Trigonometry — Practice

Question 1

Find the maximum and minimum values of $5\cos	heta+12\sin	heta+7$.

Accepted Answer

$R=\sqrt{25+144}=13$, so $5\cos	heta+12\sin	heta$ ranges over $[-13,13]$.

Adding $7$: the range is $[-6,20]$.

Maximum $20$, minimum $-6$.

Question 2

Prove that $	an20^{\circ}	an40^{\circ}	an80^{\circ}=	an60^{\circ}$.

Accepted Answer

$	an20^{\circ}	an40^{\circ}	an80^{\circ}=\dfrac{\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}}{\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}$.

The denominator is $	frac18$ by the doubling chain.

The numerator equals $\dfrac{\sqrt3}{8}$ by the corresponding sine product.

The ratio is $\sqrt3=	an60^{\circ}$.

Question 3

Find the fundamental period of $\cos^{2}x+\sin^{4}x$.

Accepted Answer

$\cos^{2}x=\dfrac{1+\cos2x}{2}$ and $\sin^{4}x=\left(\dfrac{1-\cos2x}{2}ight)^{2}$.

Both are functions of $\cos2x$, which has period $\pi$.

The period is $\pi$, not $2\pi$.

Question 4

Solve $\cos	heta+\cos2	heta+\cos3	heta=0$.

Accepted Answer

$\cos3	heta+\cos	heta=2\cos2	heta\cos	heta$, so the equation is $\cos2	heta\left(2\cos	heta+1ight)=0$.

$\cos2	heta=0$ gives $	heta=\dfrac{(2n+1)\pi}{4}$.

$\cos	heta=-	frac12$ gives $	heta=2n\pi\pm\dfrac{2\pi}{3}$.

Question 5

Solve $\cos	heta-\sin	heta=1$ for $	heta\in[0,2\pi]$, avoiding extraneous roots.

Accepted Answer

$\sqrt2\cos\left(	heta+	frac{\pi}{4}ight)=1$, so $\cos\left(	heta+	frac{\pi}{4}ight)=	frac1{\sqrt2}$.

$	heta+	frac{\pi}{4}=2n\pi\pm	frac{\pi}{4}$.

In $[0,2\pi]$ this gives $	heta=0$, $	heta=	frac{3\pi}{2}$ and $	heta=2\pi$.

Each satisfies the original, so no root is extraneous.

Question 6

Find the number of solutions of $	an	heta=	heta$ in $\left(0,	frac{5\pi}{2}ight)$.

Accepted Answer

On each interval between consecutive asymptotes, $	an	heta$ rises from $-\infty$ to $+\infty$, so it crosses the line $y=	heta$ exactly once.

The intervals inside $\left(0,	frac{5\pi}{2}ight)$ are $\left(0,	frac{\pi}{2}ight)$, $\left(	frac{\pi}{2},	frac{3\pi}{2}ight)$ and $\left(	frac{3\pi}{2},	frac{5\pi}{2}ight)$.

The first contributes only $	heta=0$, which is excluded, so there are $2$ solutions.

Question 7

Evaluate $	an^{-1}3+	an^{-1}	frac12$.

Accepted Answer

Here $xy=	frac32>1$ and both are positive, so a correction of $+\pi$ is needed.

$\dfrac{x+y}{1-xy}=\dfrac{3.5}{-0.5}=-7$.

So the sum is $\pi+	an^{-1}(-7)=\pi-	an^{-1}7$.

Numerically $1.249+0.464=1.713$ radians, and $\pi-1.429=1.713$. They agree.

Question 8

Evaluate $	an^{-1}\left(	an\dfrac{7\pi}{6}ight)$ and $\cos^{-1}\left(\cos\dfrac{4\pi}{3}ight)$.

Accepted Answer

$	an\dfrac{7\pi}{6}=	an\dfrac{\pi}{6}$, and $\dfrac{\pi}{6}$ is in $\left(-	frac{\pi}{2},	frac{\pi}{2}ight)$, so the first is $\dfrac{\pi}{6}$.

$\cos\dfrac{4\pi}{3}=\cos\dfrac{2\pi}{3}$, and $\dfrac{2\pi}{3}$ is in $[0,\pi]$, so the second is $\dfrac{2\pi}{3}$.

Question 9

Solve $\sin^{-1}x+\sin^{-1}2x=\dfrac{\pi}{3}$.

Accepted Answer

$\sin^{-1}2x=\dfrac{\pi}{3}-\sin^{-1}x$.

Taking sines: $2x=\dfrac{\sqrt3}{2}\sqrt{1-x^{2}}-\dfrac12x$, so $\dfrac{5x}{2}=\dfrac{\sqrt3}{2}\sqrt{1-x^{2}}$.

Squaring: $25x^{2}=3-3x^{2}$, so $x^{2}=\dfrac{3}{28}$ and $x=\dfrac{\sqrt3}{2\sqrt7}$.

The negative root is rejected, since it would make the left side negative.

Question 10

In a triangle with $a=7$, $b=8$, $c=9$, find $\cos A$ and the area.

Accepted Answer

$\cos A=\dfrac{64+81-49}{2\cdot8\cdot9}=\dfrac{96}{144}=\dfrac23$.

$s=12$, so $\Delta=\sqrt{12\cdot5\cdot4\cdot3}=\sqrt{720}=12\sqrt5$.

Check: $\sin A=\dfrac{\sqrt5}{3}$ and $	frac12bc\sin A=	frac12\cdot72\cdot\dfrac{\sqrt5}{3}=12\sqrt5$.

Question 11

In a triangle, $a=2$, $b=3$ and $A=30^{\circ}$. How many such triangles exist?

Accepted Answer

$\sin B=\dfrac{b\sin A}{a}=\dfrac{3\cdot\frac12}{2}=\dfrac34$.

Both $B=\sin^{-1}	frac34\approx48.6^{\circ}$ and its supplement $\approx131.4^{\circ}$ give $A+B<180^{\circ}$.

So two triangles exist. This is the ambiguous case, and it arises precisely because the sine rule cannot distinguish an angle from its supplement.

Question 12

Prove the projection formula $b=c\cos A+a\cos C$.

Accepted Answer

$b=2R\sin B=2R\sin(A+C)$.

$=2R\left(\sin A\cos C+\cos A\sin Cight)$.

$=a\cos C+c\cos A$.

Geometrically the foot of the altitude from $B$ splits $b$ into the projections of the other two sides.

Practice — Trigonometry

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