Trig identities & equations, compound angles (HL), reciprocal & inverse trig (HL)

1CoreNon-calculator3 marks

Solve $2\sin x=\sqrt3$ for $0\le x\le2\pi$. Give your answers as exact multiples of $\pi$.

$\sin x=\dfrac{\sqrt3}{2}$M1
$x=\dfrac{\pi}{3}$A1
$x=\dfrac{2\pi}{3}$ (from $\pi-\dfrac\pi3$)A1
Answer: $x=\dfrac\pi3,\ \dfrac{2\pi}3$
Examiner tip: Sine is positive in the first and second quadrants, so there are two solutions in $[0,2\pi]$: do not stop at $\dfrac\pi3$.

2StandardNon-calculator6 marks

Solve the equation $2\cos^2x+\sin x-1=0$ for $0\le x\le2\pi$. Give exact answers in terms of $\pi$.

Uses $\cos^2x=1-\sin^2x$M1
$2\sin^2x-\sin x-1=0$A1
Factorises/solves quadratic in $\sin x$: $(2\sin x+1)(\sin x-1)=0$M1
$x=\dfrac\pi2$ (from $\sin x=1$)A1
$x=\dfrac{7\pi}6$ (from $\sin x=-\tfrac12$)A1
$x=\dfrac{11\pi}6$A1
Answer: $x=\dfrac\pi2,\ \dfrac{7\pi}6,\ \dfrac{11\pi}6$
Examiner tip: Replace $\cos^2x$ so the equation is in one trig function, and never cancel a factor of $\sin x$ away: you lose solutions.

3StandardNon-calculator6 marks

The angle $\theta$ is acute and $\sin\theta=\dfrac35$.

(a) Find the exact value of $\cos\theta$. [2]

(b) Find the exact value of $\sin2\theta$. [2]

$\cos^2\theta=1-\left(\tfrac35\right)^2$ (or a 3-4-5 triangle)M1
$\cos\theta=\dfrac45$ (positive since acute)A1
$\sin2\theta=2\sin\theta\cos\theta=2\times\tfrac35\times\tfrac45$M1
$\dfrac{24}{25}$A1
$\cos2\theta=1-2\sin^2\theta$ (or $\cos^2\theta-\sin^2\theta$) with values substitutedM1
$\dfrac{7}{25}$A1
Answer: (a) $\tfrac45$ (b) $\tfrac{24}{25}$ (c) $\tfrac7{25}$
Examiner tip: State that $\cos\theta$ is positive because $\theta$ is acute, and use the double-angle formulae: $\sin2\theta$ is not $2\sin\theta$.

4StandardNon-calculator4 marks

Prove that $\dfrac{1-\cos2x}{\sin2x}\equiv\tan x$ for $0

Uses a double-angle identity for $\cos2x$, e.g. $\cos2x=1-2\sin^2x$M1
Numerator $1-\cos2x=2\sin^2x$A1
Denominator $\sin2x=2\sin x\cos x$A1
$\dfrac{2\sin^2x}{2\sin x\cos x}=\dfrac{\sin x}{\cos x}$ (cancelling $2\sin x\neq0$ on the interval) $=\tan x$ AGA1
Answer: Proved: LHS $=\tan x$
Examiner tip: Start from one side only and write every line of the working; the final line must be the target expression.

5HardNon-calculatorHL only7 marks

The angle $A$ is obtuse with $\sin A=\dfrac5{13}$, and the angle $B$ is acute with $\cos B=\dfrac35$.

(a) Find $\cos A$ and $\sin B$. [3]

(b) Find the exact value of $\sin(A+B)$. [2]

$\cos^2A=1-\left(\tfrac5{13}\right)^2$ (or 5-12-13 triangle)M1
$\cos A=-\dfrac{12}{13}$ (negative as $A$ is obtuse)A1
$\sin B=\dfrac45$A1
$\sin(A+B)=\sin A\cos B+\cos A\sin B$ with values substitutedM1
$-\dfrac{33}{65}$A1
$\cos(A+B)=\cos A\cos B-\sin A\sin B$ with values substitutedM1
$-\dfrac{56}{65}$A1
Answer: (a) $\cos A=-\tfrac{12}{13}$, $\sin B=\tfrac45$ (b) $-\tfrac{33}{65}$ (c) $-\tfrac{56}{65}$
Examiner tip: The sign of $\cos A$ is the usual lost mark: obtuse means cosine is negative.

6HardCalculatorHL only6 marks

Solve $2\tan^2x+\sec x=1$ for $0\le x\le2\pi$. Give non-exact answers to 3 significant figures.

Uses $\tan^2x=\sec^2x-1$M1
$2\sec^2x+\sec x-3=0$A1
Factorises/solves: $(2\sec x+3)(\sec x-1)=0$M1
$\sec x=1\Rightarrow x=0,\ 2\pi$ (both needed)A1
$\cos x=-\tfrac23\Rightarrow x=2.30$A1
$x=3.98$ (from $2\pi-2.301$)A1
Answer: $x=0,\ 2.30,\ 3.98,\ 2\pi$
Examiner tip: Convert $\sec x=-\tfrac32$ to $\cos x=-\tfrac23$ before using the inverse function, and remember the end-point solutions.

7HardNon-calculatorHL only4 marks

Show that $\arctan\dfrac12+\arctan\dfrac13=\dfrac\pi4$.

Lets $\alpha=\arctan\tfrac12$, $\beta=\arctan\tfrac13$ and uses $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$M1
$\dfrac{\frac12+\frac13}{1-\frac12\cdot\frac13}$A1
$=\dfrac{5/6}{5/6}=1$A1
Both $\alpha,\beta\in\left(0,\tfrac\pi4\right)$ (as $0<\tan<1$), so $0<\alpha+\beta<\tfrac\pi2$, hence $\alpha+\beta=\dfrac\pi4$ AGR1
Answer: Shown
Examiner tip: Evaluating $\tan(\alpha+\beta)=1$ is not enough on its own: say why the angle sum is in the range where $\tan^{-1}$ gives $\pi/4$.

8HardCalculator7 marks

The height, $H$ metres, of the tide in a harbour $t$ hours after midnight is modelled by $H=6+4\sin\left(\dfrac{\pi t}{6}\right)$, for $0\le t\le12$.

(a) Write down the maximum and minimum values of $H$. [2]

(b) Write down the period of the model. [1]

(c) Boats can enter the harbour only when the tide height exceeds $8$ m. Find the times between $t=0$ and $t=12$ when boats can enter, and the total time for which they can enter. [4]

Maximum $10$ mA1
Minimum $2$ mA1
Period $=\dfrac{2\pi}{\pi/6}=12$ hoursA1
Sets $6+4\sin\frac{\pi t}6=8$ (or $>8$), i.e. $\sin\dfrac{\pi t}6=\dfrac12$, or uses intersection of $H$ with $H=8$ on a GDCM1
$t=1$ (from $\frac{\pi t}6=\frac\pi6$)A1
$t=5$ (from $\frac{\pi t}6=\frac{5\pi}6$), so boats can enter for $1A1
Total time $4$ hoursA1
Answer: (a) $10$ m, $2$ m (b) $12$ h (c) between $t=1$ and $t=5$ (4 hours)
Examiner tip: Take care with the inequality: use the graph of $\sin$ to see that the values above $\tfrac12$ lie between the two solutions.

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