Further trigonometry – Year 13 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(\cos0.06\approx1-\dfrac{0.0036}2=1-0.0018=0.9982\). The calculator gives 0.998201, so the approximation is accurate to 4 decimal places.
   2. \(1-\cos4\theta\approx\dfrac{(4\theta)^2}2=8\theta^2\) and \(\theta\sin2\theta\approx\theta\cdot2\theta=2\theta^2\). The ratio is approximately \(\dfrac{8\theta^2}{2\theta^2}=\) 4.
   3. \(\sin3\theta\approx3\theta\) and \(\tan2\theta\approx2\theta\), so the numerator is \(\approx6\theta^2\) and the ratio is approximately 6.

2. Answer to Question 2

   Method: \(\sec\theta=\dfrac1{\cos\theta}\), \(\operatorname{cosec}\theta=\dfrac1{\sin\theta}\) and \(\cot\theta=\dfrac1{\tan\theta}\). Find the sine, cosine or tangent first (with its sign from the quadrant), then take the reciprocal.

   1. Use \(\cos60^\circ=\tfrac12\) and \(\sin30^\circ=\tfrac12\). \(\sec60^\circ=\dfrac1{\cos60^\circ}=\dfrac1{1/2}=2\). \(\operatorname{cosec}30^\circ=\dfrac1{\sin30^\circ}=2\). Both values are equal to 2.
   2. \(150^\circ\) is in the second quadrant, where tangent is negative. \(\cot45^\circ=\dfrac1{\tan45^\circ}=1\). \(\tan150^\circ=-\dfrac1{\sqrt3}\) so \(\cot150^\circ=-\sqrt3\). So \(\cot45^\circ=1\) and \(\cot150^\circ=-\sqrt3\).
   3. \(120^\circ\) is in the second quadrant (cosine negative) and \(240^\circ\) in the third (sine negative). \(\cos120^\circ=-\tfrac12\), so \(\sec120^\circ=-2\). \(\sin240^\circ=-\dfrac{\sqrt3}2\), so \(\operatorname{cosec}240^\circ=-\dfrac2{\sqrt3}=-\dfrac{2\sqrt3}3\). Rationalising the denominator gives the neat form \(-\dfrac{2\sqrt3}3\).

3. Answer to Question 3

   1. \(\cos x=-\tfrac12\) so \(x=\) 120°, 240°.
   2. Use \(\tan^2x=\sec^2x-1\): \(\sec^2x-1+\sec x=1\), so \(\sec^2x+\sec x-2=0\), \((\sec x+2)(\sec x-1)=0\). \(\sec x=1\) gives \(x=0^\circ\). \(\sec x=-2\) gives \(x=120^\circ,240^\circ\). Solutions: 0°, 120°, 240°.
   3. Use \(\operatorname{cosec}^2x=1+\cot^2x\): \(\cot^2x-\cot x-2=0\), \((\cot x-2)(\cot x+1)=0\). \(\cot x=2\): \(\tan x=\tfrac12\), \(x=26.6^\circ,206.6^\circ\). \(\cot x=-1\): \(\tan x=-1\), \(x=135^\circ,315^\circ\). Solutions: 26.6°, 135°, 206.6°, 315°.

4. Answer to Question 4

   Method: use the compound angle formulae \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\), \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\) and \(\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\), with exact values of \(30^\circ\), \(45^\circ\) and \(60^\circ\).

   1. Use \(\sin(A-B)\) with \(A=45^\circ\) and \(B=30^\circ\). \(\sin(45^\circ-30^\circ)=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\dfrac{\sqrt2}2\cdot\dfrac{\sqrt3}2-\dfrac{\sqrt2}2\cdot\dfrac12=\dfrac{\sqrt6-\sqrt2}4\). So \(\sin15^\circ=\dfrac{\sqrt6-\sqrt2}4\approx0.2588\).
   2. Write \(75^\circ=45^\circ+30^\circ\) and use \(\cos(A+B)\). \(\cos(45^\circ+30^\circ)=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ=\dfrac{\sqrt6}4-\dfrac{\sqrt2}4=\dfrac{\sqrt6-\sqrt2}4\). This equals \(\sin15^\circ\), as expected since \(\cos75^\circ=\sin(90^\circ-75^\circ)\).
   3. Use \(\tan60^\circ=\sqrt3\) and \(\tan45^\circ=1\), then rationalise the denominator. \(\tan(60^\circ+45^\circ)=\dfrac{\sqrt3+1}{1-\sqrt3}\). Multiply by \(\dfrac{1+\sqrt3}{1+\sqrt3}\): \(\dfrac{(\sqrt3+1)^2}{1-3}=\dfrac{4+2\sqrt3}{-2}=-(2+\sqrt3)\). The answer is negative because \(105^\circ\) is in the second quadrant.

5. Answer to Question 5

   1. \(2\sin x\cos x-\sin x=0\), so \(\sin x(2\cos x-1)=0\) (do not divide by \(\sin x\)). \(\sin x=0\) gives \(x=0^\circ,180^\circ\). \(\cos x=\tfrac12\) gives \(x=60^\circ,300^\circ\). Solutions: 0°, 60°, 180°, 300°.
   2. Use \(\cos2x=2\cos^2x-1\): \(2\cos^2x+3\cos x+1=0\), \((2\cos x+1)(\cos x+1)=0\). \(\cos x=-\tfrac12\) gives \(x=120^\circ,240^\circ\). \(\cos x=-1\) gives \(x=180^\circ\). Solutions: 120°, 180°, 240°.

6. Answer to Question 6

   1. \(R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha\). Comparing: \(R\cos\alpha=5\), \(R\sin\alpha=12\). \(R=\sqrt{25+144}=13\) and \(\tan\alpha=\dfrac{12}5\), so \(\alpha=67.38^\circ\). Hence \(f(x)=13\sin(x+67.38^\circ)\).
   2. Maximum 13, minimum \(-13\). The maximum occurs when \(x+67.38^\circ=90^\circ\), so \(x=22.62^\circ\).
   3. \(13\sin(x+67.38^\circ)=6.5\) gives \(\sin(x+67.38^\circ)=0.5\). So \(x+67.38^\circ=30^\circ,150^\circ,390^\circ\) (the values \(30^\circ\) gives a negative \(x\)). \(x=82.62^\circ\) or \(x=322.62^\circ\). Solutions: 82.6°, 322.6°.

7. Answer to Question 7

   Method: the small-angle approximations only work in radians: \(\sin\theta\approx\theta\), \(\tan\theta\approx\theta\), \(\cos\theta\approx1-\dfrac{\theta^2}{2}\). Replace each function, simplify, and the powers of \(\theta\) usually cancel.

   1. 0.08 radians is a small angle, so replace \(\tan0.08\) by \(0.08\). \(\tan\theta\approx\theta\), so \(h\approx400\times0.08=\) 32 m. So the tower is about 32 m tall.
   2. Use the calculator in radian mode. \(h=400\tan0.08=400\times0.080171=\) 32.1 m. So the true height is about 32.1 m.
   3. Divide the difference by the true value and multiply by 100. Error \(=\dfrac{32.068-32}{32.068}\times100=0.21\%\). So the approximation is very good, with an error of about 0.2%.
   4. Compare the approximation with the true value for a large angle. The approximation needs \(\theta\) close to 0. For \(\theta=1\), \(\tan1=1.557\) but the approximation gives 1, an error of over 35%. So small-angle approximations should only be used for angles close to zero.

8. Answer to Question 8

   Method: \(\sec\theta=\dfrac1{\cos\theta}\), \(\operatorname{cosec}\theta=\dfrac1{\sin\theta}\) and \(\cot\theta=\dfrac1{\tan\theta}\). Find the sine, cosine or tangent first (with its sign from the quadrant), then take the reciprocal.

   1. Use \(\operatorname{cosec}5^\circ=\dfrac1{\sin5^\circ}\) and \(\cot5^\circ=\dfrac1{\tan5^\circ}\). \(L=\dfrac{0.45}{\sin5^\circ}=5.16\) m and \(d=0.45\cot5^\circ=\dfrac{0.45}{\tan5^\circ}=5.14\) m. The two lengths are close because the angle is small.
   2. A gradient of 1 in 12 means a rise of 1 m for every 12 m across. \(\tan\theta=\tfrac1{12}\) gives \(\theta=\arctan\dfrac1{12}=4.76^\circ\). The ramp must be at most 4.76°. So the angle must be at most \(4.76^\circ\).
   3. A smaller angle gives a longer ramp, so the largest allowed angle gives the shortest ramp. The minimum length is \(L=0.45\operatorname{cosec}4.76^\circ=5.42\) m and the run is \(d=0.45\times12=5.4\) m. The 5° ramp is slightly too steep. So the ramp must be at least 5.42 m long, with a horizontal run of 5.4 m.

9. Answer to Question 9

   1. \(R=\sqrt{9+16}=5\), \(\tan\alpha=\dfrac43\), \(\alpha=53.13^\circ\). So \(h=6+5\sin(30t+53.13)^\circ\).
   2. Greatest depth \(6+5=11\) m, least depth \(6-5=1\) m.
   3. Need \(5\sin(30t+53.13)^\circ\ge3\), so \(\sin(30t+53.13)^\circ\ge0.6\). \(\sin^{-1}0.6=36.87^\circ\), so the argument lies between \(36.87^\circ\) and \(143.13^\circ\), or between \(396.87^\circ\) and \(503.13^\circ\). First interval: \(0\le30t\le90\), giving \(0\le t\le3\). Second interval: \(343.74\le30t\le360\) (the limit of the range), giving \(11.46\le t\le12\). So boats can enter from midnight until 03:00, and from about 11:28 onwards (11.46 h \(=\) 11 h 27.5 min).

10. Answer to Question 10

    1. The range is largest when \(\sin2\theta\) takes its largest value. The maximum value of \(\sin2\theta\) is 1, when \(2\theta=90^\circ\), so \(\theta=45^\circ\). Maximum range \(=\dfrac{400}{9.8}=40.8\) m. So the maximum range is about 40.8 m, at an angle of \(45^\circ\).
    2. Substitute \(R=30\), \(u=20\) and \(g=9.8\), then rearrange. \(30=\dfrac{400\sin2\theta}{9.8}\) gives \(\sin2\theta=\dfrac{294}{400}=0.735\). This is the required result.
    3. Since \(0^\circ<2\theta<180^\circ\), the sine equation has two solutions for \(2\theta\). \(2\theta=\sin^{-1}0.735=47.31^\circ\) or \(180^\circ-47.31^\circ=132.69^\circ\). So \(\theta=23.6^\circ\) (a flat kick) or \(66.4^\circ\) (a high kick). The two angles add up to \(90^\circ\), a general property of projectiles.

11. Answer to Question 11

    Method: use the compound angle formulae \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\), \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\) and \(\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\), with exact values of \(30^\circ\), \(45^\circ\) and \(60^\circ\).

    1. Substitute \(\tan A=\tfrac13\) and \(\tan B=\tfrac12\) into the formula for \(\tan(A+B)\). \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{\frac13+\frac12}{1-\frac16}=\dfrac{5/6}{5/6}=1\). This is the required result.
    2. Both angles are acute, so \(A+B\) lies between \(0^\circ\) and \(180^\circ\); its tangent is positive, so it is acute. \(\tan(A+B)=1\) with \(A+B\) acute (here \(18.43^\circ+26.57^\circ\)), so \(A+B=45^\circ\). So the path has turned through \(45^\circ\) in total.
    3. Since \(A+B=45^\circ\), use the exact value of \(\sin45^\circ\). \(\sin45^\circ=\dfrac{\sqrt2}2\). So \(\sin(A+B)=\dfrac{\sqrt2}2\).
    4. Use the formula for \(\tan(B-A)\) and multiply the top and bottom by 6. \(\tan(B-A)=\dfrac{\frac12-\frac13}{1+\frac16}=\dfrac{1/6}{7/6}=\dfrac17\). So \(\tan(B-A)=\tfrac17\).

12. Answer to Question 12

    Method: \(\sec\theta=\dfrac1{\cos\theta}\), \(\operatorname{cosec}\theta=\dfrac1{\sin\theta}\) and \(\cot\theta=\dfrac1{\tan\theta}\). Find the sine, cosine or tangent first (with its sign from the quadrant), then take the reciprocal.

    1. Use \(\operatorname{cosec}30^\circ=2\) and \(\cot30^\circ=\sqrt3\). Beam length \(=12\operatorname{cosec}30^\circ=12\times2=24\) m. Distance \(=12\cot30^\circ=12\sqrt3=20.8\) m. So the beam is 24 m long and lights a spot about 20.8 m away.
    2. Rearrange to find \(\cot\theta\), then take the reciprocal to get \(\tan\theta\). \(12\cot\theta=20\) gives \(\tan\theta=\dfrac{12}{20}=0.6\), so \(\theta=31.0^\circ\). So the beam must make about \(31.0^\circ\) with the horizontal.
    3. Dividing \(\sin^2\theta+\cos^2\theta=1\) by \(\sin^2\theta\): \(1+\cot^2\theta=\operatorname{cosec}^2\theta\). With \(\cot\theta=\dfrac{20}{12}=\dfrac53\): \(\operatorname{cosec}^2\theta=1+\dfrac{25}9=\dfrac{34}9\), so \(\operatorname{cosec}\theta=\dfrac{\sqrt{34}}3\) and the beam length is \(12\times\dfrac{\sqrt{34}}3=4\sqrt{34}=23.3\) m. So the beam length is \(4\sqrt{34}\approx23.3\) m. Check: \(\sqrt{12^2+20^2}=\sqrt{544}=23.3\) by Pythagoras.

13. Answer to Question 13

    Method: the small-angle approximations only work in radians: \(\sin\theta\approx\theta\), \(\tan\theta\approx\theta\), \(\cos\theta\approx1-\dfrac{\theta^2}{2}\). Replace each function, simplify, and the powers of \(\theta\) usually cancel.

    1. Replace \(\sin\theta\) and \(\cos\theta\) by their approximations. \(\sin\theta\approx\theta\) gives \(L\sin\theta\approx L\theta\). \(1-\cos\theta\approx\dfrac{\theta^2}2\) gives \(L(1-\cos\theta)\approx\dfrac{L\theta^2}2\). These are the required approximations.
    2. Substitute \(L=2.5\) and \(\theta=0.2\). Displacement \(\approx2.5\times0.2=0.5\) m. Rise \(\approx\dfrac{2.5\times0.04}2=0.05\) m. So the seat moves about 0.5 m sideways and rises about 5 cm.
    3. Use the calculator in radian mode for the exact values. Exact displacement \(=2.5\sin0.2=0.4967\) m (error \(\dfrac{0.5-0.4967}{0.4967}=0.67\%\)). Exact rise \(=2.5(1-\cos0.2)=0.04983\) m (error \(\dfrac{0.05-0.04983}{0.04983}=0.33\%\)). Both errors are below 1%, so the approximations are good for \(\theta=0.2\).

14. Answer to Question 14

    1. Rearrange the double angle formula \(\cos2\theta=2\cos^2\theta-1\) to make \(\cos^2\theta\) the subject. \(\cos2\theta=2\cos^2\theta-1\), so \(\cos^2\theta=\dfrac{1+\cos2\theta}2\). Hence \(I=\dfrac{I_0}2(1+\cos2\theta)\). This is the required result.
    2. Divide by \(I_0\) and take the positive square root since \(\theta\) is acute. \(\cos^2\theta=0.75\) gives \(\cos\theta=\dfrac{\sqrt3}2\) and \(\theta=30^\circ\). So \(\theta=30^\circ\).
    3. Divide by \(I_0\), take the positive square root and use the inverse cosine. \(\cos^2\theta=0.2\) gives \(\cos\theta=0.4472\), so \(\theta=63.4^\circ\). So \(\theta\approx63.4^\circ\).
    4. \(\cos^2\theta\le1\), so \(I\le I_0\); the maximum occurs at \(\theta=0^\circ\).

15. Answer to Question 15

    1. LHS \(=\dfrac1{\sin2x}+\dfrac{\cos2x}{\sin2x}=\dfrac{1+\cos2x}{\sin2x}=\dfrac{2\cos^2x}{2\sin x\cos x}=\dfrac{\cos x}{\sin x}=\cot x\). \(\blacksquare\)
    2. \(\sin3x=\sin(2x+x)=\sin2x\cos x+\cos2x\sin x=2\sin x\cos^2x+(1-2\sin^2x)\sin x\). Replace \(\cos^2x=1-\sin^2x\): \(2\sin x-2\sin^3x+\sin x-2\sin^3x=3\sin x-4\sin^3x\). \(\blacksquare\)
    3. \(3\sin x-4\sin^3x=\sin x\) gives \(2\sin x-4\sin^3x=0\), \(2\sin x(1-2\sin^2x)=0\). \(\sin x=0\): \(x=0^\circ,180^\circ\). \(\sin^2x=\tfrac12\) and \(\sin x>0\) in the range: \(\sin x=\dfrac1{\sqrt2}\), \(x=45^\circ,135^\circ\). Solutions: 0°, 45°, 135°, 180°.
    4. With \(x=15^\circ\): \(\cot15^\circ=\operatorname{cosec}30^\circ+\cot30^\circ=2+\sqrt3\). \(\blacksquare\)

16. Answer to Question 16

    1. \(t=6\): \(14+3\sin90^\circ+4\cos90^\circ=14+3+0=17\). \(t=12\): \(14+3\sin180^\circ+4\cos180^\circ=14+0-4=10\). \(t=3\): \(14+3\sin45^\circ+4\cos45^\circ=14+\dfrac{7\sqrt2}2=18.95\approx19.0\).
    2. \(R=5\), \(\tan\alpha=\dfrac43\), \(\alpha=53.13^\circ\). \(T=14+5\sin(15t+53.13)^\circ\). Maximum \(19\) °C, minimum \(9\) °C.
    3. Maximum when \(15t+53.13=90\), so \(t=2.46\) hours after 11:00, that is 13:28.
    4. \(14+5\sin(15t+53.13)^\circ\ge17\) gives \(\sin(15t+53.13)^\circ\ge0.6\). The argument lies in \([36.87^\circ,143.13^\circ]\) giving \(0\le t\le6\), and in \([396.87^\circ,503.13^\circ]\), giving \(22.92\le t\le24\) (within range). Total \(=6+(24-22.92)=7.08\) hours.

17. Answer to Question 17

    1. \(\sec^2x=1+\tan^2x\), so \(2+2\tan^2x+3\tan x-5=0\), i.e. \(2\tan^2x+3\tan x-3=0\).
    2. \(\tan x=\dfrac{-3\pm\sqrt{9+24}}4=\dfrac{-3\pm\sqrt{33}}4\), so \(\tan x=0.6861\) or \(-2.1861\). \(\tan x=0.6861\): \(x=34.5^\circ\) and \(214.5^\circ\). \(\tan x=-2.1861\): \(x=180^\circ-65.4^\circ=114.6^\circ\) and \(294.6^\circ\). Solutions: 34.5°, 114.6°, 214.5°, 294.6°.
    3. \(\sec x=\dfrac1{\cos x}\) and \(|\cos x|\le1\), so \(|\sec x|\ge1\). In the table the values are all at least 1.
    4. The tangent has period \(180^\circ\), so every value \(k\) is reached twice in \(360^\circ\). That gives two solutions for each root, four in total.

18. Answer to Question 18

    1. \(\cos A=\sqrt{1-\frac{64}{289}}=\dfrac{15}{17}\), \(\sin B=\sqrt{1-\frac9{25}}=\dfrac45\) (both positive as the angles are acute).
    2. \(\sin(A+B)=\sin A\cos B+\cos A\sin B=\dfrac8{17}\cdot\dfrac35+\dfrac{15}{17}\cdot\dfrac45=\dfrac{24+60}{85}=\dfrac{84}{85}\).
    3. \(\cos(A+B)=\cos A\cos B-\sin A\sin B=\dfrac{45}{85}-\dfrac{32}{85}=\dfrac{13}{85}\). Check: \(\dfrac{84^2+13^2}{85^2}=\dfrac{7056+169}{7225}=1\).
    4. \(\tan A=\dfrac8{15}\) and \(\tan B=\dfrac43\). \(\tan(A-B)=\dfrac{\frac8{15}-\frac43}{1+\frac8{15}\cdot\frac43}=\dfrac{-\frac{12}{15}}{\frac{77}{45}}=-\dfrac45\cdot\dfrac{45}{77}=-\dfrac{36}{77}\).
    5. \(A=28.07^\circ\) and \(B=53.13^\circ\), so \(A+B=81.2^\circ\). This is acute, consistent with \(\cos(A+B)=\frac{13}{85}>0\).

19. Answer to Question 19

    1. \(R=5\), \(\alpha=53.13^\circ\), so \(3\sin x+4\cos x=5\sin(x+53.13^\circ)\).
    2. The denominator \(7+5\sin(x+53.13^\circ)\) varies from \(2\) to \(12\). So \(B\) is greatest when the denominator is smallest: \(B_{max}=\dfrac12\), and \(B_{min}=\dfrac1{12}\).
    3. The denominator is least when \(\sin(x+53.13^\circ)=-1\), so \(x+53.13=270\), giving \(x=216.87^\circ\).
    4. \(7+5\sin(x+53.13^\circ)=4\) gives \(\sin(x+53.13^\circ)=-0.6\). So \(x+53.13=216.87^\circ\) or \(323.13^\circ\), giving \(x=163.74^\circ\) or \(x=270^\circ\). Check \(x=270\): \(7-3+0=4\). Solutions: 163.7°, 270°.

20. Answer to Question 20

    1. \(3\times36^\circ+2\times36^\circ=180^\circ\), so \(3\theta=180^\circ-2\theta\) and \(\cos3\theta=\cos(180^\circ-2\theta)=-\cos2\theta\).
    2. \(4c^3-3c=-(2c^2-1)\), so \(4c^3+2c^2-3c-1=0\).
    3. At \(c=-1\): \(-4+2+3-1=0\), so \((c+1)\) is a factor. Dividing: \(4c^3+2c^2-3c-1=(c+1)(4c^2-2c-1)\). Since \(\cos36^\circ>0\), \(c\ne-1\) so \(4c^2-2c-1=0\) and \(c=\dfrac{2\pm\sqrt{4+16}}8=\dfrac{1\pm\sqrt5}4\). The positive root gives \(\cos36^\circ=\dfrac{1+\sqrt5}4\).
    4. \(c^2=\dfrac{(1+\sqrt5)^2}{16}=\dfrac{6+2\sqrt5}{16}\), so \(\cos72^\circ=2c^2-1=\dfrac{6+2\sqrt5}8-1=\dfrac{2\sqrt5-2}8=\dfrac{\sqrt5-1}4\). Then \(\cos36^\circ-\cos72^\circ=\dfrac{1+\sqrt5-\sqrt5+1}4=\dfrac12\). \(\blacksquare\)