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Geometry and Trigonometry - Inverse Trigonometric Functions

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Domain Restriction for Invertibility: To define inverse functions, the domain of the original trigonometric functions must be restricted so they are one-to-one (bijective). For sin⁡(x)\sin(x), we restrict the domain to [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]; for cos⁡(x)\cos(x), to [0,π][0, \pi]; and for tan⁡(x)\tan(x), to (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). Visually, this selects a single monotonic 'branch' of the periodic wave that passes the horizontal line test.

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Inverse Sine (Arcsin): Defined as y=arcsin⁡(x)y = \arcsin(x) or y=sin⁡−1(x)y = \sin^{-1}(x) where the domain is x∈[−1,1]x \in [-1, 1] and the range is y∈[−π2,π2]y \in [-\frac{\pi}{2}, \frac{\pi}{2}]. The graph is a reflection of the sin⁡(x)\sin(x) curve over the line y=xy=x, appearing as an S-shaped curve passing through the origin (0,0)(0,0) and terminating at endpoints (−1,−π2)(-1, -\frac{\pi}{2}) and (1,π2)(1, \frac{\pi}{2}).

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Inverse Cosine (Arccos): Defined as y=arccos⁡(x)y = \arccos(x) or y=cos⁡−1(x)y = \cos^{-1}(x) where the domain is x∈[−1,1]x \in [-1, 1] and the range is y∈[0,π]y \in [0, \pi]. Visually, the graph is strictly decreasing, starting from the point (−1,π)(-1, \pi), crossing the y-axis at (0,π2)(0, \frac{\pi}{2}), and ending at (1,0)(1, 0).

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Inverse Tangent (Arctan): Defined as y=arctan⁡(x)y = \arctan(x) or y=tan⁡−1(x)y = \tan^{-1}(x) where the domain is all real numbers x∈Rx \in \mathbb{R} and the range is the open interval y∈(−π2,π2)y \in (-\frac{\pi}{2}, \frac{\pi}{2}). The vertical asymptotes of the tangent function become horizontal asymptotes at y=π2y = \frac{\pi}{2} and y=−π2y = -\frac{\pi}{2}. The curve passes through (0,0)(0,0) and flattens out as xx approaches ±∞\pm\infty.

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Composition Identities and Principal Values: The identity f(f−1(x))=xf(f^{-1}(x)) = x always holds within the domain of the inverse. However, f−1(f(x))=xf^{-1}(f(x)) = x only holds if xx is within the restricted principal range. If xx is outside this range, you must use the unit circle's symmetry to find the equivalent angle within the principal range (e.g., arcsin⁡(sin⁡(3π4))=π4\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}).

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Geometric Interpretation with Right Triangles: The expression θ=arcsin⁡(ac)\theta = \arcsin(\frac{a}{c}) represents an angle θ\theta such that sin⁡(θ)=ac\sin(\theta) = \frac{a}{c}. This can be visualized as a right-angled triangle with an opposite side of length aa and a hypotenuse of length cc. This allows for the evaluation of composite functions like cos⁡(arcsin⁡(x))\cos(\arcsin(x)) by finding the adjacent side length c2−a2\sqrt{c^2 - a^2} using the Pythagorean theorem.

📐Formulae

sin⁡(arcsin⁡(x))=x for x∈[−1,1]\sin(\arcsin(x)) = x \text{ for } x \in [-1, 1]

arcsin⁡(sin⁡(x))=x for x∈[−π2,π2]\arcsin(\sin(x)) = x \text{ for } x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

cos⁡(arccos⁡(x))=x for x∈[−1,1]\cos(\arccos(x)) = x \text{ for } x \in [-1, 1]

arccos⁡(cos⁡(x))=x for x∈[0,π]\arccos(\cos(x)) = x \text{ for } x \in [0, \pi]

tan⁡(arctan⁡(x))=x for x∈R\tan(\arctan(x)) = x \text{ for } x \in \mathbb{R}

arctan⁡(tan⁡(x))=x for x∈(−π2,π2)\arctan(\tan(x)) = x \text{ for } x \in (-\frac{\pi}{2}, \frac{\pi}{2})

arcsin⁡(x)+arccos⁡(x)=π2\arcsin(x) + \arccos(x) = \frac{\pi}{2}

cos⁡(arcsin⁡(x))=1−x2\cos(\arcsin(x)) = \sqrt{1 - x^2}

sin⁡(arccos⁡(x))=1−x2\sin(\arccos(x)) = \sqrt{1 - x^2}

💡Examples

Problem 1:

Evaluate the exact value of cos⁡(arcsin⁡(−12))\cos(\arcsin(-\frac{1}{2})).

Solution:

  1. Let θ=arcsin⁡(−12)\theta = \arcsin(-\frac{1}{2}). By definition, sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2} where θ∈[−π2,π2]\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}].
  2. From the unit circle or special triangles, we know sin⁡(−π6)=−12\sin(-\frac{\pi}{6}) = -\frac{1}{2}. Thus, θ=−π6\theta = -\frac{\pi}{6}.
  3. Now find cos⁡(θ)=cos⁡(−π6)\cos(\theta) = \cos(-\frac{\pi}{6}).
  4. Since cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x), cos⁡(−π6)=cos⁡(π6)=32\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}.

Explanation:

We first identify the principal angle whose sine is −12-\frac{1}{2} within the restricted range of arcsin. Then, we find the cosine of that specific angle.

Problem 2:

Simplify the expression tan⁡(arccos⁡(x))\tan(\arccos(x)) for 0<x≤10 < x \le 1.

Solution:

  1. Let θ=arccos⁡(x)\theta = \arccos(x). This implies cos⁡(θ)=x=x1\cos(\theta) = x = \frac{x}{1}.
  2. Visualize a right-angled triangle where the angle is θ\theta. The adjacent side is xx and the hypotenuse is 11.
  3. Use the Pythagorean theorem to find the opposite side: opposite=12−x2=1−x2opposite = \sqrt{1^2 - x^2} = \sqrt{1 - x^2}.
  4. By definition, tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{opposite}{adjacent}.
  5. Therefore, tan⁡(arccos⁡(x))=1−x2x\tan(\arccos(x)) = \frac{\sqrt{1 - x^2}}{x}.

Explanation:

This approach uses right-triangle trigonometry. By treating the inverse cosine as an angle in a triangle, we can determine the lengths of all sides in terms of xx and then find the tangent ratio.