Chapter 20: Inverse Trig Functions

Recall,

arccos(x)=θ0θ<πandcos(θ)=xarc\; cos(x)=\theta \Leftrightarrow 0\leq \theta <\pi\; and\; cos(\theta)=x

arcsin(x)=θπ2θ<π2andsin(θ)=xarc\; sin(x)=\theta \Leftrightarrow \frac{\pi}{2}\leq \theta <\frac{\pi}{2}\; and\; sin(\theta)=x

arctan(x)=θπ2θ<π2andtan(θ)=xarc\; tan(x)=\theta \Leftrightarrow -\frac{\pi}{2}\leq \theta <\frac{\pi}{2}\; and\; tan(\theta)=x

These are the inverse of standard trig functions.

arccot(x)=θ0<θ<πandcot(θ)=xarc\; cot(x)=\theta \Leftrightarrow 0< \theta< \pi\; and\; cot \left ( \theta \right )=x

arcsec(x)=θ0<θ<π2arc\; sec(x)=\theta \Leftrightarrow 0< \theta< \frac{\pi}{2} OR π2<θπ\frac{\pi}{2}< \theta\leq \pi and sec(θ)=xsec\left ( \theta \right )=x

arccot(x)=θπ2θ<0arc\; cot(x)=\theta \Leftrightarrow -\frac{\pi}{2}\leq \theta <0 OR 0<θπ20< \theta\leq \frac{\pi}{2} and csc(θ)=xcsc\left ( \theta \right )=x


Discuss

Where do the restrictions on θ\theta in the definitions of arcsecarcsec and arccscarccsc come from?

Answer

sec(θ)=1cos(θ)sec(\theta)=\frac{1}{cos(\theta)} and thus θπ2\theta \neq \frac{\pi}{2}

csc(θ)=1sin(θ)csc(\theta)=\frac{1}{sin(\theta)} and thus θ0\theta \neq 0


What is the slope of arctan?


Example

Calculate ddx[arctan(x)]\frac{d}{dx}[arctan(x)]  (The famous Formula Thirteen)

Find the derivative of tan(x)tan(x)

ddx[tan(x)]=ddx[sin(x)cos(x)]\frac{d}{dx}[tan(x)]=\frac{d}{dx}[\frac{sin(x)}{cos(x)}]

=cos2(x)+sin2(x)cos2(x)= \frac{cos^{2}(x)+sin^{2}(x)}{cos^{2}(x)}

=1/cos2(x)=sec2(x)=1/cos^{2}(x)=sec^{2}(x)

Apply the inverse formula

ddx[arctan(x)]=1sec2(arctan(x))\frac{d}{dx}[arctan(x)]=\frac{1}{sec^{2}(arctan(x))}

=cos2(arctan(x))=cos^{2}\left ( arctan\left ( x \right ) \right )

Find cos(arctan(x))cos(arctan(x))

θ=arctan(x)tanθ=x\theta = arctan(x)\Leftrightarrow tan\; \theta = x

Construct a triangle with unit hypothenuse that satisfies tan(θ)=xtan(\theta)=x

Thus, cos(θ)=11+x2cos\left ( \theta \right )=\frac{1}{\sqrt{1+x^{2}}}

tan(θ)=xll=xtan(\theta)=\frac{xl}{l}=x 

l2+(xl)2=1l^{2}+(xl)^{2}=1

l2(1+x2)=1l^{2}(1+x^{2})=1

l=11+x2l=\frac{1}{\sqrt{1+x^{2}}} 

ddx[arctan(x)]\frac{d}{dx}[arctan(x)] 

=cos2(arctan(x)]=cos^{2}(arctan(x)] 

=[11+x2]2=[\frac{1}{\sqrt{1+x^{2}}}]^{2} 

=1x2+1=\frac{1}{x^{2}+1}


Discuss

A formula for ddx[arcsin(x)]\frac{d}{dx}[arcsin(x)]

Find ddx[sin(x)]\frac{d}{dx}[sin(x)]

ddx[sin(x)]=cos(x))\frac{d}{dx}[sin(x)]=cos(x))

Apply the inverse formula

ddx[arcsin(x)]=1cos(arcsin(x))\frac{d}{dx}[arcsin(x)]=\frac{1}{cos(arcsin(x))}

Find cos(arcsin(x))cos(arcsin(x))

θ=arcsin(x)sin(θ)=x\theta=arcsin(x)\Leftrightarrow sin(\theta)=x

Construct a triangle with unit hypothenuse that satisfies sin(θ)=xsin(\theta) =x

l2+x2=12l^{2}+x^{2}=1^{2}

l=1x2l=\sqrt{1-x^{2}}

ddx[arcsin(x)]=1cos(arcsin(x))=1l\frac{d}{dx}[arcsin(x)]=\frac{1}{cos(arcsin(x))}=\frac{1}{l}

=11x2=1\sqrt{1-x^{2}}

Discuss

Find ddx[arccos(x)]\frac{d}{dx}[arccos(x)]

ddx[arccos(x)]=11x2\frac{d}{dx}[arccos(x)]= \frac{-1}{\sqrt{1-x^{2}}}


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