RBSE Solutions Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.2

Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2

Question 1. cos x = \( -\frac{1}{2} \), x lies in third quadrant.
Answer: Given that \( \cos x = -\frac{1}{2} \) and angle \( x \) is in the third quadrant.
We know the identity: \( \sin^2 x + \cos^2 x = 1 \)
So, \( \sin^2 x = 1 - \cos^2 x \)
\( \sin^2 x = 1 - \left( -\frac{1}{2} \right)^2 \)
\( \sin^2 x = 1 - \frac{1}{4} \)
\( \sin^2 x = \frac{4-1}{4} \)
\( \sin^2 x = \frac{3}{4} \)
\( \sin x = \pm \sqrt{\frac{3}{4}} \)
\( \sin x = \pm \frac{\sqrt{3}}{2} \)
Since \( x \) lies in the third quadrant (where \( 180^\circ \le x \le 270^\circ \) or \( \pi \le x \le \frac{3}{2}\pi \)), the sine function is negative.
Therefore, \( \sin x = -\frac{\sqrt{3}}{2} \)
Next, we find the reciprocal of \( \cos x \):
\( \sec x = \frac{1}{\cos x} = \frac{1}{-\frac{1}{2}} = -2 \)
Now, let's find the tangent of \( x \):
\( \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3} \)
The cotangent of \( x \) is the reciprocal of \( \tan x \):
\( \cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} \)
Finally, the cosecant of \( x \) is the reciprocal of \( \sin x \):
\( \csc x = \frac{1}{\sin x} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} \)
In simple words: First, we use the given cosine value and the fact that the angle is in the third quadrant to find the sine value. Then, we use these to calculate the other four trigonometric ratios (secant, tangent, cotangent, and cosecant). Remember, in the third quadrant, only tangent and cotangent are positive.

🎯 Exam Tip: Always remember the sign conventions for trigonometric functions in different quadrants to avoid common errors when determining positive or negative values.

Question 2. cot x = \( \frac{3}{4} \), x lies in third quadrant.
Answer: We are given that \( \cot x = \frac{3}{4} \) and angle \( x \) is in the third quadrant.
From this, we can find the tangent of \( x \):
\( \tan x = \frac{1}{\cot x} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \)
Now, let's find the cosecant using the identity: \( \csc^2 x = 1 + \cot^2 x \)
\( \csc^2 x = 1 + \left( \frac{3}{4} \right)^2 \)
\( \csc^2 x = 1 + \frac{9}{16} \)
\( \csc^2 x = \frac{16+9}{16} \)
\( \csc^2 x = \frac{25}{16} \)
\( \csc x = \pm \sqrt{\frac{25}{16}} \)
\( \csc x = \pm \frac{5}{4} \)
Since \( x \) is in the third quadrant, the cosecant function is negative.
Therefore, \( \csc x = -\frac{5}{4} \)
From \( \csc x \), we can find \( \sin x \):
\( \sin x = \frac{1}{\csc x} = \frac{1}{-\frac{5}{4}} = -\frac{4}{5} \)
Next, we use \( \cot x \) to find \( \cos x \):
\( \cot x = \frac{\cos x}{\sin x} \)
\( \frac{3}{4} = \frac{\cos x}{-\frac{4}{5}} \)
\( \cos x = \frac{3}{4} \times \left( -\frac{4}{5} \right) \)
\( \cos x = -\frac{3}{5} \)
Finally, we find the secant of \( x \):
\( \sec x = \frac{1}{\cos x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3} \)
In simple words: We are given the cotangent value and the quadrant. We first find the tangent. Then, using identities, we calculate cosecant, sine, cosine, and secant. Remember that in the third quadrant, only tangent and cotangent values are positive.

🎯 Exam Tip: When given one trigonometric ratio and the quadrant, always determine the signs of other ratios correctly based on the ASTC (All, Sine, Tan, Cos) rule for quadrants.

Question 3. sec x = \( \frac{13}{5} \), x lies in fourth quadrant.
Answer: We are given that \( \sec x = \frac{13}{5} \) and angle \( x \) is in the fourth quadrant.
From \( \sec x \), we can find \( \cos x \):
\( \cos x = \frac{1}{\sec x} = \frac{1}{\frac{13}{5}} = \frac{5}{13} \)
Next, we use the identity: \( \sin^2 x + \cos^2 x = 1 \)
\( \sin^2 x = 1 - \cos^2 x \)
\( \sin^2 x = 1 - \left( \frac{5}{13} \right)^2 \)
\( \sin^2 x = 1 - \frac{25}{169} \)
\( \sin^2 x = \frac{169-25}{169} \)
\( \sin^2 x = \frac{144}{169} \)
\( \sin x = \pm \sqrt{\frac{144}{169}} \)
\( \sin x = \pm \frac{12}{13} \)
Since \( x \) lies in the fourth quadrant, the sine function is negative.
Therefore, \( \sin x = -\frac{12}{13} \)
Now, let's find the cosecant of \( x \):
\( \csc x = \frac{1}{\sin x} = \frac{1}{-\frac{12}{13}} = -\frac{13}{12} \)
Next, we find the tangent of \( x \):
\( \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{12}{13}}{\frac{5}{13}} = -\frac{12}{5} \)
Finally, the cotangent of \( x \) is:
\( \cot x = \frac{1}{\tan x} = \frac{1}{-\frac{12}{5}} = -\frac{5}{12} \)
In simple words: Given secant and the quadrant, we first find cosine. Then, using identities, we calculate sine, cosecant, tangent, and cotangent. In the fourth quadrant, only cosine and secant are positive.

🎯 Exam Tip: Always remember that in the fourth quadrant, the cosine and secant values are positive, while sine, cosecant, tangent, and cotangent values are negative.

Question 4. Find the value of sin 765°.
Answer: We know that the values of \( \sin x \) repeat after an interval of \( 2\pi \) or \( 360^\circ \).
So, to find \( \sin 765^\circ \), we can write it as:
\( \sin (765^\circ) = \sin (720^\circ + 45^\circ) \)
Since \( 720^\circ = 2 \times 360^\circ \), this is equivalent to:
\( \sin (2 \times 360^\circ + 45^\circ) \)
Because of the periodic nature of sine, this simplifies to:
\( \sin 45^\circ \)
The value of \( \sin 45^\circ \) is known to be:
\( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
Thus, \( \sin 765^\circ = \frac{1}{\sqrt{2}} \). This means that after two full rotations, the angle lands in the first quadrant, where all trigonometric ratios are positive.
In simple words: To find the sine of a large angle like 765 degrees, we can subtract full circles (360 degrees) until the angle is between 0 and 360 degrees. Since 765 is 720 (two full circles) plus 45, the sine of 765 degrees is the same as the sine of 45 degrees, which is 1 over root 2.

🎯 Exam Tip: For angles greater than \( 360^\circ \), reduce them by subtracting multiples of \( 360^\circ \) (or \( 2\pi \) for radians) until the angle is within the \( 0^\circ \) to \( 360^\circ \) range.

Question 5. Find the value of tan \( \frac{19\pi}{3} \).
Answer: We need to find the value of \( \tan \frac{19\pi}{3} \).
First, we can rewrite the angle:
\( \frac{19\pi}{3} = \frac{18\pi + \pi}{3} = \frac{18\pi}{3} + \frac{\pi}{3} = 6\pi + \frac{\pi}{3} \)
Since the tangent function has a period of \( \pi \) (and \( 6\pi \) is a multiple of \( \pi \)), we can simplify:
\( \tan \left( 6\pi + \frac{\pi}{3} \right) = \tan \left( 3 \times 2\pi + \frac{\pi}{3} \right) \)
As \( 2\pi \) represents a full rotation, adding \( 6\pi \) (three full rotations) doesn't change the value of the tangent function.
So, \( \tan \left( 6\pi + \frac{\pi}{3} \right) = \tan \frac{\pi}{3} \)
The angle \( \frac{\pi}{3} \) is in the first quadrant, where all trigonometric ratios are positive.
The value of \( \tan \frac{\pi}{3} \) is:
\( \tan \frac{\pi}{3} = \sqrt{3} \)
Therefore, \( \tan \frac{19\pi}{3} = \sqrt{3} \).
In simple words: To find the tangent of \( \frac{19\pi}{3} \), we can first simplify the angle by removing full rotations of \( 2\pi \). Since \( \frac{19\pi}{3} \) is \( 6\pi + \frac{\pi}{3} \), and \( 6\pi \) is three full rotations, the tangent value is the same as \( \tan \frac{\pi}{3} \), which is \( \sqrt{3} \).

🎯 Exam Tip: For tangent functions, you can also simplify by subtracting multiples of \( \pi \) (not just \( 2\pi \)) from the angle, as tangent has a period of \( \pi \).

Question 6. Find the value of sin \( \left(-\frac{11\pi}{3}\right) \).
Answer: We need to find the value of \( \sin \left(-\frac{11\pi}{3}\right) \).
First, we use the property \( \sin(-\theta) = -\sin(\theta) \):
\( \sin \left(-\frac{11\pi}{3}\right) = -\sin \left(\frac{11\pi}{3}\right) \)
Now, let's simplify the angle \( \frac{11\pi}{3} \):
\( \frac{11\pi}{3} = \frac{12\pi - \pi}{3} = 4\pi - \frac{\pi}{3} \)
So, we have:
\( -\sin \left(4\pi - \frac{\pi}{3}\right) \)
Since \( 4\pi \) represents two full rotations, it can be ignored for the sine function:
\( -\sin \left(-\frac{\pi}{3}\right) \)
Applying \( \sin(-\theta) = -\sin(\theta) \) again:
\( - \left( -\sin \frac{\pi}{3} \right) = \sin \frac{\pi}{3} \)
The value of \( \sin \frac{\pi}{3} \) is:
\( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \)
Thus, \( \sin \left(-\frac{11\pi}{3}\right) = \frac{\sqrt{3}}{2} \). The angle \( 4\pi - \frac{\pi}{3} \) is effectively \( -\frac{\pi}{3} \) or \( \frac{5\pi}{3} \), which lies in the fourth quadrant where sine is negative, so \( -\sin(4\pi - \frac{\pi}{3}) = -(-\sin \frac{\pi}{3}) = \sin \frac{\pi}{3} \).
In simple words: To find the sine of a negative angle like \( -\frac{11\pi}{3} \), we first use the rule that \( \sin(-\theta) = -\sin(\theta) \). Then we simplify the angle \( \frac{11\pi}{3} \) by taking out full rotations (\( 4\pi \)). This leaves us with \( -\sin(-\frac{\pi}{3}) \), which simplifies to \( \sin \frac{\pi}{3} \), and its value is \( \frac{\sqrt{3}}{2} \).

🎯 Exam Tip: When dealing with negative angles or angles outside the primary range, first use trigonometric identities like \( \sin(-\theta) = -\sin(\theta) \) and then reduce the angle by adding or subtracting multiples of \( 2\pi \) (or \( 360^\circ \)).

Question 7. Find the value of cot \( \left(-\frac{15\pi}{4}\right) \).
Answer: We need to find the value of \( \cot \left(-\frac{15\pi}{4}\right) \).
First, we use the property \( \cot(-\theta) = -\cot(\theta) \):
\( \cot \left(-\frac{15\pi}{4}\right) = -\cot \left(\frac{15\pi}{4}\right) \)
Now, let's simplify the angle \( \frac{15\pi}{4} \):
\( \frac{15\pi}{4} = \frac{16\pi - \pi}{4} = 4\pi - \frac{\pi}{4} \)
So, we have:
\( -\cot \left(4\pi - \frac{\pi}{4}\right) \)
Since \( 4\pi \) represents two full rotations, it can be ignored for the cotangent function:
\( -\cot \left(-\frac{\pi}{4}\right) \)
Applying \( \cot(-\theta) = -\cot(\theta) \) again:
\( - \left( -\cot \frac{\pi}{4} \right) = \cot \frac{\pi}{4} \)
The value of \( \cot \frac{\pi}{4} \) is:
\( \cot \frac{\pi}{4} = 1 \)
Therefore, \( \cot \left(-\frac{15\pi}{4}\right) = 1 \). The angle \( 4\pi - \frac{\pi}{4} \) is equivalent to \( -\frac{\pi}{4} \) or \( \frac{7\pi}{4} \), which falls in the fourth quadrant. In the fourth quadrant, cotangent is negative. So, \( -\cot(4\pi - \frac{\pi}{4}) = -(-\cot \frac{\pi}{4}) = \cot \frac{\pi}{4} \).
In simple words: To find the cotangent of \( -\frac{15\pi}{4} \), first use the rule \( \cot(-\theta) = -\cot(\theta) \). Then, simplify the angle \( \frac{15\pi}{4} \) by removing full rotations of \( 2\pi \) (which is \( 4\pi \)). This leads to \( -\cot(-\frac{\pi}{4}) \), which simplifies to \( \cot \frac{\pi}{4} \), and its value is 1.

🎯 Exam Tip: Be careful with the signs of trigonometric functions when working with negative angles or angles in different quadrants. Use the appropriate identity for negative angles and then simplify using the periodicity of the function.