Trigonometric Functions: past paper questions & worked solutions

Expand $R\sin(2x-\alpha)=R\cos\alpha\sin2x-R\sin\alpha\cos2x$. Comparison gives $R\cos\alpha=\sqrt3$ and $R\sin\alpha=3$. Thus $R=\sqrt{3+9}=2\sqrt3$ and $\tan\alpha=\sqrt3$. As $0\lt \alpha\lt \pi/2$, $\alpha=\pi/3$ and

The denominator is always positive, between $2\sqrt3$ and $6\sqrt3$. The fraction is smallest when the denominator is largest:

The maximum denominator occurs when $\sin(6x-\pi/3)=1$:

The smallest positive value is $x=5\pi/36$, since $n=-1$ gives a negative value.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Comparing coefficients gives $R\cos\alpha=1$ and $R\sin\alpha=2$. Thus $R^2=1^2+2^2=5$ and $\tan\alpha=2$. The specified first-quadrant phase is $\alpha=\arctan2=1.1071487\ldots$ radians.

Using the exact phase $\alpha=\arctan2$, write $g(x)=3-7\sqrt5\cos(2x-\alpha)$. (i) The negative cosine coefficient means the maximum occurs when the cosine is $-1$; the maximum is $3+7\sqrt5$. (ii) $2x-\alpha=\pi+2n\pi$, so $x=(\pi+\alpha)/2+n\pi$. The $n=-1$ value is negative; $n=0$ is the smallest positive one, $x=2.1243706857\ldots$, or 2.12 to two decimal places.

Expand $R\cos(x+\alpha)=R\cos\alpha\cos x-R\sin\alpha\sin x$. Comparing coefficients gives

Hence

Thus $f(x)=25\cos(x+\alpha)$, using the unrounded $\alpha$ in subsequent calculations.

Since $-25\leq f(2x)\leq25$, the positive denominator $90-3f(2x)$ lies between $15$ and $165$. The smallest reciprocal occurs at the largest denominator:

The minimum requires $f(2x)=-25$, so

Since $0\lt \alpha\lt \pi/2$, the first positive solution uses $n=0$:

For $x\ge0$, $0\lt 4/(3x+2)\le2$. Thus $f(0)=3$ is attained, while $f(x)$ increases towards 5 without reaching it.

(i) Put $y=5-4/(3x+2)$. Rearranging gives $(5-y)(3x+2)=4$, hence

Exchange variables. (ii) The inverse domain is the range of $f$ found above.

The original function has an absolute value and angle $x/3+\pi/6$:

Then $fg(-\pi)=f(2)=5-4/(3\cdot2+2)=5-1/2=9/2$.

The principal arcsine curve increases from $(-2,-\pi/2)$ through $(0,0)$ to $(2,\pi/2)$. It is concave down on the negative-x half, concave up on the positive-x half, and vertical at both ends.

Differentiate the supplied inverse relation: $dx/dy=2\cos y$. For $-\pi/2\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2/4}$. Taking the reciprocal gives

Thus $A=4$; the endpoints have vertical tangents, not finite derivatives.

At $y=\pi/4$, $x=2\sin(\pi/4)=\sqrt2$. The gradient is $1/\sqrt{4-2}=1/\sqrt2$. Hence $y-\pi/4=(x-\sqrt2)/\sqrt2$, so $y=x/\sqrt2+\pi/4-1$.

Expanding the required form gives

Comparing coefficients with \(5\cos\theta-4\sin\theta\),

Squaring and adding, using \(R\gt 0\), gives

Also, since \(0\lt \alpha\lt \frac\pi2\),

Thus the exact form is

A stretch parallel to the \(y\)-axis with scale factor \(\boxed{\sqrt{41}}\), taking \(y=\cos\theta\) to \(y=\sqrt{41}\cos\theta\).

A translation by the vector

This moves the curve \(\alpha\) units to the left, giving \(y=\sqrt{41}\cos(\theta+\alpha)\).

Since \(\theta\) can be any real number,

Hence \(4\leq4+(f(\theta))^2\leq45\). Taking reciprocals of these positive bounds and multiplying by \(90\) gives

Both endpoints are attained: the minimum occurs when \(f(\theta)=\pm\sqrt{41}\), and the maximum when \(f(\theta)=0\).

Differentiate to obtain $y'=6x/(x^2-5)-8x$. At a stationary point in $x\gt \sqrt5$, x is nonzero, so $6/(x^2-5)=8$, giving $6=8x^2-40$ and $x^2=23/4$. The original positive domain selects $x=\sqrt{23}/2$, which exceeds $\sqrt5$. Hence $p=23$.

(a) Differentiate the squared sine with the chain rule: $y'=4-24\sin x\cos x=4-12\sin2x$. Thus $A=4$, $B=-12$. (b) Since $-1\le\sin2x\le1$, the largest derivative is attained when $\sin2x=-1$: $4-12(-1)=16$.

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Therefore $R\cos\alpha=12$, $R\sin\alpha=5$. Thus $R=\sqrt{12^2+5^2}=13$ and $\alpha=\arctan(5/12)=0.3947911197\ldots$ radians, in the required first quadrant.

Write $g(\theta)=10+13\sin(2\theta-\pi/6-\alpha)$. (i) The minimum is $10-13=-3$. (ii) For sine to equal $-1$, $2\theta-\pi/6-\alpha=-\pi/2+2\pi n$. Hence $\theta=-\pi/6+\alpha/2+\pi n$. For $n=0$ this is negative, so the smallest permitted value uses $n=1$:

Use the unrounded phase angle for the calculation.

As $\beta$ varies over the reals, $12\sin\beta-5\cos\beta$ covers $[-13,13]$. Its square therefore covers $[0,169]$, not $[169,169]$. Subtracting from 10 reverses the bounds, giving $[-159,10]$; both endpoints are attained.

Use $2\sin x\cos x=\sin2x$ and $2\sin^2x=1-\cos2x$:

Expanding $R\sin(2x+\alpha)$ and comparing coefficients gives

Therefore

Thus $h(x)=5+\frac{3\sqrt5}{2}\sin(2x+\alpha)$ with the exact angle $\alpha=\arctan(2)$.

The given interval includes two complete periods. The range is

At a minimum, $\sin(2x+\alpha)=-1$, so

On $0\le x\le2\pi$, $n=0$ and $n=1$ give the two minima. The larger coordinate is

Use $\sin2x=2\sin x\cos x$ and $2\cos^2x=1+\cos2x$:

Expanding $R\sin(2x+\alpha)$ and matching coefficients gives

Squaring and adding, $R^2=20$, so $R=2\sqrt5$ since $R\gt 0$. Also $\tan\alpha=1/2$, and $0\lt \alpha\lt \pi/2$ selects $\alpha=\arctan(1/2)=0.463647609\ldots$ radians. Thus $f(x)=2\sqrt5\sin(2x+\arctan(1/2))-1$. With the phase displayed to 3 significant figures, $f(x)\approx2\sqrt5\sin(2x+0.464)-1$.

(i) The sine term has maximum 1, hence the maximum of $f$ is $2\sqrt5-1$. (ii) Maxima occur when

The value for $k=0$ is $0.553574\ldots\gt 0$, while $k=-1$ is negative. The second positive value therefore has $k=1$:

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Thus $R\cos\alpha=8$, $R\sin\alpha=15$, so $R=\sqrt{64+225}=17$ and $\alpha=\arctan(15/8)=1.0808390005\ldots$ radians, in the required quadrant.

Using the exact phase from (a),

The denominator lies between 7 and 75, so it is always positive. (i) The smallest fraction occurs at its largest denominator: $f_{\min}=15/75=1/5$. (ii) This occurs when $x-\alpha=\pi/2+2n\pi$. The smallest positive value is $x=\alpha+\pi/2=2.6516353273\ldots$, or 2.65 to three significant figures.

The transformation $2f(x)-5$ preserves the locations of minima and maps their y coordinate to $2(1/5)-5=-23/5$.

Reflection in the x-axis changes minima of f into maxima of $-f$. The argument $2x$ halves their x coordinates, so the first positive maximum is at

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Matching coefficients gives $R\cos\alpha=1$, $R\sin\alpha=4$. Thus $R=\sqrt{1^2+4^2}=\sqrt{17}$ and $\tan\alpha=4$. Since $0\lt \alpha\lt \pi/2$, $\alpha=\arctan4=1.32581766\ldots$ radians.

The model becomes $H=24/[3+\sqrt{17}\cos(t/2-\alpha)]$. For $0\le t\le6.5$, the phase lies between $-\alpha$ and $3.25-\alpha$; throughout this interval the denominator remains positive. Its maximum $3+\sqrt{17}$ is attained at $t=2\alpha=2.651635\ldots$, inside the time interval. The height is then smallest:

To the nearest centimetre this is $3.37\ \mathrm m$ (337 cm).

For $H=10$, $3+\sqrt{17}\cos(t/2-\alpha)=24/10$, so

Let $\gamma=\arccos[-3/(5\sqrt{17})]$. In the permitted phase interval $[-\alpha,3.25-\alpha]$, only $+\gamma$ occurs: $-\gamma\lt -\alpha$, and shifts by $2\pi$ are outside it. Using the unrounded $\alpha=\arctan4$,

At $y=\pi/3$, $x=4(\sqrt3/2)^2-1=3-1=2$.

(i) Differentiating with respect to $y$ gives

(ii) For $0\lt y\lt \pi/2$, both sine and cosine are positive, and $-1\lt x\lt 3$. From $x+1=4\sin^2y$,

Consequently $dx/dy=2\sqrt{x+1}\sqrt{3-x}$. Taking the reciprocal proves the required formula on $-1\lt x\lt 3$. At $x=-1$ and $x=3$ the denominator is zero: these are endpoint vertical tangents, not finite derivatives.

At $P=(2,\pi/3)$ the tangent gradient is $1/(2\sqrt3)$, so the normal gradient is $-2\sqrt3$. Thus the normal is $y-\pi/3=-2\sqrt3(x-2)$. Setting $y=0$ gives $N=(2+\pi/(6\sqrt3),0)$. Triangle OPN has horizontal base $ON=2+\pi/(6\sqrt3)$ and perpendicular height $\pi/3$: