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Chapter 2 · Inverse Trigonometric Functions

Example 1: Principal value of sin⁡−1(12)\sin^{-1}\left(\frac{1}{\sqrt2}\right)

ExamplesPrincipal values · Section 2.2
The question

Find the principal value of sin⁡−1(12)\sin^{-1}\left(\dfrac{1}{\sqrt2}\right)⁠.

Exam answer

Let sin⁡−1(12)=y\sin^{-1}\left(\frac{1}{\sqrt2}\right) = y⁠. Then

sin⁡y=12=sin⁡π4\sin y = \frac{1}{\sqrt2} = \sin\frac{\pi}{4}

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

∴\therefore Principal value of sin⁡−1(12)\sin^{-1}\left(\dfrac{1}{\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

Infinitely many angles have sine equal to 12\frac{1}{\sqrt2}⁠: π4\frac{\pi}{4}⁠, 3π4\frac{3\pi}{4}⁠, 9π4\frac{9\pi}{4}⁠, and so on. The symbol sin⁡−1\sin^{-1} must give one answer, so it is defined to pick the one angle that lies in its principal range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. On that interval sine takes each value in [−1,1][-1,1] exactly once, so the question "which angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] has sine 12\frac{1}{\sqrt2}⁠?" has exactly one answer, π4\frac{\pi}{4}⁠.

Given, to find, and the plan
Given
The number 12\frac{1}{\sqrt2}⁠, which lies in [−1,1][-1,1]⁠, the domain of sin⁡−1\sin^{-1}⁠.
To find
The principal value: the unique yy in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with sin⁡y=12\sin y = \frac{1}{\sqrt2}⁠.
Plan
(1) Name the answer yy and turn the question into an equation sin⁡y=12\sin y = \frac{1}{\sqrt2} with a condition on yy⁠. (2) Recall a special angle whose sine is 12\frac{1}{\sqrt2}⁠. (3) Check it lies in the principal range. (4) Explain why no other angle in that range works.

Basics you need

1Sine on the unit circle

Turn through an angle θ\theta (in radians, anticlockwise from the positive xx-axis) on the circle of radius 1. The point you reach is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)⁠: sine is the height of that point. Basics: sin and cos on the unit circle

2Radians

A full turn is 2π2\pi radians, so π\pi radians is 180∘180^\circ⁠, π2\frac{\pi}{2} is 90∘90^\circ and π4\frac{\pi}{4} is 45∘45^\circ⁠. Basics: angles and radians

3The special angle π4\frac{\pi}{4}

In a right triangle with two equal legs of length 1, the hypotenuse is 12+12=2\sqrt{1^2+1^2}=\sqrt2 and both acute angles are 45∘45^\circ⁠. So sin⁡π4=oppositehypotenuse=12\sin\frac{\pi}{4} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt2}⁠. Basics: special angles

4The principal value branch of sin⁡−1\sin^{-1}

sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1} : [-1,1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. For xx in [−1,1][-1,1]⁠, sin⁡−1x\sin^{-1}x is the angle yy with sin⁡y=x\sin y = x and −π2≤y≤π2-\frac{\pi}{2}\le y\le\frac{\pi}{2}⁠. Basics: what an inverse needs

Solution

  1. Name the answer and write down what it must satisfy

    Let y=sin⁡−1(12)y = \sin^{-1}\left(\frac{1}{\sqrt2}\right)⁠. By the definition of the principal value branch (Basic 4), this means exactly two things:

    sin⁡y=12and−π2≤y≤π2.\sin y = \frac{1}{\sqrt2} \qquad\text{and}\qquad -\frac{\pi}{2} \le y \le \frac{\pi}{2}.
    An inverse function answers the question backwards: sin⁡−1\sin^{-1} is handed a sine value and must return an angle. Writing yy for the unknown angle turns the question into an equation we can solve. The second condition is not optional. Without it the equation sin⁡y=12\sin y = \frac{1}{\sqrt2} has infinitely many solutions, and "the" value would not make sense.
    The value is allowed: 12≈0.7071\frac{1}{\sqrt2}\approx 0.7071 lies in [−1,1][-1,1]⁠, the domain of sin⁡−1\sin^{-1}⁠. A number like sin⁡−12\sin^{-1}2 would not exist, because no angle has sine 2. Lesson: principal values

  2. Find an angle whose sine is 12\frac{1}{\sqrt2}

    From the 45° right triangle (Basic 3),

    sin⁡π4=12.\sin\frac{\pi}{4} = \frac{1}{\sqrt2}.

    So y=π4y = \frac{\pi}{4} satisfies the first condition.

    The value 12\frac{1}{\sqrt2} (also written 22\frac{\sqrt2}{2}⁠) is one of the special values you should recognise on sight. The ones that appear again and again are sin⁡π6=12\sin\frac{\pi}{6}=\frac12⁠, sin⁡π4=12\sin\frac{\pi}{4}=\frac{1}{\sqrt2} and sin⁡π3=32\sin\frac{\pi}{3}=\frac{\sqrt3}{2}⁠.
    In the isosceles right triangle the side opposite the 45∘45^\circ angle is 1 and the hypotenuse is 2\sqrt2⁠, so the ratio is 12\frac{1}{\sqrt2}⁠. Basics: special angles

  3. Check the second condition: is π4\frac{\pi}{4} in the principal range?

    Yes:

    −π2≤π4≤π2,-\frac{\pi}{2} \le \frac{\pi}{4} \le \frac{\pi}{2},

    since π4=45∘\frac{\pi}{4} = 45^\circ lies between −90∘-90^\circ and 90∘90^\circ⁠. Both conditions of Step 1 hold.

    This check is the whole point of the word "principal". An angle like 3π4\frac{3\pi}{4} also has sine 12\frac{1}{\sqrt2} (it is the mirror image of π4\frac{\pi}{4} in the yy-axis, at the same height), but 3π4=135∘\frac{3\pi}{4} = 135^\circ is outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so it is not the principal value.

  4. Explain why π4\frac{\pi}{4} is the only answer

    On [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] the sine function is strictly increasing: as the angle runs from −π2-\frac{\pi}{2} to π2\frac{\pi}{2}⁠, the point on the unit circle climbs the right half of the circle from the bottom (0,−1)(0,-1) to the top (0,1)(0,1)⁠, and its height rises the whole way, from −1-1 to 11⁠. A strictly increasing function never takes the same value twice. So there is exactly one angle in this interval with height 12\frac{1}{\sqrt2}⁠, and we have found it:

    sin⁡−1(12)=π4.\sin^{-1}\left(\frac{1}{\sqrt2}\right) = \frac{\pi}{4}.
    x y O π/4 P Q 1/√2 π/2 −π/2 0 thick blue arc: angles in [−π/2, π/2], the range of sin⁻¹ P: angle π/4, height (sine) 1/√2 Q: angle 3π/4, same height 1/√2, but not on the blue arc gold line: every point at height 1/√2
    The unit circle. Sine is the height of the point. The gold line at height 12\frac{1}{\sqrt2} meets the circle at PP (angle π4\frac{\pi}{4}⁠) and QQ (angle 3π4\frac{3\pi}{4}⁠), but only PP is on the thick blue half that sin⁡−1\sin^{-1} answers from.
    Uniqueness is what makes sin⁡−1\sin^{-1} a genuine function. The textbook restricts sine to [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] precisely because on that interval sine is one-one (each height occurs once) and onto [−1,1][-1,1] (every height occurs). The figure shows it: the gold line at height 12\frac{1}{\sqrt2} meets the circle twice, at PP and QQ⁠, but meets the thick blue half only once, at PP⁠.
    A function has an inverse only when it is one-one and onto (Chapter 1). Restricting the domain of sine is how we make it one-one. Lesson: why the domain must be restricted
Result

The principal value of sin⁡−1(12)\sin^{-1}\left(\frac{1}{\sqrt2}\right) is π4\frac{\pi}{4}. It is the only angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] whose sine is 12\frac{1}{\sqrt2}⁠. This agrees with the textbook's solution.

Check it with numbers

π4≈0.785398\frac{\pi}{4} \approx 0.785398 and sin⁡(0.785398)≈0.707107\sin(0.785398) \approx 0.707107⁠, while 12≈0.707107\frac{1}{\sqrt2} \approx 0.707107⁠. They agree. A calculator in radian mode gives sin⁡−1(0.707107)≈0.785398\sin^{-1}(0.707107)\approx 0.785398⁠, the same number. The rejected angle 3π4≈2.356194\frac{3\pi}{4}\approx 2.356194 also has sin⁡(2.356194)≈0.707107\sin(2.356194)\approx 0.707107⁠, but 2.356194>π2≈1.5707962.356194 > \frac{\pi}{2}\approx 1.570796⁠, so it is outside the principal range.

Common mistakes
  1. Answering in degrees. The answer is an angle in radians: π4\frac{\pi}{4}⁠, not 4545⁠. (Saying "π4\frac{\pi}{4}⁠, that is 45∘45^\circ" is fine; writing sin⁡−112=45\sin^{-1}\frac{1}{\sqrt2} = 45 is wrong.)
  2. Giving 3π4\frac{3\pi}{4}⁠, or "π4\frac{\pi}{4} and 3π4\frac{3\pi}{4}". Both have sine 12\frac{1}{\sqrt2}⁠, but sin⁡−1\sin^{-1} returns only the value in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.
  3. Reading sin⁡−1x\sin^{-1}x as 1sin⁡x\frac{1}{\sin x}⁠. The −1-1 means "inverse function", not "reciprocal". 1sin⁡x\frac{1}{\sin x} is cosec⁡x\operatorname{cosec}x⁠.
What to take away
  1. To evaluate sin⁡−1a\sin^{-1}a⁠: set y=sin⁡−1ay = \sin^{-1}a⁠, write sin⁡y=a\sin y = a with −π2≤y≤π2-\frac{\pi}{2}\le y\le\frac{\pi}{2}⁠, and find the one angle satisfying both.
  2. On the unit circle, sin⁡−1\sin^{-1} always answers with a point on the right half (from the bottom to the top). For positive aa the answer is in (0,π2]\left(0, \frac{\pi}{2}\right]⁠; for negative aa it is in [−π2,0)\left[-\frac{\pi}{2}, 0\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, 3Section 2.2 and its table of principal value branches (Lesson: principal values). The textbook's solution of Example 1 prints this range with round brackets, (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, and some PDF viewers also show a wrong symbol in place of π\pi there. Both are printing slips: the table in Section 2.2 gives the closed interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, and the endpoints do belong to the range, since sin⁡−11=π2\sin^{-1}1 = \frac{\pi}{2}⁠.
Principal value = the value in the principal rangeStep 1Section 2.2, Note 3
sin⁡π4=12\sin\frac{\pi}{4} = \frac{1}{\sqrt2}Step 2Class XI special angles; 45° right triangle
Sine is strictly increasing on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Step 4Unit circle; Section 2.2 (sine restricted to this interval is one-one and onto [−1,1][-1,1]⁠)
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