Unit 6, lesson 57

Composing Trigonometric and Inverse Functions

How to evaluate and simplify expressions such as cos(arcsin x) by turning the inverse angle into a point on the unit circle, choosing the sign from its principal interval and keeping the correct domain.

  • Theory part 113
  • Practice part 114
  • Inverse Trigonometric Functions

Suppose you know that an angle has sine 35\frac{3}{5} and you want its cosine. You do not need the angle itself. You need to know where its point sits on the unit circle, and the inverse function tells you exactly which quarter of the circle to look in. Expressions like cos⁡(arcsin⁡x)\cos(\arcsin x) turn up every time calculus differentiates an inverse function or makes a trigonometric substitution, so it pays to read them quickly and correctly.

Key terms

Nested function

Definition A nested function here is a composition f(g(x))f(g(x)) in which one of ff and gg is a trigonometric function and the other is an inverse trigonometric function, for example cos⁡(arcsin⁡x)\cos(\arcsin x) or arccos⁡(cos⁡x)\arccos(\cos x). As for any composition, its domain is the set of all xx in the domain of gg for which g(x)g(x) lies in the domain of ff.

In plain words Work from the inside out. The inner function produces a number, and the outer function must be able to accept that number.

Example To evaluate cos⁡(arcsin⁡35)\cos\left(\arcsin \frac{3}{5}\right), first let θ=arcsin⁡35\theta = \arcsin \frac{3}{5}. This is the angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with sin⁡θ=35\sin\theta = \frac{3}{5}. Then

cos⁡θ=1−(35)2=1625=45.\cos\theta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}.

The next sections explain why the positive root is the right one.

Inner value

Definition The inner value is the number the inner function hands to the outer one. Before anything else, it must lie in the real domain of the function that receives it. The functions arcsin⁡\arcsin and arccos⁡\arccos accept only inputs in [−1,1][-1, 1], and arctan⁡\arctan accepts every real number. Two consequences follow.

sin⁡(arcsin⁡x)=xandcos⁡(arccos⁡x)=xonly for −1≤x≤1,arcsin⁡(sin⁡x)=xonly for −π2≤x≤π2.\begin{gathered} \sin(\arcsin x) = x \quad \text{and} \\[0.4em] \cos(\arccos x) = x \\[0.4em] \text{only for } -1 \le x \le 1, \\[1em] \arcsin(\sin x) = x \\[0.4em] \text{only for } -\frac{\pi}{2} \le x \le \frac{\pi}{2}. \end{gathered}

Similarly arccos⁡(cos⁡x)=x\arccos(\cos x) = x only for 0≤x≤π0 \le x \le \pi, and arctan⁡(tan⁡x)=x\arctan(\tan x) = x only for −π2<x<π2-\frac{\pi}{2} < x < \frac{\pi}{2}, while tan⁡(arctan⁡x)=x\tan(\arctan x) = x for every real xx.

In plain words An inverse function undoes the original only on the piece where the original was made one-to-one. Outside that piece the two do not cancel.

Example The expression sin⁡(arcsin⁡2)\sin(\arcsin 2) is undefined, because no real angle has sine 22. For arcsin⁡(sin⁡5π6)\arcsin\left(\sin \frac{5\pi}{6}\right) the inner value is sin⁡5π6=12\sin \frac{5\pi}{6} = \frac{1}{2}, so

arcsin⁡(sin⁡5π6)=arcsin⁡12=π6,\arcsin\left(\sin \frac{5\pi}{6}\right) = \arcsin \frac{1}{2} = \frac{\pi}{6},

not 5π6\frac{5\pi}{6}, since 5π6\frac{5\pi}{6} lies outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Principal angle

Definition For a valid input uu, the principal angle is the single angle, in radians, that the inverse function returns.

θ=arcsin⁡u  ⟺  sin⁡θ=u and −π2≤θ≤π2,−1≤u≤1,θ=arccos⁡u  ⟺  cos⁡θ=u and 0≤θ≤π,−1≤u≤1,θ=arctan⁡u  ⟺  tan⁡θ=u and −π2<θ<π2,u∈R.\begin{gathered} \theta = \arcsin u \iff \sin\theta = u \\[0.4em] \text{ and } -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}, \quad -1 \le u \le 1, \\[1em] \theta = \arccos u \iff \cos\theta = u \\[0.4em] \text{ and } 0 \le \theta \le \pi, \quad -1 \le u \le 1, \\[1em] \theta = \arctan u \iff \tan\theta = u \\[0.4em] \text{ and } -\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2}, \quad u \in \mathbb{R}. \end{gathered}

In plain words Many angles share the same sine, but the inverse function always picks the one in its own fixed interval. That interval tells you which quadrants are possible.

Example The angle θ=arccos⁡(−12)\theta = \arccos\left(-\frac{1}{2}\right) must lie in [0,π][0, \pi] and have cosine −12-\frac{1}{2}. The only such angle is θ=2π3\theta = \frac{2\pi}{3}, in the second quadrant. The angle 4π3\frac{4\pi}{3} also has cosine −12-\frac{1}{2}, but it is not principal.

Coordinate ratio

Definition The point at angle θ\theta on the unit circle is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). Writing it as (x,y)(x, y), every trigonometric value is a coordinate or a ratio of coordinates,

sin⁡θ=y,cos⁡θ=x,tan⁡θ=yx (x≠0),sec⁡θ=1x (x≠0),csc⁡θ=1y (y≠0),cot⁡θ=xy (y≠0).\begin{gathered} \sin\theta = y, \qquad \cos\theta = x, \\[0.6em] \tan\theta = \frac{y}{x} \ (x \ne 0), \\[0.6em] \sec\theta = \frac{1}{x} \ (x \ne 0), \\[0.6em] \csc\theta = \frac{1}{y} \ (y \ne 0), \\[0.6em] \cot\theta = \frac{x}{y} \ (y \ne 0). \end{gathered}

In plain words Once you know the point where the angle lands, any outer trigonometric function is read straight off its coordinates.

Example Find tan⁡(arccos⁡(−35))\tan\left(\arccos\left(-\frac{3}{5}\right)\right). The angle θ\theta lies in [0,π][0, \pi] with x=cos⁡θ=−35x = \cos\theta = -\frac{3}{5}. Points with 0≤θ≤π0 \le \theta \le \pi have y≥0y \ge 0, and x2+y2=1x^2 + y^2 = 1 on the unit circle, so y=1−925=45y = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}. Then

tan⁡θ=yx=4/5−3/5=−43.\tan\theta = \frac{y}{x} = \frac{4/5}{-3/5} = -\frac{4}{3}.

Pythagorean relation

Definition Every point of the unit circle satisfies x2+y2=1x^2 + y^2 = 1, which is the identity

sin⁡2θ+cos⁡2θ=1for every real θ.\begin{gathered} \sin^2\theta + \cos^2\theta = 1 \\[0.4em] \text{for every real } \theta. \end{gathered}

Solving for one coordinate gives two possibilities, y=1−x2y = \sqrt{1 - x^2} or y=−1−x2y = -\sqrt{1 - x^2}, and similarly for xx.

In plain words One coordinate fixes the other up to sign. The circle alone cannot tell you whether the point is above or below the axis.

Example If sin⁡θ=−35\sin\theta = -\frac{3}{5}, then

cos⁡2θ=1−925=1625,\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25},

so cos⁡θ\cos\theta is 45\frac{4}{5} or −45-\frac{4}{5}. Both happen for some angles. Deciding between them is the next step.

Sign choice

Definition For every real number qq,

q2=∣q∣.\sqrt{q^2} = \lvert q \rvert.

Replacing q2\sqrt{q^2} by qq is correct only when q≥0q \ge 0. When the outer coordinate comes from a Pythagorean relation, its sign is fixed by the principal interval of the inner function. In particular

cos⁡(arcsin⁡u)=1−u2,−1≤u≤1,sin⁡(arccos⁡u)=1−u2,−1≤u≤1,\begin{aligned} \cos(\arcsin u) &= \sqrt{1 - u^2}, \\[0.4em] &\quad -1 \le u \le 1, \\[0.8em] \sin(\arccos u) &= \sqrt{1 - u^2}, \\[0.4em] &\quad -1 \le u \le 1, \end{aligned}

because cosine is nonnegative on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and sine is nonnegative on [0,π][0, \pi].

In plain words Do not guess the sign. Look at the interval the inverse function lands in and ask whether the coordinate you want can be negative there.

Example Return to sin⁡θ=−35\sin\theta = -\frac{3}{5} with θ=arcsin⁡(−35)\theta = \arcsin\left(-\frac{3}{5}\right). Its sine is negative, so this angle lies in (−π2,0)\left(-\frac{\pi}{2}, 0\right), the fourth quadrant, where the xx coordinate is positive. So

cos⁡(arcsin⁡(−35))=45.\cos\left(\arcsin\left(-\frac{3}{5}\right)\right) = \frac{4}{5}.

If instead θ=arccos⁡(−35)\theta = \arccos\left(-\frac{3}{5}\right), the angle lies in [0,π][0, \pi], and sin⁡θ=45\sin\theta = \frac{4}{5}, again positive.

Reference triangle

Definition For an angle θ\theta that is not a multiple of π2\frac{\pi}{2}, the reference triangle is the right triangle formed by the origin, a point (x,y)(x, y) on the terminal side and the foot of the perpendicular from that point to the xx axis. Its hypotenuse is r=x2+y2>0r = \sqrt{x^2 + y^2} > 0 and the trigonometric values are

sin⁡θ=yr,cos⁡θ=xr,tan⁡θ=yx.\begin{gathered} \sin\theta = \frac{y}{r}, \qquad \cos\theta = \frac{x}{r}, \\[0.6em] \tan\theta = \frac{y}{x}. \end{gathered}

Physical side lengths are ∣x∣\lvert x \rvert and ∣y∣\lvert y \rvert, which are nonnegative. When a ratio is negative, label the legs with the signed coordinates xx and yy, keep the hypotenuse rr positive, and let the quadrant decide the signs.

In plain words Draw the angle in its correct quadrant, mark the legs with signed lengths, and the outer function becomes a ratio you can read from the drawing.

Example Simplify sin⁡(arctan⁡x)\sin(\arctan x). Let θ=arctan⁡x\theta = \arctan x, so θ\theta lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) and tan⁡θ=x\tan\theta = x. In that interval the horizontal coordinate is positive, so take the signed legs 11 and xx, with hypotenuse r=1+x2r = \sqrt{1 + x^2}. Then

sin⁡(arctan⁡x)=x1+x2for every real x.\begin{gathered} \sin(\arctan x) = \frac{x}{\sqrt{1 + x^2}} \\[0.4em] \text{for every real } x. \end{gathered}

At x=−2x = -2 the angle is in the fourth quadrant and the formula gives −25-\frac{2}{\sqrt{5}}, a negative sine, as it should. At x=0x = 0 the angle is 00 and the triangle flattens into a segment, but the point (1,0)(1, 0) still lies on the terminal side, so the coordinate ratio yr=01=0\frac{y}{r} = \frac{0}{1} = 0 still gives the sine, and the formula holds there too.

Common domain

Definition A simplified formula equals a composition only on the domain of the composition. When you write f(g(x))=h(x)f(g(x)) = h(x), the equation is a statement only about those xx where the left side is defined. Whenever the formula hh accepts more inputs than the composition, state that set with the answer. When several compositions are combined, the domain is the intersection of their domains.

In plain words Simplifying the look of an expression must not change where it lives. Carry the original domain along with the new formula.

Example Two symbolic compositions, with their domains.

tan⁡(arcsin⁡x)=x1−x2,−1<x<1,cos⁡(2arcsin⁡x)=1−2x2,−1≤x≤1.\begin{aligned} \tan(\arcsin x) &= \frac{x}{\sqrt{1 - x^2}}, \\[0.4em] &\quad -1 < x < 1, \\[0.8em] \cos(2\arcsin x) &= 1 - 2x^2, \\[0.4em] &\quad -1 \le x \le 1. \end{aligned}

In the first, arcsin⁡x\arcsin x needs −1≤x≤1-1 \le x \le 1, and at x=±1x = \pm 1 the angle is ±π2\pm\frac{\pi}{2} where tangent is undefined, so the endpoints drop out. The second follows from cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta. The polynomial 1−2x21 - 2x^2 is defined everywhere, yet the composition is not. At x=3x = 3 the left side makes no sense, so the equation holds only on [−1,1][-1, 1].

Check yourself

  1. Find the exact value of sin⁡(arctan⁡(−34))\sin\left(\arctan\left(-\frac{3}{4}\right)\right).
Show the answer

The angle lies in (−π2,0)\left(-\frac{\pi}{2}, 0\right), so use the signed legs x=4x = 4 and y=−3y = -3 with hypotenuse 55. The sine is −35-\frac{3}{5}.

  1. Simplify sec⁡(arcsin⁡x)\sec(\arcsin x) and state where your formula is valid.
Show the answer

Cosine of arcsin⁡x\arcsin x is 1−x2\sqrt{1 - x^2}, which is never negative. Secant is its reciprocal, so

sec⁡(arcsin⁡x)=11−x2,−1<x<1.\begin{gathered} \sec(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}, \\[0.4em] -1 < x < 1. \end{gathered}

The endpoints are excluded because there the cosine is 00.

  1. Evaluate arccos⁡(cos⁡4π3)\arccos\left(\cos \frac{4\pi}{3}\right).
Show the answer

The inner value is cos⁡4π3=−12\cos \frac{4\pi}{3} = -\frac{1}{2}, and the angle in [0,π][0, \pi] with that cosine is 2π3\frac{2\pi}{3}. The answer is 2π3\frac{2\pi}{3}, not 4π3\frac{4\pi}{3}.

How this lesson connects