Welcome to Inverse Circular Functions: Part 2

This stage covers identities such as \(\arcsin x + \arccos x = \frac{\pi}{2}\), compositions of inverse circular functions, and sum formulas for arcsin, arccos, and arctan.

Keep the domain restrictions in mind when working with compositions and identities!

Dr Brian Brooks
Mathematics InSight

You already know that \(\cos(-x)=\cos x\) and that \(\sin(-x)=-\sin x\).

Let's see what the relationship is between \(\arccos(-x)\) and \(\arccos x\), and between \(\arcsin(-x)\) and \(\arcsin x\).

Test out some standard values of arccos and arcsin with positive and negative arguments to find a relationship between \(\arccos(-x)\) and \(\arccos x\), and between \(\arcsin(-x)\) and \(\arcsin x\).

A few different values will soon convince you that

\[\arccos x+\arccos(-x)=\pi\] \[\arcsin x+\arcsin(-x)=0\]

Firstly, I want to show you a generalization of the fact that angles in a triangle add up to \(180^\circ\) using arcsin and arccos functions.

What are

\[\arcsin x + \arccos x\]

when

\[x=0,\;x=\pm\frac{1}{2},\;\pm\frac{\sqrt 2}{2},\;\pm\frac{\sqrt 3}{2},\;\pm 1\]

What are

\[\arcsin x + \arccos x\]

when

\[x=0,\;x=\pm\frac{1}{2},\;\pm\frac{\sqrt 2}{2},\;\pm\frac{\sqrt 3}{2},\;\pm 1\]

For example: \[\arcsin\frac{\sqrt{2}}{2} + \arccos\frac{\sqrt{2}}{2} = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\]

\[\arcsin0 + \arccos0 = 0 + \frac{\pi}{2} = \frac{\pi}{2}\]

\[\arcsin1 + \arccos1 = \frac{\pi}{2} + 0 = \frac{\pi}{2}\]

\[\arcsin( - 1) + \arccos( - 1) = - \frac{\pi}{2} + \pi = \frac{\pi}{2}\]

\[\arcsin\left(-\frac{\sqrt{2}}{2}\right) + \arccos\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4} + \frac{3\pi}{4} = \frac{\pi}{2}\]

\[\arcsin\frac{1}{2} + \arccos\frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\]

\[\arcsin\left(-\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} + \frac{5\pi}{6} = \frac{\pi}{2}\]

When \(x\) is positive, what is \(\arcsin x + \arccos x\)?

When \(x\) is positive, what is \(\arcsin x + \arccos x\)?

\(\arcsin x + \arccos x = \theta + \varphi = \displaystyle{\frac{\pi}{2}}\) when \(x\) is positive.

When \(x\) is positive, we can easily see this from the right-angled triangle. The triangle doesn't really help so much for negative values of \(x\), but we can use the identities from the start of this journey or we can look at graphs for this.

Use the identities

\[\arccos x+\arccos(-x)=\pi\] \[\arcsin x+\arcsin(-x)=0\]

to find \(\arcsin x+\arccos x\) when \(x<0\).

Use the identities

\[\arccos x+\arccos(-x)=\pi\] \[\arcsin x+\arcsin(-x)=0\]

to find \(\arcsin x+\arccos x\) when \(x<0\).

\[\begin{aligned} x<0\Rightarrow -x>0\\[6pt] \Rightarrow \arcsin (-x)+\arccos (-x)=\frac{\pi}{2}\\[6pt] \Rightarrow -\arcsin x+\pi-\arccos x=\frac{\pi}{2}\\[6pt] \Rightarrow \arcsin x+\arccos x=\frac{\pi}{2} \end{aligned}\]

On the same set of axes, draw the graphs \(y = \arcsin x\) and \(y = \arccos x\)

Here are the graphs \(y = \arcsin x\) and \(y = \arccos x\).

Where do they cross?

They cross at \(\displaystyle{\left( \frac{\sqrt{2}}{2},\frac{\pi}{4}\right)}\).

The easiest way to see this is by looking at the symmetry of the diagram.

Alternatively, note that \[\sin y = \cos y \Rightarrow \tan y = 1 \Rightarrow y = \frac{\pi}{4}\].

What are the vertical distances of the pink and green points from the horizontal line of symmetry?

The vertical distances are \[\arccos x-\frac{\pi}{4}\quad\text{and}\quad \frac{\pi}{4}-\arcsin x\]

Now use the symmetry of the diagram to find \[\arccos x+\arcsin x\]

Symmetry shows that these two distances are equal, which means

\[\begin{aligned} &\arccos x-\frac{\pi}{4}=\frac{\pi}{4}-\arcsin x\\[12pt] \Rightarrow &\arccos x+\arcsin x=\frac{\pi}{2} \end{aligned} \]

Next, let's see what happens when we combine circular functions with their inverses. Some of the results might be a bit unexpected.

Find

\[\begin{aligned} &\cos\left(\arcsin \frac{1}{2}\right)\\[6pt] &\cos\left(\arcsin \left(-\frac{1}{2}\right)\right)\\[6pt] &\sin\left(\arccos \frac{1}{2}\right)\\[6pt] &\sin\left(\arccos \left(-\frac{1}{2}\right)\right) \end{aligned}\]

Find

\[\begin{aligned} &\cos\left(\arcsin \frac{1}{2}\right)\\[6pt] &\cos\left(\arcsin \left(-\frac{1}{2}\right)\right)\\[6pt] &\sin\left(\arccos \frac{1}{2}\right)\\[6pt] &\sin\left(\arccos \left(-\frac{1}{2}\right)\right) \end{aligned}\]

\[\begin{aligned} &\cos\left(\arcsin \frac{1}{2}\right)=\cos\frac{\pi}{6}=\frac{\sqrt 3}{2}\\[6pt] &\cos\left(\arcsin \left(-\frac{1}{2}\right)\right)=\cos\left(-\frac{\pi}{6}\right)=\frac{\sqrt 3}{2}\\[6pt] &\sin\left(\arccos \frac{1}{2}\right)=\sin\frac{\pi}{3}=\frac{\sqrt 3}{2}\\[6pt] &\sin\left(\arccos \left(-\frac{1}{2}\right)\right)=\sin\frac{2\pi}{3}=\frac{\sqrt 3}{2} \end{aligned}\]

We've seen what happens with some values. Now let's generalize.

Find \(\cos(\arcsin x)\) when \(x\) is positive.

Find \(\sin(\arccos x)\) when \(x\) is positive.

Right triangle with angle theta where sin theta = x

Find \(\cos(\arcsin x)\) when \(x\) is positive.

Find \(\sin(\arccos x)\) when \(x\) is positive.

From the diagram

\[\begin{aligned} &x>0\Rightarrow\cos(\arcsin x) = \cos\theta=\sqrt{1-x^2}\\[6pt] &x>0\Rightarrow\sin(\arccos x) = \cos\varphi=\sqrt{1-x^2}\\ \end{aligned}\]

Remembering that \[\begin{aligned} &\arcsin x+\arcsin(-x)=0 \text{ and}\\[6pt] &\arccos x+\arccos(-x)=\pi \end{aligned}\]

find \(\cos(\arcsin x)\) and \(\sin(\arccos x)\) when \(x\) is negative.

Remembering that \[\begin{aligned} &\arcsin x+\arcsin(-x)=0 \text{ and}\\[6pt] &\arccos x+\arccos(-x)=\pi \end{aligned}\]

find \(\cos(\arcsin x)\) and \(\sin(\arccos x)\) when \(x\) is negative.

\[\begin{aligned} &x<0\Rightarrow -x> 0 \Rightarrow \cos(\arcsin x) = \cos(-\arcsin(-x))=\cos(\arcsin(-x))=\sqrt{1-x^2}\\ &x<0\Rightarrow -x> 0 \Rightarrow \sin(\arccos x) = \sin(\pi-\arccos(-x))=\sin(\arccos(-x))=\sqrt{1-x^2} \end{aligned}\]

Find

\[\begin{aligned} &\arccos\left(\sin \frac{\pi}{6}\right)\\[6pt] &\arccos\left(\sin \left(-\frac{\pi}{6}\right)\right)\\[6pt] &\arccos\left(\sin \frac{5\pi}{6}\right)\\[6pt] &\arccos\left(\sin \left(-\frac{5\pi}{6}\right)\right) \end{aligned}\]

Find

\[\begin{aligned} &\arccos\left(\sin \frac{\pi}{6}\right)\\[6pt] &\arccos\left(\sin \left(-\frac{\pi}{6}\right)\right)\\[6pt] &\arccos\left(\sin \frac{5\pi}{6}\right)\\[6pt] &\arccos\left(\sin \left(-\frac{5\pi}{6}\right)\right) \end{aligned}\]

\[\begin{aligned} &\arccos\left(\sin \frac{\pi}{6}\right)=\arccos \frac{1}{2}=\frac{\pi}{3}\\[6pt] &\arccos\left(\sin \left(-\frac{\pi}{6}\right)\right)=\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3}\\[6pt] &\arccos\left(\sin \frac{5\pi}{6}\right)=\arccos\left(\frac{1}{2}\right)=\frac{\pi}{3}\\[6pt] &\arccos\left(\sin \left(-\frac{5\pi}{6}\right)\right)=\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3} \end{aligned}\]

Find \(\arccos(\sin\theta)\) in terms of \(\theta\) when

\[\begin{aligned} 0\le\theta\le\frac{\pi}{2}\\[6pt] \frac{\pi}{2}\le\theta\le\pi\\[6pt] -\frac{\pi}{2}\le\theta\le 0\\[6pt] -\pi\le\theta\le -\frac{\pi}{2}\\[6pt] \end{aligned}\]

Find \(\arccos(\sin\theta)\) in terms of \(\theta\) when

\[\begin{aligned} 0\le\theta\le\frac{\pi}{2}\\[6pt] \frac{\pi}{2}\le\theta\le\pi\\[6pt] -\frac{\pi}{2}\le\theta\le 0\\[6pt] -\pi\le\theta\le -\frac{\pi}{2}\\[6pt] \end{aligned}\]

\[\begin{aligned} &0\le\theta\le\frac{\pi}{2}\Rightarrow \arccos(\sin\theta) =\arccos x= \varphi=\frac{\pi}{2} - \theta\\[24pt] &\frac{\pi}{2}\le\theta\le\pi\Rightarrow \sin \theta=\sin(\pi-\theta)\\[6pt] &\phantom{\frac{\pi}{2}\le\theta\le\pi}\Rightarrow \arcsin(\sin\theta)=\arcsin(\sin(\pi-\theta))=\pi-\theta\\ &\phantom{\frac{\pi}{2}\le\theta\le\pi}\Rightarrow \arccos(\sin\theta)=\frac{\pi}{2}-\arcsin(\sin(\pi-\theta))=\frac{\pi}{2}-(\pi-\theta)=\theta-\frac{\pi}{2}\\[24pt] &-\frac{\pi}{2}\le\theta\le 0\Rightarrow \arccos(\sin\theta)=\frac{\pi}{2}-\arcsin(\sin\theta)\\[6pt] &\phantom{-\frac{\pi}{2}\le\theta\le 0\Rightarrow \arccos(\sin\theta)}=\frac{\pi}{2}-\theta\\[24pt] &-\pi\le\theta\le-\frac{\pi}{2}\Rightarrow 0\le \pi+\theta\le \frac{\pi}{2}\text{ and }\sin\theta=-\sin\left(\pi+\theta\right)\\[6pt] &\phantom{-\pi\le\theta\le-\frac{\pi}{2}}\Rightarrow \arccos(\sin\theta)=\frac{\pi}{2}-\arcsin\left(-\sin\left(\pi+\theta\right)\right)\\[6pt] &\phantom{-\pi\le\theta\le-\frac{\pi}{2} \Rightarrow\arccos(\sin\theta)}=\frac{\pi}{2}+\pi+\theta\\[6pt] &\phantom{-\pi\le\theta\le-\frac{\pi}{2}\Rightarrow\arccos(\sin\theta)}=\theta+\frac{3\pi}{2}\\[24pt] \end{aligned}\]

Find

(a) \(\sin(\arccos x)\)

(b) \(\arcsin(\cos\phi)\) for \(\phi \in [0, \pi]\)

Right triangle with angle phi where cos phi = x

Find

(a) \(\sin(\arccos x)\)

(b) \(\arcsin(\cos\phi)\) for \(\phi \in [0, \pi]\)

\[\begin{aligned} &\text{(a) } \sin(\arccos x) = \sqrt{1-x^2}, \quad |x| \leq 1\\[10pt] &\text{(b) } \arcsin(\cos\phi) = \frac{\pi}{2} - \phi \end{aligned}\]

For (b): \(\sin\bigl(\tfrac{\pi}{2}-\phi\bigr)=\cos\phi\) and \(\tfrac{\pi}{2}-\phi\in\bigl[-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr]\) for \(\phi\in[0,\pi]\). ✓

Find \(\tan(\arcsin x)\) and \(\tan(\arccos x)\).

\[\begin{aligned} &\tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}}, \quad |x| < 1\\[10pt] &\tan(\arccos x) = \frac{\sqrt{1-x^2}}{x}, \quad 0 < |x| \leq 1 \end{aligned}\]

Find \(\sin(\arctan x)\) and \(\cos(\arctan x)\).

Right triangle for sin and cos of tan inverse x

Find \(\sin(\arctan x)\) and \(\cos(\arctan x)\).

\[\begin{aligned} &\sin(\arctan x) = \frac{x}{\sqrt{1+x^2}}\\[10pt] &\cos(\arctan x) = \frac{1}{\sqrt{1+x^2}} \end{aligned}\]

Since \(\arctan x \in \bigl(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr)\), the cosine is always positive.

Using the fact that

\[\sin(A + B) = \sin A\cos B + \cos A\sin B\]

and putting \(A = \arcsin a\) and \(B = \arcsin b\), find \(\sin\left(\arcsin a + \arcsin b\right)\) in terms of \(a\) and \(b\), without using \(\sin\) or \(\cos\) in your expression.

Hence find \(\arcsin a + \arcsin b\).

Find \(\sin\left(\arcsin a + \arcsin b\right)\) and hence \(\arcsin a + \arcsin b\).

\[\begin{aligned} \sin(A + B) &= \sin A\cos B + \cos A\sin B \\[6pt] A=\arcsin& \,a\quad\text{and}\quad B=\arcsin b\\[6pt] \Rightarrow\quad\sin(\arcsin a + \arcsin b) &= \sin(\arcsin a)\cos(\arcsin b) + \cos(\arcsin a)\sin(\arcsin b) \\[6pt] &= a\sqrt{1 - b^{2}} + b\sqrt{1 - a^{2}} \\[10pt] \Rightarrow\quad\arcsin a + \arcsin b &= \arcsin\left(a\sqrt{1 - b^{2}} + b\sqrt{1 - a^{2}}\right) \end{aligned}\]

although this isn't quite right if \(\arcsin a+\arcsin b\gt\frac{\pi}{2}\) or \(\le-\frac{\pi}{2}\)

In these cases, we have \[\begin{aligned} \arcsin a + \arcsin b&=\pm\pi- \arcsin\left(a\sqrt{1 - b^{2}} + b\sqrt{1 - a^{2}}\right) \end{aligned}\]

Find \(\cos\left(\arcsin a + \arcsin b\right)\) and hence \(\arccos a + \arccos b\).

\[\begin{aligned} \cos(A+B)&=\cos(\arcsin a)\cos(\arcsin b)-ab\\[12pt] &=\sqrt{1-a^2}\sqrt{1-b^2}-ab\\[12pt] \Rightarrow \arccos a+\arccos b&=\arccos\left(\sqrt{(1-a^2)(1-b^2)}-ab\right)\\[12pt] \end{aligned}\]

Putting \(c=\cos A\) and \(d=\cos B\), find \(\cos (A+B)\) and hence find \(\arccos c + \arccos d\).

\[\begin{aligned} \cos(A+B)&=cd-\sin(\arccos c)\sin(\arccos d)\\[12pt] &=cd-\sqrt{1-c^2}\sqrt{1-d^2}\\[12pt] \Rightarrow \arccos c+\arccos d&=\arccos\left(cd-\sqrt{(1-c^2)(1-d^2)}\right)\\[12pt] \end{aligned}\]

Putting \(u=\tan A\) and \(v=\tan B\), find \(\tan (A+B)\) and hence find \(\arctan u + \arctan v\).

\[\begin{aligned} &\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{u+v}{1-uv}\\[12pt] \Rightarrow &\arctan u+\arctan v=\arctan\frac{u+v}{1-uv} \end{aligned}\]

Well done on completing Part 2!

You have explored identities, compositions, and sum formulas for the inverse circular functions — including the beautiful result that \(\arcsin x + \arccos x = \frac{\pi}{2}\).

Part 3 continues with derivatives and integrals of the inverse circular functions.

Dr Brian Brooks
Mathematics InSight

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Well done on completing Part 2!