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Chapter 2 · Inverse Trigonometric Functions

Example 6: The value of sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\frac{3\pi}{5}\right)

Examplessin⁻¹(sin x) outside the principal range · Miscellaneous Examples
The question

Find the value of sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right)⁠.

Exam answer

Here 3π5∉[−π2,π2]\frac{3\pi}{5} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. Using sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta⁠,

sin⁡−1(sin⁡3π5)=sin⁡−1(sin⁡(π−3π5))=sin⁡−1(sin⁡2π5)\begin{aligned} &\sin^{-1}\left(\sin\frac{3\pi}{5}\right) \\ &= \sin^{-1}\left(\sin\left(\pi - \frac{3\pi}{5}\right)\right) \\ &= \sin^{-1}\left(\sin\frac{2\pi}{5}\right) \end{aligned}

As 2π5∈[−π2,π2]\frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, the principal range of sin⁡−1\sin^{-1}⁠,

∴ sin⁡−1(sin⁡3π5)=2π5\therefore\ \sin^{-1}\left(\sin\dfrac{3\pi}{5}\right) = \dfrac{2\pi}{5}⁠.

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

The one idea this problem is built on

sin⁡−1\sin^{-1} can only ever output an angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. So sin⁡−1(sin⁡x)\sin^{-1}(\sin x) is not "xx" in general: it is the angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] that has the same sine as xx. Since 3π5\frac{3\pi}{5} is bigger than π2\frac{\pi}{2}⁠, the answer cannot be 3π5\frac{3\pi}{5}⁠. Its mirror image in the yy-axis, π−3π5=2π5\pi - \frac{3\pi}{5} = \frac{2\pi}{5}⁠, has the same sine and does lie in the range.

Given, to find, and the plan
Given
The angle 3π5\frac{3\pi}{5} (that is, 108∘108^\circ⁠).
To find
sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\frac{3\pi}{5}\right)⁠: the unique angle yy in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with sin⁡y=sin⁡3π5\sin y = \sin\frac{3\pi}{5}⁠.
Plan
(1) See why the tempting answer 3π5\frac{3\pi}{5} is impossible. (2) Understand why sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x fails outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. (3) Use sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta) = \sin\theta to find an angle in the range with the same sine. (4) Conclude, and state the general rule.

Basics you need

1The principal range of sin⁡−1\sin^{-1}

sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1} : [-1,1]\to\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. Every output of sin⁡−1\sin^{-1} lies between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}⁠, whatever the input. Lesson: principal values

2Sine is not one-one on R\R

Many angles share a sine: sin⁡π6=sin⁡5π6=sin⁡13π6=12\sin\frac{\pi}{6} = \sin\frac{5\pi}{6} = \sin\frac{13\pi}{6} = \frac12⁠. That is why the domain of sine must be restricted before it can be inverted. Basics: what an inverse needs

3Supplementary angles have the same sine

On the unit circle, π−θ\pi-\theta is the mirror image of θ\theta in the yy-axis: same height. So sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta) = \sin\theta⁠. Basics: reference angles

4Comparing angles that are fractions of π\pi

abπ\frac{a}{b}\pi is compared with π2\frac{\pi}{2} by comparing ab\frac{a}{b} with 12\frac12⁠. In degrees: 3π5=108∘\frac{3\pi}{5} = 108^\circ and 2π5=72∘\frac{2\pi}{5} = 72^\circ⁠. Basics: angles and radians

Solution

  1. Why the tempting answer 3π5\frac{3\pi}{5} is impossible

    The textbook's solution first applies the rule sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x to get 3π5\frac{3\pi}{5}⁠, and then rejects that value. Here is why it must be rejected:

    35>12⟹3π5>π2.\frac{3}{5} > \frac12 \quad\Longrightarrow\quad \frac{3\pi}{5} > \frac{\pi}{2}.

    So 3π5∉[−π2,π2]\frac{3\pi}{5}\notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. But every value of sin⁡−1\sin^{-1} lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (Basic 1). An output of sin⁡−1\sin^{-1} can never equal 3π5\frac{3\pi}{5}⁠.

    The rule sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x is true, but only for x∈[−π2,π2]x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (Section 2.3 states it with that condition). Using it without checking the condition is the trap this example is designed to expose.
    35=0.6>0.5=12\frac35 = 0.6 > 0.5 = \frac12⁠; in degrees, 108∘>90∘108^\circ > 90^\circ⁠.

  2. Why sin⁡−1(sin⁡x)≠x\sin^{-1}(\sin x) \ne x outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

    Follow what the two functions do. First sin⁡\sin turns the angle xx into a number, its sine. Then sin⁡−1\sin^{-1} is asked: "which angle has this sine?" Because sine is not one-one (Basic 2), many angles qualify, and sin⁡−1\sin^{-1} is defined to answer with the one in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. So

    sin⁡−1(sin⁡x)=the angle in [−π2,π2]with the same sine as x.\begin{gathered} \sin^{-1}(\sin x) = \text{the angle in } \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \\ \text{with the same sine as } x. \end{gathered}

    If xx is already in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, that angle is xx itself. If not, it is some other angle, and the information about which of the many angles we started from has been lost in the first step.

    π/2 −π/2 −π π/2 π 3π/2 2π (3π/5, 2π/5) dashed: y = x y = π − x here
    Equal scales. The graph of y=sin⁡−1(sin⁡x)y = \sin^{-1}(\sin x) (blue) agrees with y=xy = x (gold) only for −π2≤x≤π2-\frac{\pi}{2}\le x\le\frac{\pi}{2}⁠. Then it folds back as y=π−xy = \pi - x⁠, so it never leaves [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. At x=3π5x = \frac{3\pi}{5} its height is 2π5\frac{2\pi}{5}⁠.
    This is the same reason a square root cannot undo squaring for negative numbers: (−3)2=3\sqrt{(-3)^2} = 3⁠, not −3-3⁠, because squaring forgot the sign. Sine "forgets" which of its many angles produced a given value, and sin⁡−1\sin^{-1} can only return the principal one.
    A function undoes another only on a set where the first is one-one; sine is one-one on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. Lesson: why the domain must be restricted

  3. Find the angle in the range with the same sine

    Use sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta) = \sin\theta (Basic 3) with θ=3π5\theta = \frac{3\pi}{5}⁠:

    sin⁡3π5=sin⁡(π−3π5)=sin⁡2π5.\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}.

    And 2π5\frac{2\pi}{5} is in the principal range, because 0<25<120 < \frac25 < \frac12 gives

    −π2<0<2π5<π2.-\frac{\pi}{2} < 0 < \frac{2\pi}{5} < \frac{\pi}{2}.
    x y O P Q 2π/5 π/2 −π/2 thick blue arc: [−π/2, π/2], the range of sin⁻¹ Q: angle 3π/5 (108°), on the left half P: angle 2π/5 (72°), mirror image of Q in the y-axis gold line: P and Q have the same height, sin(3π/5) = sin(2π/5)
    QQ at angle 3π5\frac{3\pi}{5} and PP at angle 2π5\frac{2\pi}{5} are mirror images in the yy-axis, so they have the same height (sine). Only PP is on the thick right half, from which sin⁡−1\sin^{-1} answers.
    We want an angle that keeps the sine but moves into [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. Reflecting in the yy-axis does exactly that for an angle in the second quadrant: it keeps the height and moves the point to the first quadrant. Other identities, such as sin⁡(θ−2π)=sin⁡θ\sin(\theta - 2\pi) = \sin\theta⁠, keep the sine too but give −7π5-\frac{7\pi}{5}⁠, still outside the range.
    π−3π5=5π5−3π5=2π5\pi - \frac{3\pi}{5} = \frac{5\pi}{5} - \frac{3\pi}{5} = \frac{2\pi}{5}⁠.

  4. Conclude

    By Step 3,

    sin⁡−1(sin⁡3π5)=sin⁡−1(sin⁡2π5)=2π5,\begin{aligned} \sin^{-1}\left(\sin\frac{3\pi}{5}\right) &= \sin^{-1}\left(\sin\frac{2\pi}{5}\right) \\ &= \frac{2\pi}{5}, \end{aligned}

    where the last equality is the rule sin⁡−1(sin⁡t)=t\sin^{-1}(\sin t) = t⁠, now used legitimately because t=2π5t = \frac{2\pi}{5} is in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. It is the only possible answer: sine is strictly increasing on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so no other angle there has this sine.

    The first equality only replaces the number sin⁡3π5\sin\frac{3\pi}{5} by the equal number sin⁡2π5\sin\frac{2\pi}{5}⁠; sin⁡−1\sin^{-1} of equal numbers is equal. The second equality is where the range condition is checked, and this time it is satisfied.
    sin⁡−1(sin⁡t)=t\sin^{-1}(\sin t) = t for t∈[−π2,π2]t\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (Section 2.3). Lesson: sin⁻¹(sin x) and the folding rules

  5. The general rule behind the answer

    The same reasoning works for every xx with π2≤x≤3π2\frac{\pi}{2}\le x\le\frac{3\pi}{2}⁠. Then π−x\pi - x lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and has the same sine as xx⁠, so

    sin⁡−1(sin⁡x)=π−xfor π2≤x≤3π2.\begin{gathered} \sin^{-1}(\sin x) = \pi - x \\ \text{for } \frac{\pi}{2}\le x\le\frac{3\pi}{2}. \end{gathered}

    With x=3π5x = \frac{3\pi}{5} this gives π−3π5=2π5\pi - \frac{3\pi}{5} = \frac{2\pi}{5} again. This is the downward-sloping piece of the zigzag in the first figure.

    Knowing the rule makes similar questions quick, but always confirm which piece of the zigzag your angle falls on. For example sin⁡−1(sin⁡2π3)=π−2π3=π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \pi - \frac{2\pi}{3} = \frac{\pi}{3} uses the same piece, which is Exercise 2.2, Q10.
    If π2≤x≤3π2\frac{\pi}{2}\le x\le\frac{3\pi}{2}⁠, subtracting from π\pi reverses the inequalities: −π2≤π−x≤π2-\frac{\pi}{2}\le \pi - x\le\frac{\pi}{2}⁠.
Result sin⁡−1(sin⁡3π5)=2π5.\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \frac{2\pi}{5}.

The value is not 3π5\frac{3\pi}{5}⁠, because 3π5\frac{3\pi}{5} is outside the principal range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. This agrees with the textbook.

Check it with numbers

3π5≈1.884956\frac{3\pi}{5}\approx1.884956 and sin⁡(1.884956)≈0.951057\sin(1.884956)\approx0.951057⁠. In radian mode, sin⁡−1(0.951057)≈1.256637\sin^{-1}(0.951057)\approx1.256637⁠, and 2π5≈1.256637\frac{2\pi}{5}\approx1.256637⁠. Equal. The tempting answer 1.8849561.884956 is larger than π2≈1.570796\frac{\pi}{2}\approx1.570796⁠, so a calculator could never return it from sin⁡−1\sin^{-1}⁠.

Common mistakes
  1. Cancelling sin⁡−1\sin^{-1} and sin⁡\sin on sight. sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x only when x∈[−π2,π2]x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. Check the angle first.
  2. Using π+x\pi + x or x−πx - \pi⁠. These change the sign of the sine: sin⁡(3π5−π)=−sin⁡3π5\sin\left(\frac{3\pi}{5} - \pi\right) = -\sin\frac{3\pi}{5}⁠. The angle with the same sine is π−x\pi - x⁠.
  3. Mixing up with sin⁡(sin⁡−1x)=x\sin\left(\sin^{-1}x\right) = x⁠. That one is true for every x∈[−1,1]x\in[-1,1]⁠, with no extra condition. Only the order "sin⁡−1\sin^{-1} of sin⁡\sin" needs the range check.
What to take away
  1. sin⁡−1(sin⁡x)\sin^{-1}(\sin x) means "the angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with the same sine as xx".
  2. For xx in [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]⁠, that angle is π−x\pi - x⁠. For other xx⁠, first add or subtract a multiple of 2π2\pi to bring xx into [−π2,3π2]\left[-\frac{\pi}{2}, \frac{3\pi}{2}\right]⁠.
  3. The same thinking handles cos⁡−1(cos⁡x)\cos^{-1}(\cos x) (range [0,π][0,\pi]⁠) and tan⁡−1(tan⁡x)\tan^{-1}(\tan x) (range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠).

Every fact used, and where it comes from

FactUsed inSource
sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1} : [-1,1]\to\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Steps 1, 2Section 2.2 and its table of principal value branches
sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for x∈[−π2,π2]x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Steps 1, 4Section 2.3 (Lesson: the folding rules)
Sine is one-one only on a restricted intervalStep 2Section 2.2 (Lesson: why the domain must be restricted)
sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta) = \sin\thetaSteps 3, 5Class XI allied angles
sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x) = \pi - x on [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]Step 5Derived in Step 5 (the textbook uses this idea in Example 6 without stating it)
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