Inverse Trigonometric Function flash cards

1. 1.State the principal value branch (range) of $\sin^{-1}x$, $\cos^{-1}x$, and $\tan^{-1}x$ along with their domains.

   $\sin^{-1}x:[-1,1]\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$; $\cos^{-1}x:[-1,1]\to[0,\pi]$; $\tan^{-1}x:\mathbb{R}\to\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

   Hint: Think of restricted domains needed to make trig functions bijective.

2. 2.What are the fundamental identities relating pairs of inverse trigonometric functions of the same variable $x$?

   $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\ (x\in[-1,1])$, $\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\ (x\in\mathbb{R})$, $\sec^{-1}x+\text{cosec}^{-1}x=\frac{\pi}{2}\ (|x|\ge1)$.

   Hint: All three pairs sum to the same constant.

3. 3.Give the formula for $\tan^{-1}x+\tan^{-1}y$ and state the condition under which $\pi$ is added or subtracted.

   $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$ if $xy<1$; add $\pi$ if $x>0,\ y>0,\ xy>1$; subtract $\pi$ if $x<0,\ y<0,\ xy>1$.

   Hint: Check the sign of $xy-1$ and the signs of $x,y$ before applying the direct formula.

4. 4.What is the value of $\sin^{-1}x + \sin^{-1}y$ when $x,y\in[0,1]$ and $x^2+y^2>1$, in terms of $\sin^{-1}$?

   $\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$.

   Hint: The usual sum formula for $\sin^{-1}x+\sin^{-1}y$ needs a $\pi-$ correction here.

5. 5.How do you simplify $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ for $x>0$ using the substitution $x=\tan\theta$?

   It equals $\frac{1}{2}\tan^{-1}x$, obtained by substituting $x=\tan\theta\ \left(\theta\in\left(0,\frac{\pi}{2}\right)\right)$ so that $\sqrt{1+x^2}=\sec\theta$, and using $\frac{\sec\theta-1}{\tan\theta}=\tan\frac{\theta}{2}$.

   Hint: Convert to half-angle form using $\sec\theta - 1$ over $\tan\theta$.

6. 6.Evaluate $\sin^{-1}\left(\sin\dfrac{13\pi}{7}\right)$, carefully accounting for the range restriction of $\sin^{-1}$.

   Since $\sin^{-1}(\sin\theta)=\theta$ only for $\theta\in[-\pi/2,\pi/2]$, first reduce the angle: $\dfrac{13\pi}{7}=2\pi-\dfrac{\pi}{7}$, so $\sin\dfrac{13\pi}{7}=\sin\left(-\dfrac{\pi}{7}\right)$. Since $-\dfrac{\pi}{7}\in[-\pi/2,\pi/2]$, we get $\sin^{-1}\left(\sin\dfrac{13\pi}{7}\right)=-\dfrac{\pi}{7}$.

   Hint: Reduce the angle modulo $2\pi$ so it lands inside $[-\pi/2,\pi/2]$.

7. 7.Evaluate $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$.

   The range of $\cos^{-1}$ is $[0,\pi]$, but $\dfrac{7\pi}{6}\notin[0,\pi]$. Since cosine is even, $\cos\dfrac{7\pi}{6}=\cos\left(2\pi-\dfrac{7\pi}{6}\right)=\cos\dfrac{5\pi}{6}$, and $\dfrac{5\pi}{6}\in[0,\pi]$. Hence $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)=\dfrac{5\pi}{6}$.

   Hint: Use $\cos(2\pi-\theta)=\cos\theta$ to bring the angle into $[0,\pi]$.

8. 8.Find $\tan^{-1}\left(\tan\dfrac{5\pi}{4}\right)$ and explain why the answer is not $\dfrac{5\pi}{4}$.

   $\tan^{-1}(\tan\theta)=\theta$ only when $\theta\in(-\pi/2,\pi/2)$. Since $\dfrac{5\pi}{4}=\pi+\dfrac{\pi}{4}$ lies outside this interval, use the period $\pi$ of tangent: $\tan\dfrac{5\pi}{4}=\tan\dfrac{\pi}{4}$, and $\dfrac{\pi}{4}\in(-\pi/2,\pi/2)$. So $\tan^{-1}\left(\tan\dfrac{5\pi}{4}\right)=\dfrac{\pi}{4}$.

   Hint: $\tan$ has period $\pi$ — subtract a multiple of $\pi$ to land in $(-\pi/2,\pi/2)$.

9. 9.Find the exact piecewise form of $f(x)=\sin^{-1}(2x\sqrt{1-x^2})$ on $[-1,1]$, and state where it equals $2\sin^{-1}x$.

   Let $\theta=\sin^{-1}x\in[-\pi/2,\pi/2]$, so $\cos\theta\ge0$ and $\sin2\theta=2x\sqrt{1-x^2}$, giving $f(x)=\sin^{-1}(\sin2\theta)$. This equals $2\theta$ only if $2\theta\in[-\pi/2,\pi/2]$, i.e. $x\in\left[-\tfrac1{\sqrt2},\tfrac1{\sqrt2}\right]$. For $x\in\left(\tfrac1{\sqrt2},1\right]$, $2\theta\in(\pi/2,\pi]$ so $f(x)=\pi-2\sin^{-1}x$; for $x\in\left[-1,-\tfrac1{\sqrt2}\right)$, $f(x)=-\pi-2\sin^{-1}x$.

   Hint: Compare $\sin(2\sin^{-1}x)$ to the given expression, then check where $2\sin^{-1}x$ leaves $[-\pi/2,\pi/2]$.

10. 10.For what values of $x$ does $\cos^{-1}(2x^2-1)=2\cos^{-1}x$ actually hold?

    Let $\theta=\cos^{-1}x\in[0,\pi]$, so $x=\cos\theta$ and $2x^2-1=\cos2\theta$, giving $\cos^{-1}(2x^2-1)=\cos^{-1}(\cos2\theta)$. This equals $2\theta$ only if $2\theta\in[0,\pi]$, i.e. $\theta\in[0,\pi/2]$, i.e. $x\in[0,1]$. For $x\in[-1,0)$, $2\theta\in(\pi,2\pi]$ and instead $\cos^{-1}(2x^2-1)=2\pi-2\theta$.

    Hint: Substitute $\theta=\cos^{-1}x$ and check when $2\theta$ stays inside $[0,\pi]$.

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