Class 12 Maths, Worked in Full Past papersGet updates
Chapter 2 · Inverse Trigonometric Functions

Exercise 2.1 · Q10: Principal value of cosec⁡−1(−2)\operatorname{cosec}^{-1}\left(-\sqrt2\right)

Exercise 2.1Principal values · Section 2.2
The question

Find the principal value of cosec⁡−1(−2)\operatorname{cosec}^{-1}\left(-\sqrt2\right)⁠.

Exam answer

Let cosec⁡−1(−2)=y\operatorname{cosec}^{-1}\left(-\sqrt2\right) = y⁠. Then

cosec⁡y=−2=−cosec⁡π4=cosec⁡(−π4)\begin{aligned} \operatorname{cosec} y &= -\sqrt2 = -\operatorname{cosec}\frac{\pi}{4} \\ &= \operatorname{cosec}\left(-\frac{\pi}{4}\right) \end{aligned}

Range of principal value branch of cosec⁡−1\operatorname{cosec}^{-1} is [−π2,π2]−{0}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}⁠, and −π4∈[−π2,π2]−{0}-\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}⁠.

∴\therefore Principal value of cosec⁡−1(−2)\operatorname{cosec}^{-1}\left(-\sqrt2\right) is −π4-\dfrac{\pi}{4}⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

cosec⁡y=−2\operatorname{cosec} y = -\sqrt2 is the same statement as sin⁡y=−12\sin y = -\frac{1}{\sqrt2}⁠, because cosec is 1sin⁡\frac{1}{\sin}⁠. So we need the one angle in the principal range [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \{0\} whose sine is −12-\frac{1}{\sqrt2}⁠. The reference angle is π4\frac{\pi}{4}⁠, and a negative value in this range means a negative angle: the answer is −π4-\frac{\pi}{4}⁠.

Given, to find, and the plan
Given
The number −2≈−1.414-\sqrt2 \approx -1.414⁠, which lies in the domain R−(−1,1)\R - (-1,1) of cosec⁡−1\operatorname{cosec}^{-1}⁠, since −2≤−1-\sqrt2 \le -1⁠.
To find
The unique yy in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, y≠0y\ne0⁠, with cosec⁡y=−2\operatorname{cosec} y = -\sqrt2⁠.
Plan
(1) State the principal range. (2) Rewrite the condition in terms of sine. (3) Find the reference angle. (4) Use the sign to place the answer. (5) Check, and exclude the other angles with the same cosecant.

Basics you need

1Principal range of cosec⁡−1\operatorname{cosec}^{-1}

Domain R−(−1,1)\R - (-1,1)⁠, range [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \{0\}⁠. It is the range of sin⁡−1\sin^{-1} with 00 removed, because cosec⁡0=1sin⁡0\operatorname{cosec} 0 = \frac{1}{\sin 0} is undefined. Lesson: principal values

2cosec is 1/sin⁡1/\sin

cosec⁡y=1sin⁡y\operatorname{cosec} y = \dfrac{1}{\sin y}⁠, defined wherever sin⁡y≠0\sin y \ne 0⁠. So cosec⁡y=c\operatorname{cosec} y = c exactly when sin⁡y=1c\sin y = \frac1c (for c≠0c\ne0⁠). Basics: tan, cot, sec and cosec

3The special angle π4\frac{\pi}{4}

sin⁡π4=12\sin\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, so cosec⁡π4=2\operatorname{cosec}\frac{\pi}{4} = \sqrt2⁠. Basics: special angles and reference angles

4sin is odd

sin⁡(−α)=−sin⁡α\sin(-\alpha) = -\sin\alpha for every α\alpha⁠: reflecting a point in the xx-axis flips the sign of its height. Basics: sin and cos on the unit circle

Solution

  1. Name the unknown and its range

    Let y=cosec⁡−1(−2)y = \operatorname{cosec}^{-1}\left(-\sqrt2\right)⁠. By definition of the principal value branch,

    cosec⁡y=−2,−π2≤y≤π2,y≠0.\operatorname{cosec} y = -\sqrt2, \qquad -\frac{\pi}{2} \le y \le \frac{\pi}{2}, \qquad y \ne 0.

    The input is allowed, since ∣−2∣=2≥1|-\sqrt2| = \sqrt2 \ge 1⁠.

    The domain of cosec⁡−1\operatorname{cosec}^{-1} is R−(−1,1)\R - (-1,1) because ∣sin⁡y∣≤1|\sin y| \le 1 forces ∣cosec⁡y∣≥1|\operatorname{cosec} y| \ge 1⁠. The range is the sine branch [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] with the single point 00 removed, where cosec does not exist.
    Principal branch of cosec⁡−1\operatorname{cosec}^{-1}⁠: domain R−(−1,1)\R - (-1,1)⁠, range [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \{0\} (Section 2.2). Lesson: principal values

  2. Turn cosec into sine

    Since cosec⁡y=1sin⁡y\operatorname{cosec} y = \frac{1}{\sin y}⁠, the condition cosec⁡y=−2\operatorname{cosec} y = -\sqrt2 is equivalent to

    sin⁡y=1−2=−12.\sin y = \frac{1}{-\sqrt2} = -\frac{1}{\sqrt2}.
    Sine values are the ones we know by heart and can read off the unit circle (the height of the point). Taking reciprocals is reversible here, because neither side is zero, so no solution is gained or lost by the switch.
    cosec⁡y=c  ⟺  sin⁡y=1c\operatorname{cosec} y = c \iff \sin y = \frac1c when c≠0c \ne 0⁠. Basics: tan, cot, sec and cosec

  3. Find the reference angle

    Drop the sign: the acute angle with sin⁡α=12\sin\alpha = \frac{1}{\sqrt2} is

    α=π4,since sin⁡π4=12.\alpha = \frac{\pi}{4}, \qquad\text{since } \sin\frac{\pi}{4} = \frac{1}{\sqrt2}.
    Equivalently, cosec⁡π4=2\operatorname{cosec}\frac{\pi}{4} = \sqrt2⁠. The size of the value fixes the reference angle; the sign is dealt with next.
    sin⁡π6=12\sin\frac{\pi}{6} = \frac12⁠, sin⁡π4=12\sin\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, sin⁡π3=32\sin\frac{\pi}{3} = \frac{\sqrt3}{2}⁠. Basics: special angles and reference angles

  4. Use the sign to place the answer

    On the principal range, sin⁡y>0\sin y > 0 for 0<y≤π20 < y \le \frac{\pi}{2} and sin⁡y<0\sin y < 0 for −π2≤y<0-\frac{\pi}{2} \le y < 0⁠. Our sine is negative, so yy lies in [−π2,0)\left[-\frac{\pi}{2}, 0\right)⁠, the fourth-quadrant part. There the angle with reference angle α\alpha is −α-\alpha⁠, and

    sin⁡(−π4)=−sin⁡π4=−12.\sin\left(-\frac{\pi}{4}\right) = -\sin\frac{\pi}{4} = -\frac{1}{\sqrt2}.

    Hence

    y=−π4.y = -\frac{\pi}{4}.
    This is the sign rule for sin⁡−1\sin^{-1} and cosec⁡−1\operatorname{cosec}^{-1}⁠: the range is symmetric about 00⁠, so a negative input gives the negative of the reference angle. In short, cosec⁡−1(−x)=−cosec⁡−1x\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}x for every xx with ∣x∣≥1|x| \ge 1⁠. This is different from cos⁡−1\cos^{-1}⁠, where a negative input gives π−α\pi - \alpha⁠.
    sin⁡(−α)=−sin⁡α\sin(-\alpha) = -\sin\alpha (Basic 4). Basics: sin and cos on the unit circle

  5. Check it, and exclude every other candidate

    Check. −π2≤−π4≤π2-\frac{\pi}{2} \le -\frac{\pi}{4} \le \frac{\pi}{2} and −π4≠0-\frac{\pi}{4} \ne 0 ✓. Also cosec⁡(−π4)=1−1/2=−2\operatorname{cosec}\left(-\frac{\pi}{4}\right) = \frac{1}{-1/\sqrt2} = -\sqrt2 ✓.

    Other angles with the same cosecant. These are the angles with sin⁡y=−12\sin y = -\frac{1}{\sqrt2}⁠, namely

    y=−π4+2nπory=−3π4+2nπ,n∈Z.\begin{aligned} y &= -\frac{\pi}{4} + 2n\pi \quad\text{or}\\ y &= -\frac{3\pi}{4} + 2n\pi, \qquad n\in\Z. \end{aligned}

    (The second family comes from sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta with θ=−π4\theta = -\frac{\pi}{4}⁠, since π+π4\pi + \frac{\pi}{4} is the same point as −3π4-\frac{3\pi}{4}⁠.) The closest of the second family is −3π4<−π2-\frac{3\pi}{4} < -\frac{\pi}{2}⁠, outside the range; the closest others, 5π4\frac{5\pi}{4} and 7π4\frac{7\pi}{4}⁠, exceed π2\frac{\pi}{2}⁠, and every remaining solution is further away still. So −π4-\frac{\pi}{4} is the only candidate in the principal range.

    O −π/4 −3π/4 −π/2 π/2 0 y = −1/√2 Range of cosec⁻¹: [−π/2, π/2] without 0 Blue: y = −π/4, the answer. sin y = −1/√2, so cosec y = −√2. Rose: −3π/4 has the same sine, but it lies in the left half: not in range. sin y is the y-coordinate, so the point is on the gold line. Hollow 0: not in range.
    The principal range of cosec⁡−1\operatorname{cosec}^{-1} is the closed right half of the unit circle with the point at angle 00 removed (green, hollow dot). Points with sin⁡y=−12\sin y = -\frac{1}{\sqrt2} lie on the horizontal line y=−12y = -\frac{1}{\sqrt2} (gold), meeting the circle at −π4-\frac{\pi}{4} (blue, in range) and −3π4-\frac{3\pi}{4} (rose, not in range).
    Equal sines mean equal heights, so the candidates lie on one horizontal line. Such a line cuts the circle at a point and its mirror image in the yy-axis, one on the right half and one on the left half. The right half is the principal range, so exactly one survives.
    On [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \{0\}⁠, cosec is one-one, so it can be inverted there. Lesson: branches
Result cosec⁡−1(−2)=−π4.\operatorname{cosec}^{-1}\left(-\sqrt2\right) = -\frac{\pi}{4}.

(Textbook answer: −π4-\frac{\pi}{4}⁠.)

Check it with numbers

−π4≈−0.7853982-\frac{\pi}{4} \approx -0.7853982⁠, and sin⁡(−0.7853982)≈−0.7071068\sin(-0.7853982) \approx -0.7071068⁠, so the cosecant of −0.7853982-0.7853982 is ≈−1.4142136\approx -1.4142136⁠, while −2≈−1.4142136-\sqrt2 \approx -1.4142136 ✓. The rival −3π4≈−2.3561945-\frac{3\pi}{4} \approx -2.3561945 gives the same cosec, but it is below −π2≈−1.5707963-\frac{\pi}{2} \approx -1.5707963 ✓.

Common mistakes
  1. Answering 3π4\frac{3\pi}{4} by copying the cos⁡−1\cos^{-1} rule π−α\pi - \alpha⁠. Its sine is +12+\frac{1}{\sqrt2}⁠, so its cosecant is +2+\sqrt2⁠, the wrong sign; and 5π4\frac{5\pi}{4}⁠, which has the right cosecant, is outside the range. For cosec⁡−1\operatorname{cosec}^{-1} the range is [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \{0\}⁠, so a negative input gives a negative angle.
  2. Answering 7π4\frac{7\pi}{4}⁠. It has the right cosecant, but it is far outside the principal range.
  3. Thinking cosec⁡−1(−2)=1sin⁡−1(−2)\operatorname{cosec}^{-1}(-\sqrt2) = \frac{1}{\sin^{-1}(-\sqrt2)}⁠. The −1-1 is not a reciprocal, and sin⁡−1(−2)\sin^{-1}(-\sqrt2) does not even exist, since −2<−1-\sqrt2 < -1⁠. The useful link is cosec⁡−1x=sin⁡−11x\operatorname{cosec}^{-1}x = \sin^{-1}\frac1x for ∣x∣≥1|x| \ge 1⁠.
What to take away
  1. Convert cosec to sine (and sec to cos) by taking reciprocals, then use the familiar sine values.
  2. For sin⁡−1\sin^{-1} and cosec⁡−1\operatorname{cosec}^{-1}⁠: negative input gives −α-\alpha⁠. For cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} and cot⁡−1\cot^{-1}⁠: negative input gives π−α\pi - \alpha⁠.
  3. Always confirm the answer is in the range, including the gap at 00 for cosec⁡−1\operatorname{cosec}^{-1}⁠.

Every fact used, and where it comes from

FactUsed inSource
cosec⁡−1\operatorname{cosec}^{-1}⁠: domain R−(−1,1)\R - (-1,1)⁠, range [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right] - \{0\}Steps 1, 5Section 2.2 and its table of principal value branches (Lesson: principal values)
cosec⁡x=1sin⁡x\operatorname{cosec} x = \frac{1}{\sin x}Step 2Section 2.2 (definition of the cosec function)
sin⁡π4=12\sin\frac{\pi}{4} = \frac{1}{\sqrt2}Step 3Class XI trigonometry (special angles)
sin⁡(−α)=−sin⁡α\sin(-\alpha) = -\sin\alpha⁠; sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\thetaSteps 4, 5Class XI trigonometry (allied angles)
Principal value means the value in the principal rangeStep 1Note, points 2 and 3, Section 2.2
All numerical valuesCheckRecomputed in the site's verification script
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →