) or \( \frac{3\pi}{2} \).

🎯 Exam Tip: Remember the principal values for \( \sin x = 0 \) (\( x = 0 \)) and \( \sin x = -1 \) (\( x = -\frac{\pi}{2} \)) and then use the general solution formula to find all possible values for n.

 

Question 2. Find the period and amplitude of
(i) y = sin 7x
(ii) y = -sin (\frac{1}{3}x)
(iii) y = 4 sin (-2x)
Answer:
For a general sine function \( y = a \sin(bx - c) + d \):
Amplitude = \( |a| \)
Period = \( \frac{2\pi}{|b|} \)

(i) For \( y = \sin 7x \):
Here, \( a = 1 \) and \( b = 7 \).
Amplitude = \( |1| = 1 \)
Period = \( \frac{2\pi}{|7|} = \frac{2\pi}{7} \)

(ii) For \( y = -\sin (\frac{1}{3}x) \):
Here, \( a = -1 \) and \( b = \frac{1}{3} \).
Amplitude = \( |-1| = 1 \)
Period = \( \frac{2\pi}{|\frac{1}{3}|} = 2\pi \times 3 = 6\pi \)

(iii) For \( y = 4 \sin (-2x) \):
Here, \( a = 4 \) and \( b = -2 \).
Amplitude = \( |4| = 4 \)
Period = \( \frac{2\pi}{|-2|} = \frac{2\pi}{2} = \pi \)
The period represents how often the wave repeats, and amplitude is half the distance from the highest to the lowest point of the wave.
In simple words: For a sine wave, the 'a' number tells you how tall the wave is (amplitude), and the 'b' number helps you find how long it takes for one full wave cycle (period).

🎯 Exam Tip: Always take the absolute value of 'a' for amplitude and 'b' for the period calculation, as these represent distances and durations which are always positive.

 

Question 3. Sketch the graph of y = sin (\frac{1}{3}x) for \( 0 \le x \le 6\pi \)
Answer:
First, we find the period of \( y = \sin(\frac{1}{3}x) \). Here, \( b = \frac{1}{3} \).
Period = \( \frac{2\pi}{|b|} = \frac{2\pi}{|\frac{1}{3}|} = 6\pi \).
This means the graph completes one full cycle over the interval \( 0 \le x \le 6\pi \).

We can create a table of values for key points within this interval:

The graph of \( y = \sin(\frac{1}{3}x) \) from \( 0 \) to \( 6\pi \) looks like a standard sine wave, stretched horizontally to complete one full cycle over this wider range.

In simple words: To draw this graph, first find how long it takes for one full wave cycle. Then plot points at the start, peak, middle, bottom, and end of the cycle to get the shape of the wave.

🎯 Exam Tip: When sketching graphs of trigonometric functions, always determine the period and amplitude first, then mark the intercepts and extreme points clearly on the axes.

 

Question 4. Find the value of
(i) \( \sin^{-1} (\sin(\frac{2\pi}{3})) \)
(ii) \( \sin^{-1} (\sin(\frac{5\pi}{4})) \)
Answer:
(i) We need to find the value of \( \sin^{-1} (\sin(\frac{2\pi}{3})) \).
The principal value branch of \( \sin^{-1} x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
Since \( \frac{2\pi}{3} \) is not in this range, we use the identity \( \sin(\pi - \theta) = \sin \theta \).
So, \( \sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3}) \).
Now, \( \sin^{-1} (\sin(\frac{2\pi}{3})) = \sin^{-1} (\sin(\frac{\pi}{3})) \).
Since \( \frac{\pi}{3} \) is in the principal value branch \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), the value is \( \frac{\pi}{3} \).

(ii) We need to find the value of \( \sin^{-1} (\sin(\frac{5\pi}{4})) \).
Again, \( \frac{5\pi}{4} \) is not in the principal value branch \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
We use the identity \( \sin(\pi + \theta) = -\sin \theta \).
So, \( \sin(\frac{5\pi}{4}) = \sin(\pi + \frac{\pi}{4}) = -\sin(\frac{\pi}{4}) \).
Now, \( \sin^{-1} (\sin(\frac{5\pi}{4})) = \sin^{-1} (-\sin(\frac{\pi}{4})) \).
Using the property \( \sin^{-1}(-x) = -\sin^{-1}(x) \), we get:
\( -\sin^{-1} (\sin(\frac{\pi}{4})) \).
Since \( \frac{\pi}{4} \) is in the principal value branch \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), the value is \( -\frac{\pi}{4} \).
The key is to convert the angle inside the sine function to an equivalent angle within the principal range of \( \sin^{-1} \).
In simple words: When you have sine inverse of sine, the answer is the angle itself, but only if that angle is between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \). If it's outside this range, you must use sine rules to change the angle to an equivalent one that fits.

🎯 Exam Tip: Always remember the principal value branch for inverse trigonometric functions. For \( \sin^{-1}x \), it's \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), and you must adjust the angle within the sine function to fall into this range before finding the inverse sine.

 

Question 5. For what value of x does \( \sin x = \sin^{-1} x \)?
Answer:
We need to find the value of x for which the equation \( \sin x = \sin^{-1} x \) holds true.
Consider the graphs of \( y = \sin x \) and \( y = \sin^{-1} x \).
The domain of \( y = \sin x \) is \( (-\infty, \infty) \), and its range is \( [-1, 1] \).
The domain of \( y = \sin^{-1} x \) is \( [-1, 1] \), and its range is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
For the equation to have a solution, x must be in the intersection of their domains, which is \( [-1, 1] \). Also, the values of \( \sin x \) and \( \sin^{-1} x \) must be equal.
We know that \( \sin x \le x \) for \( x > 0 \) and \( \sin x \ge x \) for \( x < 0 \).
Similarly, \( \sin^{-1} x \ge x \) for \( x > 0 \) and \( \sin^{-1} x \le x \) for \( x < 0 \).
The only point where \( \sin x = x \) and \( \sin^{-1} x = x \) is at \( x = 0 \).
Let's check: \( \sin(0) = 0 \) and \( \sin^{-1}(0) = 0 \).
Therefore, the only value of x for which \( \sin x = \sin^{-1} x \) is \( x = 0 \). This is the unique point where both functions intersect and are equal to x itself.
In simple words: The only number for which the sine of the number is equal to the inverse sine of the number is zero. You can see this by drawing the graphs of both functions.

🎯 Exam Tip: When solving equations involving functions and their inverses, consider their domains and ranges. Often, graphical analysis can reveal the number and location of solutions, especially for transcendental equations.

 

Question 6. Find the domain of the following
(i) \( f(x) = \sin^{-1} (\frac{x^2+1}{2x}) \)
(ii) \( g(x) = 2 \sin^{-1} (2x - 1) - \frac{\pi}{4} \)
Answer:
(i) For the function \( f(x) = \sin^{-1} (\frac{x^2+1}{2x}) \) to be defined, the argument of \( \sin^{-1} \) must be in the interval \( [-1, 1] \).
So, we must have \( -1 \le \frac{x^2+1}{2x} \le 1 \).
This inequality can be split into two parts:
Part 1: \( \frac{x^2+1}{2x} \ge -1 \)
\( \frac{x^2+1}{2x} + 1 \ge 0 \)
\( \frac{x^2+1+2x}{2x} \ge 0 \)
\( \frac{(x+1)^2}{2x} \ge 0 \)
Since \( (x+1)^2 \ge 0 \) for all real x, for the fraction to be non-negative, \( 2x \) must be positive. So, \( 2x > 0 \implies x > 0 \).
If \( x=0 \), the expression is undefined. If \( x=-1 \), the numerator is 0, making the fraction 0, which satisfies \( \ge 0 \), but \( 2x = -2 \) makes the denominator negative, so it doesn't satisfy \( \ge 0 \). Thus, \( x \ge -1 \) is part of the numerator's effect.
Considering \( (x+1)^2 \ge 0 \) always, we need \( 2x > 0 \), so \( x > 0 \). However, if \( x = -1 \), the numerator is 0, and the inequality \( 0 \ge 0 \) holds if \( x \ne 0 \). Let's be more precise. If \( x > 0 \), then \( \frac{(x+1)^2}{2x} \ge 0 \) is true. If \( x < 0 \), then \( \frac{(x+1)^2}{2x} \le 0 \). So for Part 1, we require \( x > 0 \) or \( x = -1 \) (since \( (x+1)^2 \) is zero). But if \( x=-1 \), \( 2x = -2 \), making the overall fraction negative. So we need \( x>0 \) AND \( x=-1 \) (if possible). This part holds for \( x > 0 \). It also holds for \( x = -1 \) because \( \frac{0}{-2} = 0 \ge -1 \). So, for part 1, we have \( x \in [-1, 0) \cup (0, \infty) \).

Part 2: \( \frac{x^2+1}{2x} \le 1 \)
\( \frac{x^2+1}{2x} - 1 \le 0 \)
\( \frac{x^2+1-2x}{2x} \le 0 \)
\( \frac{(x-1)^2}{2x} \le 0 \)
Since \( (x-1)^2 \ge 0 \) for all real x, for the fraction to be non-positive, \( 2x \) must be negative. So, \( 2x < 0 \implies x < 0 \).
If \( x=1 \), the numerator is 0, making the fraction 0, which satisfies \( \le 0 \), and \( 2x=2 \) makes the denominator positive. So, for part 2, we have \( x < 0 \) or \( x = 1 \). So, for part 2, we have \( x \in (-\infty, 0) \cup \{1\} \).

We need to satisfy both parts simultaneously. The intersection of \( (x \in [-1, 0) \cup (0, \infty)) \) and \( (x \in (-\infty, 0) \cup \{1\}) \) is \( x \in [-1, 0) \cup \{1\} \).
Wait, let's recheck the reasoning from the source's steps for clarity. The source splits the inequality for positive and negative x. If \( x > 0 \): \( x^2+1 \ge -2x \implies x^2+2x+1 \ge 0 \implies (x+1)^2 \ge 0 \). This is always true for \( x>0 \). \( x^2+1 \le 2x \implies x^2-2x+1 \le 0 \implies (x-1)^2 \le 0 \). This is only true if \( (x-1)^2 = 0 \), so \( x = 1 \). So for \( x > 0 \), the only solution is \( x = 1 \). If \( x < 0 \): Multiplying by \( 2x \) reverses the inequality signs. \( x^2+1 \le -2x \implies x^2+2x+1 \le 0 \implies (x+1)^2 \le 0 \). This is only true if \( (x+1)^2 = 0 \), so \( x = -1 \). \( x^2+1 \ge 2x \implies x^2-2x+1 \ge 0 \implies (x-1)^2 \ge 0 \). This is always true for \( x<0 \). So for \( x < 0 \), the only solution is \( x = -1 \). If \( x = 0 \), the expression is undefined. Combining these, the domain is \( \{-1, 1\} \). The source's answer \( [-1, 1] \) seems incorrect based on its own derived steps. Let's re-examine the original solution for any subtle points. The inequality \( -1 \le \frac{x^2+1}{2x} \le 1 \) means the fraction must be between -1 and 1. If \( x > 0 \): \( -2x \le x^2+1 \le 2x \). \( x^2+2x+1 \ge 0 \implies (x+1)^2 \ge 0 \) (Always true). \( x^2-2x+1 \le 0 \implies (x-1)^2 \le 0 \) (Only true for \( x=1 \)). So for \( x>0 \), the solution is \( x=1 \). If \( x < 0 \): (Multiplying by \( 2x \) reverses signs)
\( -2x \ge x^2+1 \ge 2x \) (original inequality was \( -1 \le \frac{x^2+1}{2x} \le 1 \)) So, \( x^2+1 \le -2x \) and \( x^2+1 \ge 2x \). \( x^2+2x+1 \le 0 \implies (x+1)^2 \le 0 \) (Only true for \( x=-1 \)). \( x^2-2x+1 \ge 0 \implies (x-1)^2 \ge 0 \) (Always true). So for \( x<0 \), the solution is \( x=-1 \). If \( x=0 \), the expression is undefined. Therefore, the domain is \( \{-1, 1\} \). The source's solution \( [-1, 1] \) is mathematically incorrect here; the only values are -1 and 1. I will present the correct derivation which leads to \(\{-1, 1\}\). This is an example where the source calculation leads to a different final answer than stated. I must follow the calculation. (i) For the function \( f(x) = \sin^{-1} (\frac{x^2+1}{2x}) \) to be defined, the argument must satisfy \( -1 \le \frac{x^2+1}{2x} \le 1 \). We analyze two cases based on the sign of \( x \): Case 1: \( x > 0 \)
Multiply the inequality by \( 2x \) (which is positive, so inequality signs remain the same):
\( -2x \le x^2+1 \le 2x \)
This gives us two separate inequalities:
(a) \( -2x \le x^2+1 \implies 0 \le x^2+2x+1 \implies 0 \le (x+1)^2 \). This is true for all real \( x \).
(b) \( x^2+1 \le 2x \implies x^2-2x+1 \le 0 \implies (x-1)^2 \le 0 \). This is only true when \( (x-1)^2 = 0 \), which means \( x = 1 \).
Combining (a) and (b) for \( x > 0 \), the only solution is \( x = 1 \). Case 2: \( x < 0 \)
Multiply the inequality by \( 2x \) (which is negative, so inequality signs reverse):
\( -2x \ge x^2+1 \ge 2x \)
This also gives two separate inequalities:
(c) \( x^2+1 \le -2x \implies x^2+2x+1 \le 0 \implies (x+1)^2 \le 0 \). This is only true when \( (x+1)^2 = 0 \), which means \( x = -1 \).
(d) \( x^2+1 \ge 2x \implies x^2-2x+1 \ge 0 \implies (x-1)^2 \ge 0 \). This is true for all real \( x \).
Combining (c) and (d) for \( x < 0 \), the only solution is \( x = -1 \). Case 3: \( x = 0 \)
The expression \( \frac{x^2+1}{2x} \) is undefined when \( x=0 \). Therefore, the domain of \( f(x) \) is \( \{-1, 1\} \). This type of function is only defined at specific points.
(ii) For the function \( g(x) = 2 \sin^{-1} (2x - 1) - \frac{\pi}{4} \) to be defined, the argument of \( \sin^{-1} \) must be in the interval \( [-1, 1] \).
So, we must have \( -1 \le 2x - 1 \le 1 \).
Add 1 to all parts of the inequality:
\( -1 + 1 \le 2x - 1 + 1 \le 1 + 1 \)
\( 0 \le 2x \le 2 \)
Divide all parts by 2:
\( \frac{0}{2} \le \frac{2x}{2} \le \frac{2}{2} \)
\( 0 \le x \le 1 \)
The term \( -\frac{\pi}{4} \) is a constant and does not affect the domain of the function. Therefore, the domain of \( g(x) \) is \( [0, 1] \).
In simple words: For \( \sin^{-1} \) to work, the number inside it must be between -1 and 1. We solve these inequalities to find the allowed values for x. In the first part, x can only be -1 or 1. In the second part, x can be any number from 0 to 1.

🎯 Exam Tip: Always remember that the domain of \( \sin^{-1}(y) \) is \( [-1, 1] \). Set the argument (the expression inside \( \sin^{-1} \)) between -1 and 1 to find the valid x-values. Be careful when multiplying or dividing inequalities by expressions containing variables, as their sign can reverse the inequality.

 

Question 7. Find the value of \( \sin^{-1}(\sin(\frac{5\pi}{9}) \cos(\frac{\pi}{9}) + \cos(\frac{5\pi}{9}) \sin(\frac{\pi}{9})) \)
Answer:
The expression inside \( \sin^{-1} \) is in the form \( \sin A \cos B + \cos A \sin B \), which is the formula for \( \sin(A+B) \).
Here, \( A = \frac{5\pi}{9} \) and \( B = \frac{\pi}{9} \).
So, \( \sin(\frac{5\pi}{9}) \cos(\frac{\pi}{9}) + \cos(\frac{5\pi}{9}) \sin(\frac{\pi}{9}) = \sin(\frac{5\pi}{9} + \frac{\pi}{9}) \)
\( = \sin(\frac{6\pi}{9}) \)
\( = \sin(\frac{2\pi}{3}) \)
Now we need to find \( \sin^{-1}(\sin(\frac{2\pi}{3})) \).
The principal value branch of \( \sin^{-1} x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
Since \( \frac{2\pi}{3} \) is not in this range, we need to find an equivalent angle within the principal branch.
We use the identity \( \sin(\pi - \theta) = \sin \theta \).
So, \( \sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3}) \).
Therefore, \( \sin^{-1}(\sin(\frac{2\pi}{3})) = \sin^{-1}(\sin(\frac{\pi}{3})) \).
Since \( \frac{\pi}{3} \) is in the principal value branch \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), the value is \( \frac{\pi}{3} \).
This problem combines the sine addition formula with the concept of principal values for inverse trigonometric functions.
In simple words: First, use the sine addition formula to simplify the inside part. Then, find the inverse sine of the result. Remember to adjust the angle to be in the main range for inverse sine, which is between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

🎯 Exam Tip: Always look for trigonometric identities to simplify expressions before applying inverse functions. Also, confirm the resulting angle lies within the principal value branch of the inverse trigonometric function.