Trigonometry: inverse circular functions; trigonometric equations and general values

Let $\theta=\tan^{-1}x$, so $\tan\theta=x$ with $-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}$.

In a right triangle with perpendicular $x$ and base $1$, the hypotenuse is $\sqrt{1+x^2}$, so

$\sin\theta=\frac{x}{\sqrt{1+x^2}}.$

Since $\theta$ lies in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, the principal range of $\sin^{-1}$, we get

$\tan^{-1}x=\sin^{-1}\frac{x}{\sqrt{1+x^2}}.$

Group the terms $\sin\theta+\sin3\theta$ and $\cos\theta+\cos3\theta$:

$\sin3\theta+\sin\theta=2\sin2\theta\cos\theta,\qquad \cos3\theta+\cos\theta=2\cos2\theta\cos\theta.$

The equation becomes

$2\sin2\theta\cos\theta+\sin2\theta=2\cos2\theta\cos\theta+\cos2\theta$

$\Rightarrow\;\sin2\theta\,(2\cos\theta+1)=\cos2\theta\,(2\cos\theta+1)$

$\Rightarrow\;(2\cos\theta+1)(\sin2\theta-\cos2\theta)=0.$

Case 1: $2\cos\theta+1=0\Rightarrow\cos\theta=-\dfrac12=\cos\dfrac{2\pi}{3}$, so

$\theta=2n\pi\pm\frac{2\pi}{3},\quad n\in\mathbb{Z}.$

Case 2: $\sin2\theta=\cos2\theta\Rightarrow\tan2\theta=1=\tan\dfrac{\pi}{4}$, so $2\theta=n\pi+\dfrac{\pi}{4}$, i.e.

$\theta=\frac{n\pi}{2}+\frac{\pi}{8},\quad n\in\mathbb{Z}.$

General solution: $\theta=2n\pi\pm\dfrac{2\pi}{3}$ or $\theta=\dfrac{n\pi}{2}+\dfrac{\pi}{8}$, $n\in\mathbb{Z}$.

Let $\theta=\sin^{-1}x$, so $\sin\theta=x$ with $-\dfrac\pi2\le\theta\le\dfrac\pi2$. In this range $\cos\theta\ge0$, so

$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}.$

Then

$\sin(2\sin^{-1}x)=\sin2\theta=2\sin\theta\cos\theta=2x\sqrt{1-x^2}.\ \blacksquare$

(b) $\dfrac{\pi}{6}$

$\sin\dfrac{5\pi}{6}=\dfrac12$, and the principal value of $\sin^{-1}$ lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$, so $\sin^{-1}\left(\dfrac12\right)=\dfrac\pi6$.

Let $A=\tan^{-1}\dfrac12$ and $B=\tan^{-1}\dfrac13$, so $\tan A=\dfrac12$, $\tan B=\dfrac13$.

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac12+\frac13}{1-\frac16}=\frac{5/6}{5/6}=1.$

Since $A,B>0$ and both are less than $\frac\pi4$, we have $0<A+B<\frac\pi2$, so $A+B=\dfrac\pi4$. Hence

$\tan^{-1}\frac12+\tan^{-1}\frac13=\frac\pi4.$

Now the second part:

$\tan^{-1}\frac12+\tan^{-1}\frac13+\tan^{-1}\frac14=\frac\pi4+\tan^{-1}\frac14.$

Let $C=\tan^{-1}\dfrac14$. Then

$\tan\left(\frac\pi4+C\right)=\frac{1+\frac14}{1-\frac14}=\frac{5/4}{3/4}=\frac53.$

Since $\frac\pi4+C<\frac\pi2$,

$\tan^{-1}\frac12+\tan^{-1}\frac13+\tan^{-1}\frac14=\tan^{-1}\frac53\approx1.030\text{ rad}\ (\approx59.04^\circ).$

$2\cos^2\theta-3\cos\theta+1=0\ \Rightarrow\ (2\cos\theta-1)(\cos\theta-1)=0.$

So $\cos\theta=\dfrac12$ or $\cos\theta=1$.

For $0^\circ\le\theta\le360^\circ$:

• $\cos\theta=\dfrac12$: $\theta=60^\circ,\ 300^\circ$
• $\cos\theta=1$: $\theta=0^\circ,\ 360^\circ$

Solutions: $\theta=0^\circ,\ 60^\circ,\ 300^\circ,\ 360^\circ$.

General solution:

• $\cos\theta=\dfrac12=\cos\dfrac\pi3 \Rightarrow \theta=2n\pi\pm\dfrac\pi3$
• $\cos\theta=1 \Rightarrow \theta=2n\pi$

$\theta=2n\pi\ \text{ or }\ \theta=2n\pi\pm\frac\pi3,\qquad n\in\mathbb Z$

In degrees: $\theta=360^\circ n$ or $\theta=360^\circ n\pm60^\circ$.

Using $\sin2x=2\sin x\cos x$:

$2\sin x\cos x+\cos x=0\ \Rightarrow\ \cos x\,(2\sin x+1)=0.$

So $\cos x=0$ or $\sin x=-\dfrac12$.

Solutions in $0^\circ\le x\le360^\circ$

• $\cos x=0$: $x=90^\circ,\ 270^\circ$.
• $\sin x=-\dfrac12$: sine is negative in the third and fourth quadrants, so $x=180^\circ+30^\circ=210^\circ$ and $x=360^\circ-30^\circ=330^\circ$.

$x=90^\circ,\ 210^\circ,\ 270^\circ,\ 330^\circ$

General solution

• $\cos x=0$: $x=(2n+1)\dfrac\pi2$
• $\sin x=-\dfrac12=\sin\left(-\dfrac\pi6\right)$: $x=n\pi+(-1)^{n}\left(-\dfrac\pi6\right)=n\pi+(-1)^{n+1}\dfrac\pi6$

$x=(2n+1)\frac\pi2\ \text{ or }\ x=n\pi+(-1)^{n+1}\frac\pi6,\qquad n\in\mathbb Z$

In degrees: $x=(2n+1)90^\circ$ or $x=180^\circ n+(-1)^{n+1}30^\circ$.