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NSWMaths Extension 1Quick questions

Trigonometric Functions (ME-T1, T2, T3)

Quick questions on Inverse trigonometric functions: definitions, principal branches, domains, ranges and graphs

12short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is why we need to restrict?
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sin⁡x\sin x, cos⁡x\cos x and tan⁡x\tan x are periodic and not one-to-one over R\mathbb{R}. To define inverses, we choose principal branches on which each is one-to-one.
What is arcsin⁡\arcsin (or sin⁡−1\sin^{-1})?
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Restrict sin⁡x\sin x to x∈[−π2,π2]x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]. On this interval sin⁡\sin is strictly increasing from −1-1 to 11.
What is arccos⁡\arccos (or cos⁡−1\cos^{-1})?
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Restrict cos⁡x\cos x to x∈[0,π]x \in [0, \pi]. On this interval cos⁡\cos is strictly decreasing from 11 to −1-1.
What is arctan⁡\arctan (or tan⁡−1\tan^{-1})?
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Restrict tan⁡x\tan x to x∈(−π2,π2)x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right). On this interval tan⁡\tan is strictly increasing from −∞-\infty to +∞+\infty.
What is identities?
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The complementary identity links arcsin⁡\arcsin and arccos⁡\arccos:
What is composing trig with inverse trig?
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For xx in the appropriate range,
What is composition with reduction?
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Find arcsin⁡ ⁣(sin⁡5π6)\arcsin\!\left( \sin \frac{5 \pi}{6} \right).
What is inverse-trig sum?
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Show that arctan⁡1+arctan⁡2+arctan⁡3=π\arctan 1 + \arctan 2 + \arctan 3 = \pi.
What is out-of-domain input?
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arcsin⁡(2)\arcsin(2) is undefined because 2∉[−1,1]2 \not\in [-1, 1]. Always check the argument is in range.
What is forgetting the principal-value restriction?
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arcsin⁡(sin⁡π)=0\arcsin(\sin \pi) = 0, not π\pi, because the range of arcsin⁡\arcsin is [−π2,π2]\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
What is treating arccos⁡\arccos as odd?
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arccos⁡\arccos is neither even nor odd. The correct identity is arccos⁡(−x)=π−arccos⁡x\arccos(-x) = \pi - \arccos x.
What is sign errors in arctan⁡\arctan sum identity?
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The identity arctan⁡x+arctan⁡y=arctan⁡ ⁣(x+y1−xy)\arctan x + \arctan y = \arctan\!\left( \frac{x + y}{1 - x y} \right) requires xy<1x y < 1. If xy>1x y > 1 and both positive, add π\pi to the right side; if both negative, subtract π\pi. :::

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