Trigonometric Functions: past paper questions & worked solutions

The period of $\tan x$ is $\pi$.

In $-\pi<x<\pi$, the line $y=(\pi+2)x$ meets $y=\tan x$ once at the origin and once on each side of the origin, giving 3 roots.

For $y=\pi x$, the interval $-2\pi<x<2\pi$ has the 3 central roots plus one further root on each neighbouring branch, giving 5 roots.

There are 99 positive outer roots, 99 negative outer roots, and 3 central roots, so the total is $99+99+3=201$.

At $P(30,-3)$, $x-30=0$, so $\cos0^\circ=1$. Therefore $A=-3$.

Since $A=-3$, maximum points occur when $\cos(x-30)^\circ=-1$.

$x-30=180+360n$, so $x=210+360n$. The maximum shown at $Q$ is two full periods after the first maximum, so $x=210+720=930$.

The maximum y-value is $3$, hence $Q=(930,3)$.

For $y=\sin\left(\frac{x}{12}\right)$, one full sine cycle needs $\frac{x}{12}$ to increase by $2\pi$, so the period is $24\pi$.

The sine function is $-1$ when its angle is $\frac{3\pi}{2}$. Therefore $\frac{x}{12}=\frac{3\pi}{2}$, so $x=18\pi$ and $M=(18\pi,-1)$.

If the smallest positive solution of $\sin\left(\frac{x}{12}\right)=k$ is $\alpha$, then the closest negative solution is one half-period before the opposite sine position: $-12\pi-\alpha$.

Since $\cos\theta=\sin\left(\frac\pi2-\theta\right)$, the corresponding smallest positive cosine solution is $6\pi-\alpha$.

A sine curve reaches its minimum when its angle is $\frac{3\pi}{2}$. Therefore $$x+\frac\pi8=\frac{3\pi}{2},$$ so $$x=\frac{12\pi}{8}-\frac\pi8=\frac{11\pi}{8}.$$

For $\sin u=\sin\left(\frac\pi8\right)$, $$u=\frac\pi8+2n\pi\quad\text{or}\quad u=\frac{7\pi}{8}+2n\pi.$$ With $u=x+\frac\pi8$, this gives $$x=2n\pi\quad\text{or}\quad x=\frac{3\pi}{4}+2n\pi.$$

For $0\le x<101\pi$, each family has $n=0,1,\ldots,50$, giving 51 roots from each family. Hence the total number of roots is $$51+51=102.$$

Because $y=-5\sin\left(x+\frac\pi8\right)$, the minimum value $-5$ occurs when the sine value is $1$. Thus $$x+\frac\pi8=\frac\pi2,$$ so $x=\frac{3\pi}{8}$. Therefore $$Q=\left(\frac{3\pi}{8},-5\right).$$

For $f(x)=\cos 2x^\circ$, the first minimum occurs when $2x=180^\circ$, so $x=90^\circ$ and $y=-1$. Therefore $Q=(90,-1)$.

The point $R(k,0)$ is the x-intercept after the next maximum. This occurs when $2x=450^\circ$, so $x=225^\circ$. Hence $k=225$.

A horizontal line $y=p$ gives exactly two intersections on $0\le x\le225$ when it cuts the two negative branches only, so $-1<p<0$.

There are also exactly two solutions when $p=1$, at $x=0$ and $x=180$. Therefore $-1<p<0$ or $p=1$.

For $y=4\cos x^\circ$, the maximum value is $4$ and the minimum value is $-4$. The minimum immediately to the left of the y-axis is at $x=-180^\circ$, so $P=(-180,-4)$.

The x-intercepts of $4\cos x^\circ$ occur at $x=90^\circ,270^\circ,450^\circ,\ldots$. The point shown to the right of the next maximum is $Q=(450,0)$.

For $C_2$, $y=4\cos x^\circ+k$. Its minimum is $-4+k$. Since the minimum value is $-1$, $-4+k=-1$, so $k=3$.

The maximum value of $C_2$ is $4+3=7$. The maximum point with the smallest positive x-coordinate occurs at $x=360^\circ$, so $R=(360,7)$.

The first maximum of $y=3\sin x$ occurs at $x=90^\circ$, with $y=3$. So $P=(90^\circ,3)$.

After the second maximum, the curve returns to the x-axis at $x=540^\circ$. So $Q=(540^\circ,0)$.

For $C_2$, $y=3\sin x+k$. Its maximum is $3+k=10$, so $k=7$.

The minimum value is $-3+7=4$. The smallest positive x-coordinate for a sine minimum is $270^\circ$, so $R=(270^\circ,4)$.

For $y=4\cos x$, the x-intercepts occur at $x=90^\circ+180^\circ n$. The labelled point $P$ is the right-hand intercept after $180^\circ$, so $P=(270^\circ,0)$.

The minimum value is $-4$, reached at $x=180^\circ+360^\circ n$. The labelled point on the left is $Q=(-180^\circ,-4)$.

$4\cos x=3$ has two solutions every $360^\circ$. The interval $0<x\le18000^\circ$ contains $50$ full periods, so there are $100$ solutions.

$5+4\cos x=1$ gives $\cos x=-1$. This occurs once every $360^\circ$. In $-720^\circ<x\le720^\circ$, the solutions are $-540^\circ,-180^\circ,180^\circ,540^\circ$, so there are $4$.

$4\cos x-3=1$ gives $\cos x=1$. In $-1080^\circ\le x\le1080^\circ$, the multiples of $360^\circ$ are $-1080,-720,-360,0,360,720,1080$, so there are $7$.

For $y=12\sin x$, maxima occur when $x=\dfrac{\pi}{2}+2\pi n$, with value $12$.

The maximum point $P$ shown is the second positive maximum, so $P=(\dfrac{5\pi}{2},12)$.

For $y=12\sin x+k$, the maximum value is $12+k$. Given the maximum is $3$, $k=-9$.

The minimum value is $-12-9=-21$, and it first occurs at $x=\dfrac{3\pi}{2}$. So the point is $(\dfrac{3\pi}{2},-21)$.

For $y=12\sin(x+B)$, any minimum has value $-12$, so $A=-12$.

At the minimum $x=\dfrac{\pi}{4}$, so $\dfrac{\pi}{4}+B=\dfrac{3\pi}{2}$. Hence $B=\dfrac{5\pi}{4}$.

For $y=5\cos x$, the minimum value is $-5$ and it occurs at $x=180^\circ+360^\circ n$.

The plotted point $P$ is the displayed minimum at $x=540^\circ$, so $P=(540^\circ,-5)$.

For $y=5\cos x-2$, the maximum value is $5-2=3$. The smallest positive x-coordinate for a maximum shown is $x=360^\circ$, so $Q=(360^\circ,3)$.

For $y=-5\cos x$, a maximum occurs where $\cos x=-1$, first at $x=180^\circ$. The maximum value is $5$, so $Q=(180^\circ,5)$.

For $y=3\cos\left(\dfrac{x}{n}\right)^\circ$, the first positive zero occurs when $\dfrac{x}{n}=90^\circ$.

Since $P=(270,0)$, $\dfrac{270}{n}=90$, so $n=3$.

The period is $360n=360(3)=1080^\circ$.

A minimum occurs when $\dfrac{x}{3}=180^\circ+360^\circ k$. The point $Q$ shown is the next displayed minimum after one full period, so $x=1620$ and $y=-3$.

For $y=2\sin x^\circ+k$, the points give $2\sin a+k=\dfrac{12}{5}$ and $-2\sin a+k=-\dfrac35$.

Adding gives $2k=\dfrac{12}{5}-\dfrac35=\dfrac95$, so $k=\dfrac{9}{10}$.

The shown dashed asymptote is the right-hand positive asymptote before $2\pi$, so $l$ has equation $x=\dfrac{3\pi}{2}$.

For $y=\dfrac1x+1$, the vertical asymptote is $x=0$ and the horizontal asymptote is $y=1$. The right branch lies above $y=1$ and the left branch lies below $y=1$.

The number of solutions of $\tan x=\dfrac1x+1$ is the number of intersections of the two graphs. From the sketch on $-2\pi\le x\le2\pi$, there are $5$ intersections.

From the repeated tangent branches and the reciprocal curve, there is one intersection in each positive interval between consecutive tangent asymptotes. From $0$ to $40\pi$, this gives $40$ solutions.

Extending the same intersection pattern across the stated interval gives $14$ intersections, so there are $14$ solutions.

For $y=-4\cos x$, minimum points occur at $x=2\pi n$ with value $-4$.

The displayed minimum to the left of the y-axis is $P=(-2\pi,-4)$.

The point $Q$ is the positive x-intercept after the maximum at $x=\pi$, so $Q=(\dfrac{3\pi}{2},0)$.

For $y=-4\cos x+k$, the maximum value is $4+k$. Given the maximum is $11$, $k=7$.

The minimum value is $-4+7=3$. The first positive minimum is at $x=2\pi$, so the point is $(2\pi,3)$.

$-4\cos x=5-\dfrac{10}{\pi}x$ is the intersection of the cosine curve with the straight line $y=5-\dfrac{10}{\pi}x$.

The line passes through $(0,5)$ and meets the displayed curve once, at the x-axis point, so there is 1 solution.

$\sin(180^\circ-\alpha)=\sin\alpha=p$, so $2\sin(180^\circ-\alpha)=2p$.

$\sin(\alpha-180^\circ)=-\sin(180^\circ-\alpha)=-p$.

$\sin(180^\circ+\alpha)=-\sin\alpha=-p$, so $3+\sin(180^\circ+\alpha)=3-p$.

The graph has amplitude 1 and period $180^\circ$. It crosses the x-axis at $0^\circ,90^\circ,180^\circ$, has a maximum at $45^\circ$, and a minimum at $135^\circ$.

Since $\sin\alpha=p$ and $0<2x<360^\circ$, the possible values are $2x=\alpha$ or $2x=180^\circ-\alpha$.

Therefore $x=\dfrac{\alpha}{2}$ or $x=90^\circ-\dfrac{\alpha}{2}$.

The graph has amplitude $10$, period $2\pi$, and value $-10$ at $x=0$, so $f(x)=-10\cos x$.

The graph $y=10-x^2$ is a downward-opening parabola with vertex $(0,10)$ and x-intercepts at $x=\pm\sqrt{10}$.

On the given sketch, the parabola meets $f(x)$ twice. Outside the central part the parabola is below the range of $f(x)$, so there are $2$ solutions in $-100\pi\le x\le100\pi$.

The graph $y=\tan x$ has zeros at integer multiples of $\pi$ and vertical asymptotes at $x=\dfrac\pi2+n\pi$.

In $-2\pi\le x\le2\pi$, the graph $y=\tan x$ intersects $f(x)$ four times.

The interval $-100\pi\le x\le100\pi$ is $50$ times as long as $-2\pi\le x\le2\pi$, so the number of solutions is $50\times4=200$.

The first graph has amplitude 3, period $2\pi$, and maximum at $x=0$, so $f(x)=3\cos x$.

$y=f\left(x+\dfrac\pi4\right)=3\cos\left(x+\dfrac\pi4\right)$, a shift left by $\dfrac\pi4$.

$\cos\left(x+\dfrac\pi4\right)=0$ gives $x=\dfrac\pi4+n\pi$. In $-2\pi<x<2\pi$, these are $-\dfrac{7\pi}{4}$, $-\dfrac{3\pi}{4}$, $\dfrac\pi4$, and $\dfrac{5\pi}{4}$.

At $x=0$, $y=3\cos\dfrac\pi4=\dfrac{3\sqrt2}{2}$.

The second graph has amplitude 1, period $\pi$, and passes through the origin with positive gradient, so $g(x)=\sin2x$.

$y=g(x)-2=\sin2x-2$, so the graph is shifted down by 2. Its y-intercept is $(0,-2)$.

$6x-\dfrac{27}{4}-x^2=-(x^2-6x)-\dfrac{27}{4}=-(x-3)^2+9-\dfrac{27}{4}$.

$6x-\dfrac{27}{4}-x^2=\dfrac94-(x-3)^2$.

The maximum point is $P=(3,\dfrac94)$.

$\dfrac94-(x-3)^2=0$ gives $x=3\pm\dfrac32$. Hence $A$ has $x=\dfrac32$ and $B$ has $x=\dfrac92$.

The distance between consecutive zeros of $\cos(kx)$ is $\dfrac\pi{k}$.

Since $x_B-x_A=\dfrac92-\dfrac32=3$, $\dfrac\pi{k}=3$, so $k=\dfrac\pi3$. The period is $\dfrac{2\pi}{k}=6$.

The line through $P(3,\dfrac94)$ and $B(\dfrac92,0)$ has gradient $-\dfrac32$, so $L$ is $y=-\dfrac32x+\dfrac{27}{4}$.

The region lies between $A$ and $B$, so $\dfrac32\le x\le\dfrac92$.

The lower boundary is $C_2$, so $y\ge\cos\left(\dfrac\pi3x\right)$.

The region is below both $C_1$ and the line segment $L$, so $y\le\dfrac94-(x-3)^2$ and $y\le-\dfrac32x+\dfrac{27}{4}$.