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IB Mathematics: Applications and Interpretation HL · 12.10: Geometry & Trigonometry · 12.10.11
The unit circle and trigonometric equations Problem set B Worked solutions
Skills
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A point P lies on a circle with centre O at the origin and radius 6.5 cm. The line OP makes an angle of 310° with the positive x-axis, measured anticlockwise. Find the coordinates of P.
- \(x=r\cos\theta=6.5\cos310^\circ=4.178\)
- \(y=r\sin\theta=6.5\sin310^\circ=-4.979\)
- \(P=(4.18,\;-4.98)\)
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The angle θ is between 270° and 360° and \(\cos\theta=\frac{12}{37}\). Find the exact value of \(\sin\theta\).
- \(\sin^2\theta=1-\left(\frac{12}{37}\right)^2=\frac{1225}{1369}\)
- \(\sin\theta=\pm\frac{35}{37}\)
- θ is in the fourth quadrant, where sine is negative, so \(\sin\theta=-\frac{35}{37}\)
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The angle θ is between 180° and 270° and \(\sin\theta=-0.28\).
(a) Find \(\cos\theta\).
- \(\cos^2\theta=1-(-0.28)^2=0.9216\)
- \(\cos\theta=\pm0.96\); θ is in the third quadrant, so \(\cos\theta=-0.96\)
(b) Find \(\tan\theta\).
- \(\tan\theta=\frac{-0.28}{-0.96}=\frac{7}{24}\)
- Tangent is positive in the third quadrant
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Given that sin 37° = 0.602 and cos 37° = 0.799 (3 s.f.), use the unit circle to answer the following.
(a) Write down another angle between 0° and 360° with the same sine as 37°.
- Same sine: \(180^\circ-37^\circ=143^\circ\)
(b) Write down another angle between 0° and 360° with the same cosine as 37°.
- Same cosine: \(360^\circ-37^\circ=323^\circ\)
(c) cos 1.3 = 0.267 (radians). Write down another angle between 0 and 2π with the same cosine.
- Same cosine in radians: \(2\pi-1.3=4.983\)
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Solve each equation for \(0^\circ\le x\le360^\circ\).
(a) \(\sin x=0.46\)
- \(x=\sin^{-1}0.46=27.4^\circ\)
- \(180^\circ-27.4^\circ=152.6^\circ\)
(b) \(\cos x=-0.62\)
- \(x=\cos^{-1}(-0.62)=128.3^\circ\)
- \(360^\circ-128.3^\circ=231.7^\circ\)
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Solve \(6\sin 2x=2.5\) for \(0\le x\le\pi\). Give the solutions in radians to 3 decimal places.
- Radian mode. Graph \(y=6\sin\,2x\) and \(y=2.5\) for \(0\le x\le\pi\)
- Intersections: \(x=0.215\), \(x=1.356\)
In context
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The depth of water in a harbour is \(d=6-2.4\cos\left(\frac{\pi t}{6.2}\right)\) metres, where t is the time in hours after low tide, for \(0\le t\le12.4\).
(a) Find the depth of water 2 hours after low tide.
- \(d=6-2.4\cos\frac{2\pi}{6.2}=\) \(4.730\)
- 4.73 m
(b) Find the times when the depth is 7 m.
- Graph the model and the line in radian mode
- \(t=3.948\) and \(t=8.452\)
(c) A ship can enter the harbour when the depth is more than 7 m. For how long in one tide cycle can it enter?
- \(8.452-3.948=4.504\)
- 4.50
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The tip P of a fan blade moves round a circle with centre the origin and radius 25 cm. At t = 0 it is at (25, 0). It moves anticlockwise through 0.7 radians every second. Use radian mode.
(a) Find the coordinates of P after 3 seconds.
- Angle after 3 s: \(\theta=0.7\times3=2.1\) radians
- \(x=25\cos2.1=-12.621,\;y=25\sin2.1=21.580\)
(b) Find the first time at which the y-coordinate of P is 15 cm.
- \(25\sin(0.7t)=15\), so \(\sin(0.7t)=0.6\)
- \(0.7t=\sin^{-1}0.6=0.6435\), \(t=0.919\) s
Stretch
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The point P(0.8, y) lies on the unit circle, in the fourth quadrant. The angle θ is measured anticlockwise from the positive x-axis, with 0 ≤ θ ≤ 2π.
(a) Find y.
- \(x^2+y^2=1\), so \(y^2=1-0.64=0.36\)
- Fourth quadrant: y is negative, \(y=-0.6\)
(b) Find θ in radians.
- \(\cos^{-1}0.8=0.6435\), the angle in the first quadrant
- \(\theta=2\pi-0.6435=5.6397\)
(c) Find tan θ.
- \(\tan\theta=\frac{-0.6}{0.8}=-\frac34\)
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Consider the equation \(2\sin x=5\cos x\), for \(0^\circ\le x\le360^\circ\).
(a) Show that the equation can be written as \(\tan x=\frac{5}{2}\).
- Divide both sides by \(2\cos x\): \(\frac{\sin x}{\cos x}=\frac{5}{2}\)
- \(\tan x=\frac{5}{2}\)
(b) Hence solve the equation.
- \(x=\tan^{-1}\frac{5}{2}=68.2^\circ\)
- \(x=68.2^\circ+180^\circ=248.2^\circ\)
jmaths.xyz · AI HL · 12.10.11 The unit circle and trigonometric equations