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© Boardworks Ltd 2005 1 of 58 S4 Further trigonometry KS4 Mathematics

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© Boardworks Ltd 2005 2 of 58 A A A A A A S4.1 Sin, cos and tan of any angle S4 Further trigonometry Contents S3.4 Area of a triangle using ½ ab sin C S3.5 The sine rule S4.3 Graphs of trigonometric functions S4.2 Sin, cos and tan of 30°, 45° and 60° S4.6 The cosine rule

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© Boardworks Ltd 2005 3 of 58 The opposite and adjacent sides Suppose we have a right-angled triangle with hypotenuse h and acute angle θ. θ h a) Write an expression for the length of the opposite side in terms of h and θ. b) Write an expression for the length of the adjacent side in terms of h and θ.

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© Boardworks Ltd 2005 4 of 58 The opposite and adjacent sides Suppose we have a right-angled triangle with hypotenuse h and acute angle θ. θ h a) sin θ = opp hyp opp = hyp × sin θ opp = h sin θ b) cos θ = adj hyp adj = hyp × cos θ adj = h cos θ

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© Boardworks Ltd 2005 5 of 58 The opposite and adjacent sides So, for any right-angled triangle with hypotenuse h and acute angle θ. We can label the opposite and adjacent sides as follows: θ h h sin θ h cos θ We can write, tan θ = h sin θ h cos θ tan θ = sin θ cos θ opposite adjacent

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© Boardworks Ltd 2005 6 of 58 The sine of any angle

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© Boardworks Ltd 2005 7 of 58 Sine of angles in the second quadrant We have seen that the sine of angles in the first and second quadrants are positive. The sine of angles in the third and fourth quadrants are negative. In the second quadrant, 90° < θ < 180°. sin θ = sin (180° – θ ) For example, sin 130° =sin (180° – 130°) = sin 50° = 0.766 (to 3 sig. figs)

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© Boardworks Ltd 2005 8 of 58 Sine of angles in the third quadrant In the third quadrant, 180° < θ < 270°. sin θ = –sin ( θ – 180°) For example, sin 220° =– sin (220° – 180°) = – sin 40° = – 0.643 (to 3 sig. figs) Verify, using a scientific calculator, that sin 220° = –sin 40°

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© Boardworks Ltd 2005 9 of 58 Sine of angles in the fourth quadrant In the fourth quadrant, 270° θ > –90° sin θ = –sin(360° – θ ) or sin – θ = –sin θ For example, sin 300° =–sin (360° – 300°) = –sin 60° = –0.866 (to 3 sig. figs) sin –35° =–sin 35° = –0.574 (to 3 sig. figs)

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© Boardworks Ltd 2005 10 of 58 The cosine of any angle

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© Boardworks Ltd 2005 11 of 58 Cosine of angles in the second quadrant We have seen that the cosines of angles in the first and fourth quadrants are positive. The cosines of angles in the second and third quadrants are negative. In the second quadrant, 90° < θ < 180°. cos θ = –cos (180° – θ ) For example, cos 100° =–cos (180° – 100°) = –cos 80° = –0.174 (to 3 sig. figs)

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© Boardworks Ltd 2005 12 of 58 Cosine of angles in the third quadrant In the third quadrant, 180° < θ < 270°. cos θ = –cos ( θ – 180°) For example, cos 250° =–cos (250° – 180°) = –cos 70° = –0.342 (to 3 sig. figs.) Verify, using a scientific calculator, that cos 250° = –cos 70°

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© Boardworks Ltd 2005 13 of 58 Sine of angles in the fourth quadrant In the fourth quadrant, 270° θ > –90° cos θ = cos(360° – θ ) or cos – θ = cos θ For example, cos 317° =cos (360° – 317°) = cos 43° = 0.731 (to 3 sig. figs.) cos –28° =cos 28° = 0.883 (to 3 sig. figs.)

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© Boardworks Ltd 2005 14 of 58 The tangent of any angle

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© Boardworks Ltd 2005 15 of 58 The tangent of any angle

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© Boardworks Ltd 2005 16 of 58 Tangent of angles in the second quadrant We have seen that the tangent of angles in the first and third quadrants are positive. The tangent of angles in the second and fourth quadrants are negative. In the second quadrant, 90° < θ < 180°. tan θ = –tan (180° – θ ) For example, tan 116° =–tan (180° – 116°) = –tan 64° = –2.05 (to 3 sig. figs)

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© Boardworks Ltd 2005 17 of 58 Tangent of angles in the third quadrant In the third quadrant, 180° < θ < 270°. tan θ = tan ( θ – 180°) For example, tan 236° =tan (236° – 180°) = tan 56° = 1.48 (to 3 sig. figs) Verify, using a scientific calculator, that tan 236° = tan 56°

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© Boardworks Ltd 2005 18 of 58 Tangent of angles in the fourth quadrant In the fourth quadrant, 270° θ > –90° tan θ = –tan(360° – θ ) or tan – θ = –tan θ For example, tan 278° =–tan (360° – 278°) = –tan 82° = –7.12 (to 3 sig. figs) tan –16° =–tan 16° = –0.287 (to 3 sig. figs)

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© Boardworks Ltd 2005 19 of 58 Sin, cos and tan of angles between 0° and 360° The sin, cos and tan of angles in the first quadrant are positive. In the second quadrant: sin θ = sin (180° – θ ) cos θ = –cos (180° – θ ) tan θ = –tan (180° – θ ) sin θ = sin (180° – θ ) cos θ = –cos (180° – θ ) tan θ = –tan (180° – θ ) In the third quadrant: sin θ = –sin ( θ – 180°) cos θ = –cos ( θ – 180°) tan θ = tan ( θ – 180°) sin θ = –sin ( θ – 180°) cos θ = –cos ( θ – 180°) tan θ = tan ( θ – 180°) In the fourth quadrant: sin θ = –sin (360° – θ ) cos θ = cos (360° – θ ) tan θ = –tan(180° – θ ) sin θ = –sin (360° – θ ) cos θ = cos (360° – θ ) tan θ = –tan(180° – θ )

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© Boardworks Ltd 2005 20 of 58 3 rd quadrant2 nd quadrant1 st quadrant4 th quadrant Tangent is positive T T Sine is positive S S All are positive A A Remember CAST We can use CAST to remember in which quadrant each of the three ratios are positive. Cosine is positive C C

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© Boardworks Ltd 2005 21 of 58 Positive or negative?

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© Boardworks Ltd 2005 22 of 58 Find the equivalent ratio

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© Boardworks Ltd 2005 23 of 58 Solving equations in θ

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© Boardworks Ltd 2005 24 of 58 A A A A A A S4.2 Sin, cos and tan of 30°, 45° and 60° Contents S4 Further trigonometry S4.3 Graphs of trigonometric functions S3.4 Area of a triangle using ½ ab sin C S4.6 The cosine rule S3.5 The sine rule S4.1 Sin, cos and tan of any angle

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© Boardworks Ltd 2005 25 of 58 Sin, cos and tan of 45° A right-angled isosceles triangle has two acute angels of 45°. 45° Suppose the equal sides are of unit length. 1 1 Using Pythagoras’ theorem, = 2 22 We can use this triangle to write exact values for sin, cos and tan 45°: sin 45° = 1 22 cos 45° = 1 22 tan 45° = 1 The hypotenuse = 1² + 1²

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© Boardworks Ltd 2005 26 of 58 Sin, cos and tan of 30° Suppose we have an equilateral triangle of side length 2. We can use this triangle to write exact values for sin, cos and tan 30°: sin 30° = 1 2 cos 30° = 33 2 tan 30° = 22 2 60° 2 30° 1 If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. Using Pythagoras’ theorem, = 3 33 The height of the triangle = 2² – 1² 1 33

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© Boardworks Ltd 2005 27 of 58 Sin, cos and tan of 60° Suppose we have an equilateral triangle of side length 2. 2 60° 30° 1 If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. Using Pythagoras’ theorem, = 3 33 The height of the triangle = 2² – 1² We can also use this triangle to write exact values for sin, cos and tan 60°: sin 60° = cos 60° = 33 2 tan 60° = 1 2 33

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© Boardworks Ltd 2005 28 of 58 Sin, cos and tan of 30°, 45° and 60° The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 30° sin cos tan 45°60° 1 22 1 22 1 2 33 2 1 33 2 1 2 Use this table to write the exact value of sin 150°: sin 150° = 1 2 33 1 33

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© Boardworks Ltd 2005 29 of 58 Sin, cos and tan of 30°, 45° and 60° The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 30° sin cos tan 45°60° 1 22 1 22 1 2 33 2 1 33 2 1 2 Use this table to write the exact value of cos 135°: –1–1 22 cos 135° = 33 1 33

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© Boardworks Ltd 2005 30 of 58 Sin, cos and tan of 30°, 45° and 60° The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 30° sin cos tan 45°60° 1 22 1 22 1 2 33 2 33 1 1 33 33 2 1 2 Use this table to write the exact value of tan 120°: tan 120° = –3–3

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© Boardworks Ltd 2005 31 of 58 Sin, cos and tan of 30°, 45° and 60° Write the following ratios exactly: 1) cos 300° = 1 2 3) tan 240° = 5) cos –30° = 7) sin 210° = 2) tan 315° = 4) sin –330° = 6) tan –135° = 8) cos 315° = –1 33 1 2 33 2 1 –1–1 2 1 22

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© Boardworks Ltd 2005 32 of 58 A A A A A A S4.3 Graphs of trigonometric functions Contents S3.4 Area of a triangle using ½ ab sin C S4.6 The cosine rule S3.5 The sine rule S4 Further trigonometry S4.1 Sin, cos and tan of any angle S4.2 Sin, cos and tan of 30°, 45° and 60°

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© Boardworks Ltd 2005 33 of 58 The graph of sin θ

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© Boardworks Ltd 2005 34 of 58 The graph of cos θ

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© Boardworks Ltd 2005 35 of 58 The graph of tan θ

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© Boardworks Ltd 2005 36 of 58 Transforming trigonometric graphs

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© Boardworks Ltd 2005 37 of 58 A A A A A A S3.4 Area of a triangle using ½ ab sin C Contents S4.6 The cosine rule S3.5 The sine rule S4 Further trigonometry S4.1 Sin, cos and tan of any angle S4.3 Graphs of trigonometric functions S4.2 Sin, cos and tan of 30°, 45° and 60°

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© Boardworks Ltd 2005 38 of 58 The area of a triangle Remember, b h Area of a triangle = bh 1 2

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© Boardworks Ltd 2005 39 of 58 The area of a triangle Suppose that instead of the height of a triangle, we are given the base, one of the sides and the included angle. For example, What is the area of triangle ABC? A B C 7 cm 4 cm 47° Let’s call the height of the triangle h. h We can find h using the sine ratio. h 4 = sin 47° h = 4 sin 47° Area of triangle ABC = ½ × base × height = ½ × 7 × 4 sin 47° = 10.2 cm 2 (to 1 d.p.)

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© Boardworks Ltd 2005 40 of 58 The area of a triangle using ½ ab sin C The area of a triangle is equal to half the product of two of the sides and the sine of the included angle. A BC c a b Area of triangle ABC = ab sin C 1 2

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© Boardworks Ltd 2005 41 of 58 The area of a triangle using ½ ab sin C

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© Boardworks Ltd 2005 42 of 58 A A A A A A S3.5 The sine rule Contents S4.6 The cosine rule S3.4 Area of a triangle using ½ ab sin C S4 Further trigonometry S4.1 Sin, cos and tan of any angle S4.3 Graphs of trigonometric functions S4.2 Sin, cos and tan of 30°, 45° and 60°

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© Boardworks Ltd 2005 43 of 58 The sine rule Consider any triangle ABC, C AB ba If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles, ACD and BDC. D sin A = h a is the side opposite A and b is the side opposite B. h b h = b sin A sin B = h a h = a sin B b sin A = a sin B So,

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© Boardworks Ltd 2005 44 of 58 The sine rule b sin A = a sin B Dividing both sides of the equation by sin A and then by sin B we have: b sin B = a sin A If we had dropped a perpendicular from A to BC we would have found that: b sin C = c sin B Rearranging: b sin B = c sin C

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© Boardworks Ltd 2005 45 of 58 The sine rule For any triangle ABC, C AB b c a a sin A = b sin B = c sin C or sin Asin Bsin C a = b = c

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© Boardworks Ltd 2005 46 of 58 Using the sine rule to find side lengths If we are given two angles in a triangle and the length of a side opposite one of the angles, we can use the sine rule to find the length of the side opposite the other angle. For example, Find the length of side a Using the sine rule, a sin 118° = 7 sin 39° a = 7 sin 118° sin 39° a = 9.82 (to 2 d.p.) a 7 cm 118° 39° A B C

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© Boardworks Ltd 2005 47 of 58 Using the sine rule to find side lengths

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© Boardworks Ltd 2005 48 of 58 Using the sine rule to find angles If we are given two side lengths in a triangle and the angle opposite one of the given sides, we can use the sine rule to find the angle opposite the other given side. For example, Find the angle at B Using the sine rule, sin B 8 = 6 sin 46° sin B = 8 sin 46° 6 B = 73.56° (to 2 d.p.) sin –1 B = 8 sin 46° 6 6 cm 46° B 8 cm A C

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© Boardworks Ltd 2005 49 of 58 Finding the second possible value Suppose that in the last example we had not been given a diagram but had only been told that AC = 8 cm, CB = 6 cm and that the angle at A = 46°. 6 cm 46° B 8 cm A C There is a second possible value for the angle at B. Instead of this triangle … … we could have this triangle. Remember, sin θ = sin (180° – θ ) So for every acute solution, there is a corresponding obtuse solution. B = 73.56° (to 2 d.p.) or B = 180° – 73.56° = 106.44° (to 2 d.p.) 46° 6 cm B

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© Boardworks Ltd 2005 50 of 58 Using the sine rule to find angles

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© Boardworks Ltd 2005 51 of 58 A A A A A A S4.6 The cosine rule Contents S3.5 The sine rule S3.4 Area of a triangle using ½ ab sin C S4 Further trigonometry S4.1 Sin, cos and tan of any angle S4.3 Graphs of trigonometric functions S4.2 Sin, cos and tan of 30°, 45° and 60°

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© Boardworks Ltd 2005 52 of 58 The cosine rule Consider any triangle ABC. C AB b a If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles, ACD and BDC. D h a is the side opposite A and b is the side opposite B. c is the side opposite C. If we call the length AD x, then the length BD can be written as c – x. c – x x

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© Boardworks Ltd 2005 53 of 58 The cosine rule C AB b a Using Pythagoras’ theorem in triangle ACD, D h b 2 = x 2 + h 2 c – x x Also, cos A = x b In triangle BCD, x = b cos A a 2 = ( c – x ) 2 + h 2 a 2 = c 2 – 2 cx + x 2 + h 2 1 2 Substituting and, 12 a 2 = c 2 – 2 cb cos A + b 2 a 2 = b 2 + c 2 – 2 bc cos A a 2 = c 2 – 2 cx + x 2 + h 2 This is the cosine rule.

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© Boardworks Ltd 2005 54 of 58 The cosine rule For any triangle ABC, A BC c a b a 2 = b 2 + c 2 – 2 bc cos A or cos A = b 2 + c 2 – a 2 2 bc

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© Boardworks Ltd 2005 55 of 58 Using the cosine rule to find side lengths If we are given the length of two sides in a triangle and the size of the angle between them, we can use the cosine rule to find the length of the other side. For example, Find the length of side a. B C A 7 cm 4 cm 48° a a 2 = b 2 + c 2 – 2 bc cos A a 2 = 7 2 + 4 2 – 2 × 7 × 4 × cos 48° a 2 = 27.53 (to 2 d.p.) a = 5.25 cm (to 2 d.p.)

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© Boardworks Ltd 2005 56 of 58 Using the cosine rule to find side lengths

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© Boardworks Ltd 2005 57 of 58 Using the cosine rule to find angles If we are given the lengths of all three sides in a triangle, we can use the cosine rule to find the size of any one of the angles in the triangle. For example, Find the size of the angle at A. 4 cm 8 cm 6 cm A B C cos A = b 2 + c 2 – a 2 2 bc cos A = 4 2 + 6 2 – 8 2 2 × 4 × 6 cos A = –0.25 A = cos –1 –0.25 A = 104.48° (to 2 d.p.) This is negative so A must be obtuse.

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© Boardworks Ltd 2005 58 of 58 Using the cosine rule to find angles

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