GSEB Class 11 Maths Solutions Chapter 3 Trigonometric Functions Exercise 3.2

Find the values of the other five trigonometric functions in questions 1 to 5:

Question 1. cos x = \( -\frac{1}{2} \), x lies in third quadrant.
Answer: Given that \( \cos x = -\frac{1}{2} \) and x lies in the third quadrant.
In the third quadrant, the values of \( \sin x \), \( \cos x \), \( \sec x \), and \( \operatorname{cosec} x \) are negative, while \( \tan x \) and \( \cot x \) are positive.
Let \( OM = -1 \) and \( OP = 2 \).
Using the Pythagorean identity, \( MP = -\sqrt{OP^2 - OM^2} \) (since MP is negative in the third quadrant).
\( MP = -\sqrt{2^2 - (-1)^2} = -\sqrt{4 - 1} = -\sqrt{3} \)
Now, we can find the other trigonometric functions:
\( \sin x = \frac{MP}{OP} = \frac{-\sqrt{3}}{2} \)
\( \tan x = \frac{MP}{OM} = \frac{-\sqrt{3}}{-1} = \sqrt{3} \)
\( \cot x = \frac{OM}{MP} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} \)
\( \sec x = \frac{OP}{OM} = \frac{2}{-1} = -2 \)
\( \operatorname{cosec} x = \frac{OP}{MP} = \frac{2}{-\sqrt{3}} = -\frac{2}{\sqrt{3}} \)
In simple words: We are given the cosine value and the quadrant. In the third quadrant, sine and tangent are positive. We use a triangle to calculate the missing side length. Then, we find all the other trigonometric ratios using the sides, making sure to apply the correct positive or negative sign for each function in that specific quadrant.

Exam Tip: Remember to always check the quadrant of x to apply the correct positive or negative signs to the trigonometric ratios. For third quadrant, only tan and cot are positive.

Question 2. sin x = \( \frac{3}{5} \), x lies in second quadrant.
Answer: Given that \( \sin x = \frac{3}{5} \) and x lies in the second quadrant.
In the second quadrant, \( \sin x \) and \( \operatorname{cosec} x \) are positive, while \( \cos x \), \( \tan x \), \( \cot x \), and \( \sec x \) are negative.
Let \( MP = 3 \) and \( OP = 5 \).
Using the Pythagorean identity, \( OM = -\sqrt{OP^2 - MP^2} \) (since OM is negative in the second quadrant).
\( OM = -\sqrt{5^2 - 3^2} = -\sqrt{25 - 9} = -\sqrt{16} = -4 \)
Now, we can find the other trigonometric functions:
\( \cos x = \frac{OM}{OP} = \frac{-4}{5} \)
\( \tan x = \frac{MP}{OM} = \frac{3}{-4} = -\frac{3}{4} \)
\( \cot x = \frac{OM}{MP} = \frac{-4}{3} \)
\( \sec x = \frac{OP}{OM} = \frac{5}{-4} = -\frac{5}{4} \)
\( \operatorname{cosec} x = \frac{OP}{MP} = \frac{5}{3} \)
In simple words: We are given the sine value and the quadrant. In the second quadrant, sine is positive, and cosine and tangent are negative. We find the missing side length using the Pythagorean theorem. Then, we calculate the remaining trigonometric ratios, being careful to use the correct positive or negative signs based on the quadrant.

Exam Tip: For the second quadrant, only \( \sin \) and \( \operatorname{cosec} \) values are positive. All other trigonometric ratios are negative.

Question 3. cot x = \( \frac{3}{4} \), x lies in third quadrant.
Answer: Given that \( \cot x = \frac{3}{4} \) and x lies in the third quadrant.
In the third quadrant, \( \tan x \) and \( \cot x \) are positive, while \( \sin x \), \( \cos x \), \( \sec x \), and \( \operatorname{cosec} x \) are negative.
Since \( \cot x = \frac{OM}{MP} = \frac{3}{4} \) and x is in the third quadrant, both OM and MP must be negative.
So, let \( OM = -3 \) and \( MP = -4 \).
Using the Pythagorean identity, \( OP = \sqrt{OM^2 + MP^2} \).
\( OP = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
Now, we can find the other trigonometric functions:
\( \sin x = \frac{MP}{OP} = \frac{-4}{5} \)
\( \cos x = \frac{OM}{OP} = \frac{-3}{5} \)
\( \tan x = \frac{MP}{OM} = \frac{-4}{-3} = \frac{4}{3} \)
\( \sec x = \frac{OP}{OM} = \frac{5}{-3} = -\frac{5}{3} \)
\( \operatorname{cosec} x = \frac{OP}{MP} = \frac{5}{-4} = -\frac{5}{4} \)
In simple words: We know the cotangent value and the quadrant. In the third quadrant, only cotangent and tangent are positive. Because it's the third quadrant, both the x and y coordinates (OM and MP) are negative. We use these negative lengths to find the hypotenuse, then calculate all the other trigonometric functions, making sure to apply the correct positive or negative signs.

Exam Tip: For problems involving cotangent or tangent in the third quadrant, ensure both OM and MP have negative signs since both x and y coordinates are negative in this quadrant.

Question 4. sec x = \( \frac{13}{5} \), x lies in fourth quadrant.
Answer: Given that \( \sec x = \frac{13}{5} \) and x lies in the fourth quadrant.
In the fourth quadrant, \( \cos x \) and \( \sec x \) are positive, while \( \sin x \), \( \tan x \), \( \cot x \), and \( \operatorname{cosec} x \) are negative.
Since \( \sec x = \frac{OP}{OM} = \frac{13}{5} \), let \( OP = 13 \) and \( OM = 5 \).
Using the Pythagorean identity, \( MP = -\sqrt{OP^2 - OM^2} \) (since MP is negative in the fourth quadrant).
\( MP = -\sqrt{13^2 - 5^2} = -\sqrt{169 - 25} = -\sqrt{144} = -12 \)
Now, we can find the other trigonometric functions:
\( \sin x = \frac{MP}{OP} = \frac{-12}{13} \)
\( \cos x = \frac{OM}{OP} = \frac{5}{13} \)
\( \tan x = \frac{MP}{OM} = \frac{-12}{5} \)
\( \cot x = \frac{OM}{MP} = \frac{5}{-12} = -\frac{5}{12} \)
\( \operatorname{cosec} x = \frac{OP}{MP} = \frac{13}{-12} = -\frac{13}{12} \)
In simple words: We are given the secant value and the quadrant. In the fourth quadrant, secant is positive, and sine and tangent are negative. We determine the side lengths based on the secant ratio and use the Pythagorean theorem to find the missing side. Then, we calculate the remaining trigonometric ratios, applying the correct positive or negative signs for the fourth quadrant.

Exam Tip: When \( \sec x \) is given, remember it is the reciprocal of \( \cos x \). In the fourth quadrant, x-coordinates are positive and y-coordinates are negative.

Question 5. tan x = \( -\frac{5}{12} \), x lies in the second quadrant.
Answer: Given that \( \tan x = -\frac{5}{12} \) and x lies in the second quadrant.
In the second quadrant, \( \sin x \) and \( \operatorname{cosec} x \) are positive, while \( \cos x \), \( \sec x \), \( \tan x \), and \( \cot x \) are negative.
Since \( \tan x = \frac{MP}{OM} = -\frac{5}{12} \) and x is in the second quadrant, MP is positive and OM is negative.
So, let \( MP = 5 \) and \( OM = -12 \).
Using the Pythagorean identity, \( OP = \sqrt{MP^2 + OM^2} \).
\( OP = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)
Now, we can find the other trigonometric functions:
\( \sin x = \frac{MP}{OP} = \frac{5}{13} \)
\( \cos x = \frac{OM}{OP} = \frac{-12}{13} \)
\( \cot x = \frac{OM}{MP} = \frac{-12}{5} \)
\( \sec x = \frac{OP}{OM} = \frac{13}{-12} = -\frac{13}{12} \)
\( \operatorname{cosec} x = \frac{OP}{MP} = \frac{13}{5} \)
In simple words: We know the tangent value and the quadrant. In the second quadrant, tangent is negative, which means the y-coordinate (MP) is positive and the x-coordinate (OM) is negative. We find the hypotenuse using these values and then calculate the rest of the trigonometric ratios, making sure to apply the proper positive or negative signs for functions in the second quadrant.

Exam Tip: For problems involving tangent or cotangent in the second quadrant, remember that the x-coordinate (OM) is negative and the y-coordinate (MP) is positive.

Questions?

Find the values of the following trigonometric ratios:

Question 6. sin 765°
Answer: To find the value of \( \sin 765^\circ \), we can express \( 765^\circ \) in the form \( (n \times 360^\circ + \theta) \) or \( (n \times 90^\circ + \theta) \).
\( 765^\circ = 2 \times 360^\circ + 45^\circ \)
Since trigonometric functions repeat every \( 360^\circ \), \( \sin(2 \times 360^\circ + 45^\circ) = \sin 45^\circ \).
We know that \( \sin 45^\circ = \frac{1}{\sqrt{2}} \).
Therefore, \( \sin 765^\circ = \frac{1}{\sqrt{2}} \).
In simple words: To find the sine of 765 degrees, we can take away full circles of 360 degrees until we get an angle we know. Since 765 is 2 full circles plus 45 degrees, the sine of 765 degrees is the same as the sine of 45 degrees, which is 1 over root 2.

Exam Tip: To simplify angles larger than \( 360^\circ \), divide by \( 360^\circ \) and use the remainder. For angles up to \( 360^\circ \), use the identities involving \( 90^\circ \) or \( 180^\circ \).

Question 7. cosec (- 1410°)
Answer: To find the value of \( \operatorname{cosec} (-1410^\circ) \):
First, we use the identity \( \operatorname{cosec}(-\theta) = -\operatorname{cosec} \theta \).
So, \( \operatorname{cosec}(-1410^\circ) = -\operatorname{cosec}(1410^\circ) \).
Now, we simplify \( 1410^\circ \):
\( 1410^\circ = 3 \times 360^\circ + 330^\circ \)
So, \( \operatorname{cosec}(1410^\circ) = \operatorname{cosec}(330^\circ) \).
We can write \( 330^\circ \) as \( (360^\circ - 30^\circ) \).
\( \operatorname{cosec}(330^\circ) = \operatorname{cosec}(360^\circ - 30^\circ) \).
Since \( (360^\circ - \theta) \) is in the fourth quadrant, and cosecant is negative in the fourth quadrant:
\( \operatorname{cosec}(360^\circ - 30^\circ) = -\operatorname{cosec}(30^\circ) \).
We know that \( \operatorname{cosec}(30^\circ) = 2 \).
So, \( -\operatorname{cosec}(30^\circ) = -2 \).
Therefore, \( \operatorname{cosec}(-1410^\circ) = -(-2) = 2 \).
In simple words: First, negative cosecant means cosecant of the positive angle, but with a minus sign in front. Then, for the large positive angle (1410 degrees), we subtract full circles of 360 degrees until we get a smaller angle (330 degrees). Since 330 degrees is like 30 degrees short of a full circle, and cosecant is negative there, we find cosecant of 30 and make it negative. Finally, we apply the initial minus sign to get the answer.

Exam Tip: When dealing with negative angles, always apply the negative angle identity first. Then, simplify large angles by finding their co-terminal angle within \( 0^\circ \) to \( 360^\circ \).

Question 8. tan \( \frac{19\pi}{3} \)
Answer: To find the value of \( \tan \left(\frac{19\pi}{3}\right) \):
We can express \( \frac{19\pi}{3} \) in the form \( (2n\pi + \theta) \).
\( \frac{19\pi}{3} = \frac{18\pi + \pi}{3} = \frac{18\pi}{3} + \frac{\pi}{3} = 6\pi + \frac{\pi}{3} \)
Since tangent has a period of \( \pi \), and \( 6\pi \) is an even multiple of \( \pi \), \( \tan(6\pi + \frac{\pi}{3}) = \tan\left(\frac{\pi}{3}\right) \).
We know that \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \).
Therefore, \( \tan\left(\frac{19\pi}{3}\right) = \sqrt{3} \).
In simple words: To calculate the tangent of \( \frac{19\pi}{3} \), we separate the angle into full circles (multiples of \( 2\pi \)) and a remaining part. Since \( \frac{19\pi}{3} \) is \( 6\pi + \frac{\pi}{3} \), and tangent repeats every \( \pi \), it's the same as the tangent of \( \frac{\pi}{3} \), which is root 3.

Exam Tip: For angles in radians, convert to the form \( (2n\pi + \theta) \) or \( (n\pi + \theta) \) depending on the period of the trigonometric function. Tangent and cotangent have a period of \( \pi \).

Question 9. sin \( \left(\frac{-11\pi}{3}\right) \)
Answer: To find the value of \( \sin \left(-\frac{11\pi}{3}\right) \):
First, we use the identity \( \sin(-\theta) = -\sin \theta \).
So, \( \sin \left(-\frac{11\pi}{3}\right) = -\sin \left(\frac{11\pi}{3}\right) \).
Now, we simplify \( \frac{11\pi}{3} \):
\( \frac{11\pi}{3} = \frac{12\pi - \pi}{3} = 4\pi - \frac{\pi}{3} \)
So, \( \sin \left(\frac{11\pi}{3}\right) = \sin \left(4\pi - \frac{\pi}{3}\right) \).
Since sine has a period of \( 2\pi \), \( \sin(4\pi - \frac{\pi}{3}) = \sin(-\frac{\pi}{3}) \).
Again using \( \sin(-\theta) = -\sin \theta \):
\( \sin(-\frac{\pi}{3}) = -\sin\left(\frac{\pi}{3}\right) \).
We know that \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).
So, \( -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \).
Therefore, \( \sin \left(-\frac{11\pi}{3}\right) = - \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} \).
In simple words: First, for sine of a negative angle, we take the sine of the positive angle and put a minus sign in front. Then, we simplify the large angle \( \frac{11\pi}{3} \) by removing full cycles of \( 2\pi \). This simplifies to \( -\frac{\pi}{3} \). Applying the negative angle rule again gives \( -\left(-\sin \frac{\pi}{3}\right) \), which simplifies to \( \sin \frac{\pi}{3} \), which is root 3 over 2.

Exam Tip: For negative angles like \( \sin(-\theta) \), first convert it to \( -\sin(\theta) \). Then, simplify the positive angle by adding or subtracting multiples of \( 2\pi \) to get an angle in the standard range \( 0 \) to \( 2\pi \).

Question 10. cot \( \left(\frac{-15\pi}{4}\right) \)
Answer: To find the value of \( \cot \left(-\frac{15\pi}{4}\right) \):
First, we use the identity \( \cot(-\theta) = -\cot \theta \).
So, \( \cot \left(-\frac{15\pi}{4}\right) = -\cot \left(\frac{15\pi}{4}\right) \).
Now, we simplify \( \frac{15\pi}{4} \):
\( \frac{15\pi}{4} = \frac{16\pi - \pi}{4} = 4\pi - \frac{\pi}{4} \)
So, \( \cot \left(\frac{15\pi}{4}\right) = \cot \left(4\pi - \frac{\pi}{4}\right) \).
Since cotangent has a period of \( \pi \), \( \cot(4\pi - \frac{\pi}{4}) = \cot(-\frac{\pi}{4}) \).
Again using \( \cot(-\theta) = -\cot \theta \):
\( \cot(-\frac{\pi}{4}) = -\cot\left(\frac{\pi}{4}\right) \).
We know that \( \cot\left(\frac{\pi}{4}\right) = 1 \).
So, \( -\cot\left(\frac{\pi}{4}\right) = -1 \).
Therefore, \( \cot \left(-\frac{15\pi}{4}\right) = -(-1) = 1 \).
In simple words: For cotangent of a negative angle, we take the cotangent of the positive angle and add a minus sign. Then, we simplify the large angle \( \frac{15\pi}{4} \) by subtracting full cycles of \( 2\pi \) (or multiples of \( \pi \) for cotangent). This simplifies to \( -\frac{\pi}{4} \). Applying the negative angle rule for cotangent again, it becomes \( -\left(-\cot \frac{\pi}{4}\right) \), which simplifies to \( \cot \frac{\pi}{4} \), which is 1.

Exam Tip: Remember that \( \tan \) and \( \cot \) have a period of \( \pi \), not \( 2\pi \). This means \( \tan(\theta + n\pi) = \tan \theta \) for any integer \( n \).