Samacheer Kalvi Class 12 Maths Solutions Chapter 4 Inverse Trigonometric Functions Exercise 4.2

Question 1. Find all values of x such that
(j) \( -6\pi \le x \le 6\pi \) and \( \cos x = 0 \)
(ii) \( -5\pi \le x \le 5\pi \) and \( \cos x = -1 \)
Answer:
(j) When \( \cos x = 0 \), the general solution for x is given by \( x = (2n+1)\frac{\pi}{2} \), where \( n \) is an integer. We need to find the values of \( n \) such that \( -6\pi \le (2n+1)\frac{\pi}{2} \le 6\pi \).
Multiplying by \( \frac{2}{\pi} \) across the inequality gives:
\( -12 \le 2n+1 \le 12 \)
Subtracting 1 from all parts:
\( -13 \le 2n \le 11 \)
Dividing by 2:
\( -6.5 \le n \le 5.5 \)
So, the integer values for \( n \) are \( n = -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \). These values ensure that x remains within the specified range for the cosine function to be zero.

(ii) When \( \cos x = -1 \), the general solution for x is given by \( x = (2n+1)\pi \), where \( n \) is an integer. We need to find the values of \( n \) such that \( -5\pi \le (2n+1)\pi \le 5\pi \).
Dividing by \( \pi \) across the inequality gives:
\( -5 \le 2n+1 \le 5 \)
Subtracting 1 from all parts:
\( -6 \le 2n \le 4 \)
Dividing by 2:
\( -3 \le n \le 2 \)
So, the integer values for \( n \) are \( n = -3, -2, -1, 0, 1, 2 \). These are the specific integer multipliers for \( \pi \) that make \( \cos x = -1 \) within the given interval.
In simple words: For each case, we find the general form of x when cosine is 0 or -1. Then, we use the given range for x to find all the possible whole number values for 'n' that fit.

🎯 Exam Tip: Remember to always include the general solution for trigonometric equations before applying the given interval. Pay close attention to the range of the given interval and ensure you correctly solve for 'n'.

Question 2. State the reason for \( \cos^{-1}(\cos(-\frac{\pi}{6})) \neq -\frac{\pi}{6} \)
Answer: The principal value branch for the inverse cosine function, \( \cos^{-1}x \), is defined as the interval \( [0, \pi] \). This means that the output of \( \cos^{-1} \) must always lie within this range. Even though \( \cos(-\theta) = \cos(\theta) \), the argument of the inverse function must respect its defined range.
For the given expression, \( \cos^{-1}(\cos(-\frac{\pi}{6})) \):
First, we use the trigonometric identity \( \cos(-\theta) = \cos(\theta) \).
So, \( \cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) \).
Then the expression becomes \( \cos^{-1}(\cos(\frac{\pi}{6})) \).
Since \( \frac{\pi}{6} \) lies within the principal value branch \( [0, \pi] \) (as \( 0 \le \frac{\pi}{6} \le \pi \)), we have \( \cos^{-1}(\cos(\frac{\pi}{6})) = \frac{\pi}{6} \).
However, \( -\frac{\pi}{6} \) does not lie within the principal value branch \( [0, \pi] \). Therefore, \( \cos^{-1}(\cos(-\frac{\pi}{6})) \) cannot be equal to \( -\frac{\pi}{6} \). The principal value is always a positive angle in this case.
In simple words: The answer for \( \cos^{-1} \) must be between 0 and \( \pi \) (or 0 and 180 degrees). Since \( -\frac{\pi}{6} \) is a negative angle, it's outside this allowed range, so it cannot be the correct principal value.

🎯 Exam Tip: Always remember the principal value ranges for all inverse trigonometric functions. For \( \cos^{-1}x \), it's \( [0, \pi] \); for \( \sin^{-1}x \) and \( \tan^{-1}x \), it's \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

Question 3. Is \( \cos^{-1} (-x) = \pi - \cos^{-1} x \) true? Justify your answer.
Answer: Yes, the identity \( \cos^{-1} (-x) = \pi - \cos^{-1} x \) is true. This is a fundamental property of the inverse cosine function, particularly useful when dealing with negative arguments.
To justify this, let \( y = \cos^{-1}(-x) \).
This implies that \( \cos y = -x \).
For \( y \) to be in the principal value range of \( \cos^{-1} \), we know \( y \in [0, \pi] \).
Now, we can rewrite \( -x \) as \( \cos y \). So, \( x = -\cos y \).
Using the trigonometric identity \( -\cos \theta = \cos(\pi - \theta) \), we can write:
\( x = \cos(\pi - y) \)
Since \( y \in [0, \pi] \), it follows that \( (\pi - y) \) will also be in the interval \( [0, \pi] \).
Therefore, we can take the inverse cosine of both sides:
\( \cos^{-1} x = \pi - y \)
Now, substitute back \( y = \cos^{-1}(-x) \):
\( \cos^{-1} x = \pi - \cos^{-1}(-x) \)
Rearranging this equation to solve for \( \cos^{-1}(-x) \), we get:
\( \cos^{-1}(-x) = \pi - \cos^{-1} x \)
This proves the given statement is true for all \( x \in [-1, 1] \).
In simple words: This equation is true. If you take the inverse cosine of a negative number, it's the same as \( \pi \) (180 degrees) minus the inverse cosine of the positive version of that number.

🎯 Exam Tip: Remember this identity for \( \cos^{-1}(-x) \), as well as similar identities for \( \sin^{-1}(-x) = -\sin^{-1}x \) and \( \tan^{-1}(-x) = -\tan^{-1}x \). These identities are crucial for simplifying expressions and solving equations involving negative inputs for inverse trigonometric functions.

Question 4. Find the principal value of \( \cos^{-1}(\frac{1}{2}) \)
Answer: To find the principal value of \( \cos^{-1}(\frac{1}{2}) \), we need to determine the angle \( y \) such that \( \cos y = \frac{1}{2} \).
The principal value branch for \( \cos^{-1}x \) is \( [0, \pi] \). This means the angle \( y \) must be between 0 radians and \( \pi \) radians (or 0 and 180 degrees, inclusive).
We know from standard trigonometric values that \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).
Since \( \frac{\pi}{3} \) (which is 60 degrees) falls within the defined principal value range \( [0, \pi] \), it is the correct principal value.
Therefore, the principal value of \( \cos^{-1}(\frac{1}{2}) \) is \( \frac{\pi}{3} \). This represents the unique angle in the principal range whose cosine is \( \frac{1}{2} \).
In simple words: We are looking for an angle between 0 and 180 degrees whose cosine is \( \frac{1}{2} \). That angle is 60 degrees, or \( \frac{\pi}{3} \) radians.

🎯 Exam Tip: Always state the principal value range for the inverse trigonometric function before finding the value. Knowing common trigonometric values for special angles \( (\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \pi) \) is essential for quick calculations.

Question 5. Find the value of
(i) \( 2\cos^{-1}(\frac{1}{2}) + \sin^{-1}(\frac{1}{2}) \)
(ii) \( \cos^{-1}(\frac{1}{2}) + \sin^{-1}(-1) \)
(iii) \( \cos^{-1} (\cos\frac{\pi}{7}\cos\frac{\pi}{17} - \sin\frac{\pi}{7}\sin\frac{\pi}{17}) \)
Answer:
(i) To find the value of \( 2\cos^{-1}(\frac{1}{2}) + \sin^{-1}(\frac{1}{2}) \):
First, find the principal value of \( \cos^{-1}(\frac{1}{2}) \). We know \( \cos(\frac{\pi}{3}) = \frac{1}{2} \), and \( \frac{\pi}{3} \in [0, \pi] \). So, \( \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \).
Next, find the principal value of \( \sin^{-1}(\frac{1}{2}) \). We know \( \sin(\frac{\pi}{6}) = \frac{1}{2} \), and \( \frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \). So, \( \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \).
Now, substitute these values into the expression:
\( 2(\frac{\pi}{3}) + \frac{\pi}{6} = \frac{2\pi}{3} + \frac{\pi}{6} \)
To add these, find a common denominator, which is 6:
\( \frac{4\pi}{6} + \frac{\pi}{6} = \frac{4\pi+\pi}{6} = \frac{5\pi}{6} \).
This calculation shows how to combine inverse trigonometric values after finding their principal angles.

(ii) To find the value of \( \cos^{-1}(\frac{1}{2}) + \sin^{-1}(-1) \):
We already found \( \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \).
Next, find the principal value of \( \sin^{-1}(-1) \). We need an angle \( \theta \) such that \( \sin \theta = -1 \) and \( \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \). We know that \( \sin(-\frac{\pi}{2}) = -1 \). So, \( \sin^{-1}(-1) = -\frac{\pi}{2} \).
Now, substitute these values into the expression:
\( \frac{\pi}{3} + (-\frac{\pi}{2}) = \frac{\pi}{3} - \frac{\pi}{2} \)
Find a common denominator, which is 6:
\( \frac{2\pi}{6} - \frac{3\pi}{6} = \frac{2\pi-3\pi}{6} = -\frac{\pi}{6} \).

(iii) To find the value of \( \cos^{-1} (\cos\frac{\pi}{7}\cos\frac{\pi}{17} - \sin\frac{\pi}{7}\sin\frac{\pi}{17}) \):
We recognize the expression inside the parenthesis as the trigonometric identity for \( \cos(A+B) = \cos A \cos B - \sin A \sin B \).
Here, \( A = \frac{\pi}{7} \) and \( B = \frac{\pi}{17} \).
So, \( \cos\frac{\pi}{7}\cos\frac{\pi}{17} - \sin\frac{\pi}{7}\sin\frac{\pi}{17} = \cos(\frac{\pi}{7} + \frac{\pi}{17}) \).
The expression becomes \( \cos^{-1}(\cos(\frac{\pi}{7} + \frac{\pi}{17})) \).
First, add the angles: \( \frac{\pi}{7} + \frac{\pi}{17} = \frac{17\pi + 7\pi}{119} = \frac{24\pi}{119} \).
Now, we have \( \cos^{-1}(\cos(\frac{24\pi}{119})) \).
Since \( \frac{24\pi}{119} \) is approximately \( 0.2\pi \), which is between \( 0 \) and \( \pi \) (i.e., within the principal value branch of \( \cos^{-1} \)), then \( \cos^{-1}(\cos(\frac{24\pi}{119})) = \frac{24\pi}{119} \). This demonstrates the application of compound angle formulas in inverse trigonometry.
In simple words: For each part, we find the individual inverse trigonometric values and then add or subtract them. For part (iii), we use a rule to simplify the angles inside the cosine function before finding the inverse cosine.

🎯 Exam Tip: Always determine the principal value for each inverse trigonometric term separately before combining them. For expressions like part (iii), look for standard trigonometric identities (like sum/difference formulas) that can simplify the inner part of the inverse function.

Question 6. Find the domain of
(i) \( f(x) = \sin^{-1} (\frac{|x|-2}{3}) + \cos^{-1} (\frac{1-|x|}{4}) \)
(ii) \( g(x) = \sin^{-1} x + \cos^{-1} x \)
Answer:
(i) To find the domain of \( f(x) = \sin^{-1} (\frac{|x|-2}{3}) + \cos^{-1} (\frac{1-|x|}{4}) \), we must ensure that the arguments of both inverse trigonometric functions are within their valid range of \( [-1, 1] \).
For the \( \sin^{-1} \) term:
\( -1 \le \frac{|x|-2}{3} \le 1 \)
Multiply all parts by 3:
\( -3 \le |x|-2 \le 3 \)
Add 2 to all parts:
\( -1 \le |x| \le 5 \)
Since \( |x| \) must always be non-negative, the condition \( -1 \le |x| \) is automatically satisfied. So we have \( 0 \le |x| \le 5 \). This implies \( -5 \le x \le 5 \). (Condition 1)

For the \( \cos^{-1} \) term:
\( -1 \le \frac{1-|x|}{4} \le 1 \)
Multiply all parts by 4:
\( -4 \le 1-|x| \le 4 \)
Subtract 1 from all parts:
\( -5 \le -|x| \le 3 \)
Multiply all parts by -1 and reverse the inequality signs:
\( -3 \le |x| \le 5 \)
Again, since \( |x| \) must be non-negative, the condition \( -3 \le |x| \) is always true. So we have \( 0 \le |x| \le 5 \). This implies \( -5 \le x \le 5 \). (Condition 2)
The domain of \( f(x) \) is the intersection of Condition 1 and Condition 2.
Domain \( = [-5, 5] \cap [-5, 5] = [-5, 5] \). This ensures that both parts of the function are well-defined for all values within this interval.

(ii) To find the domain of \( g(x) = \sin^{-1} x + \cos^{-1} x \):
The domain of \( \sin^{-1} x \) is \( [-1, 1] \).
The domain of \( \cos^{-1} x \) is \( [-1, 1] \).
For the function \( g(x) \) to be defined, \( x \) must be in the domain of both \( \sin^{-1} x \) and \( \cos^{-1} x \).
Therefore, the domain of \( g(x) \) is the intersection of their individual domains:
Domain \( = [-1, 1] \cap [-1, 1] = [-1, 1] \). This function is actually constant, equal to \( \frac{\pi}{2} \) over its entire domain.
In simple words: For part (i), we solve the absolute value inequalities for both parts of the function and find where their solutions overlap. For part (ii), we know both \( \sin^{-1}x \) and \( \cos^{-1}x \) are only defined for x values between -1 and 1, so their sum is also defined in that same range.

🎯 Exam Tip: When finding the domain of a sum of functions, always find the domain of each individual component first. The final domain will be the intersection of all these individual domains. Remember that \( |x| \ge 0 \) is a fundamental property when solving absolute value inequalities.

Question 7. For what value of x, the inequality \( \frac{\pi}{2} < \cos^{-1} (3x - 1) < \pi \) holds?
Answer: We are given the inequality \( \frac{\pi}{2} < \cos^{-1} (3x - 1) < \pi \).
To solve this, we apply the cosine function to all parts of the inequality. Since the cosine function is a decreasing function on the interval \( [0, \pi] \) (which is the principal value range for \( \cos^{-1} \)), we must reverse the inequality signs when applying cosine.
So, we get:
\( \cos(\frac{\pi}{2}) > 3x - 1 > \cos(\pi) \)
Substitute the known values for \( \cos(\frac{\pi}{2}) \) and \( \cos(\pi) \):
\( 0 > 3x - 1 > -1 \)
This can be broken down into two separate inequalities:
1) \( 3x - 1 < 0 \)
\( 3x < 1 \)
\( x < \frac{1}{3} \)
2) \( 3x - 1 > -1 \)
\( 3x > 0 \)
\( x > 0 \)
Combining these two conditions, the inequality holds for \( 0 < x < \frac{1}{3} \). These values also ensure that \( 3x-1 \) is within the domain \( [-1, 1] \) for \( \cos^{-1} \).
In simple words: To solve this, we use the cosine function on all parts. Because cosine decreases between 0 and 180 degrees, we flip the greater/less than signs. Then we solve for x, making sure x stays between 0 and 1/3.

🎯 Exam Tip: Always remember to reverse the inequality signs when applying a decreasing function (like cosine on \( [0, \pi] \) or \( \sin^{-1} \) on \( [-1, 1] \)) to an inequality. Also, ensure the argument of the inverse trigonometric function remains within its valid domain.

Question 8. Find the value of
(i) \( \cos[\cos^{-1}(\frac{4}{5}) + \sin^{-1}(\frac{4}{5})] \)
(ii) \( \cos^{-1}(\cos \frac{4\pi}{3}) + \cos^{-1}(\cos \frac{5\pi}{4}) \)
Answer:
(i) To find the value of \( \cos[\cos^{-1}(\frac{4}{5}) + \sin^{-1}(\frac{4}{5})] \):
We use the important identity: \( \cos^{-1}y + \sin^{-1}y = \frac{\pi}{2} \), which holds for any \( y \in [-1, 1] \).
Since \( \frac{4}{5} \) is within the interval \( [-1, 1] \), we can apply this identity.
So, \( \cos^{-1}(\frac{4}{5}) + \sin^{-1}(\frac{4}{5}) = \frac{\pi}{2} \).
Substituting this back into the original expression:
\( \cos(\frac{\pi}{2}) \)
We know that \( \cos(\frac{\pi}{2}) = 0 \). This identity significantly simplifies complex inverse trigonometric expressions.

(ii) To find the value of \( \cos^{-1}(\cos \frac{4\pi}{3}) + \cos^{-1}(\cos \frac{5\pi}{4}) \):
For the first term, \( \cos^{-1}(\cos \frac{4\pi}{3}) \):
The angle \( \frac{4\pi}{3} \) is outside the principal value branch \( [0, \pi] \) for \( \cos^{-1} \).
We can find an equivalent value within the range by using trigonometric properties: \( \cos(\frac{4\pi}{3}) = \cos(\pi + \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2} \).
So, \( \cos^{-1}(\cos \frac{4\pi}{3}) = \cos^{-1}(-\frac{1}{2}) \).
The angle \( \theta \) in \( [0, \pi] \) such that \( \cos \theta = -\frac{1}{2} \) is \( \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).

For the second term, \( \cos^{-1}(\cos \frac{5\pi}{4}) \):
The angle \( \frac{5\pi}{4} \) is also outside the principal value branch \( [0, \pi] \).
Using trigonometric properties: \( \cos(\frac{5\pi}{4}) = \cos(\pi + \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \).
So, \( \cos^{-1}(\cos \frac{5\pi}{4}) = \cos^{-1}(-\frac{\sqrt{2}}{2}) \).
The angle \( \phi \) in \( [0, \pi] \) such that \( \cos \phi = -\frac{\sqrt{2}}{2} \) is \( \phi = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
Now, add the two results:
\( \frac{2\pi}{3} + \frac{3\pi}{4} \)
To add these, find a common denominator, which is 12:
\( \frac{8\pi}{12} + \frac{9\pi}{12} = \frac{8\pi+9\pi}{12} = \frac{17\pi}{12} \). This process shows how to adjust angles to fit the principal value range before applying inverse functions.
In simple words: For part (i), we use a special rule that says \( \cos^{-1}y \) plus \( \sin^{-1}y \) always equals \( \frac{\pi}{2} \), and the cosine of \( \frac{\pi}{2} \) is 0. For part (ii), we first adjust the angles \( \frac{4\pi}{3} \) and \( \frac{5\pi}{4} \) to find their equivalent values in the allowed range for \( \cos^{-1} \), then we add those results.

🎯 Exam Tip: Always check if the angle inside \( \cos^{-1}(\cos\theta) \) is within the principal value range \( [0, \pi] \). If not, use trigonometric identities (like \( \cos(\pi+\theta) = -\cos\theta \) or \( \cos(2\pi-\theta) = \cos\theta \)) to find an equivalent angle within the range before simplifying.