Higher Maths Revision Notes Trigonometry Get Started

Trigonometry solve trigonometric equations in a given interval know and apply the addition formulae apply trigonometric formulae in the solution of geometric problems solve trigonometric equations involving addition formulae and double angle formulae know and apply the double angle formulae

Test Yourself? y = 0·7 The graph shows the solution of sin x = 0·7, 0 ≤ x ≤ 2π. One of the solutions comes directly from the calculator: x = sin –1 (0·7) = 0·775 radians ( to 3 d.p.) The other comes from the fact that the sine wave is symmetrical about x = π / 2 … if x is an answer then so is π – x: π – x = π – sin –1 (0·7) = 2·37 radians ( to 3 d.p.) The graph shows the solution of cos x = 0·7, 0 ≤ x ≤ 2π. One of the solutions comes directly from the calculator: x = cos –1 (0·7) = 0·795 radians ( to 3 d.p.) The other comes from the fact that the cos wave is symmetrical about x = π … if x is an answer then so is 2π – x: 2π – x = 2π – cos –1 (0·7) = 2·37 radians ( to 3 d.p.)

Standard formulae will be given. sin (A + B) = sin A cos B + cos A sin B sin (A – B) = sin A cos B – cos A sin B cos (A + B) = cos A cos B – sin A sin B cos (A – B) = cos A cos B + sin A sin B Test Yourself? 1 1 √2 π/4π/4 2 1 √3 π/3π/3 π/6π/6 Half-square and half-equilateral triangle should be known

Standard formulae will be given. sin 2A = 2 sin A cos A cos 2A = cos 2 A – sin 2 A = 2 cos 2 A – 1 = 1 – 2 sin 2 A Test Yourself? Appreciate that variations exist sin 4A = 2 sin 2A cos 2A cos 4A = cos 2 2A – sin 2 2A = 2 cos 2 2A – 1 = 1 – 2 sin 2 2A sin A = 2 sin A / 2 cos A / 2 cos A = cos 2 A / 2 – sin 2 A / 2 = 2 cos 2 A / 2 – 1 = 1 – 2 sin 2 A / 2

Test Yourself? To solve equations of the form p sin x + q cos x = r The left-hand side must be expressed in the form k sin (x + a) To solve equations of the form p sin 2x + q cos x = 0 expand the term in 2x: 2p sin x cos x + q cos x = 0 take out the common factor: cos x(2p sin x + q) = 0 solve the two equations: cos x = 0 and 2p sin x + q = 0 To solve equations of the form p cos 2x + q cos x = 0 expand the term in 2x … choosing the form which suits the other term : p [2cos 2 x – 1] + q cos x = 0 express as a quadratic equation:2p cos 2 x + q cos x – p = 0 solve the quadratic for two values of cos x propagate solutions for both values of cos x. To solve equations of the form p cos 2x + q sin x = 0 expand the term in 2x … choosing the form which suits the other term : p [1 – 2sin 2 x] + q sin x = 0 express as a quadratic equation:p + q cos x – 2p sin 2 x = 0 solve the quadratic for two values of cos x propagate solutions for both values of cos x.

Remember basic geometric facts: The sum of the angles of a triangle is 180˚ So, if A, B, and C are the three angles of a triangle C = 180 – (A + B) sin C = sin(180 – (A + B)) = sin (A + B) If an acute angle P has a sine of a / b then a right angle can be drawn thus: the third side can be calculated by Pythagoras viz. √(b 2 – a 2 ) the cosine and tangent can be read from the triangle. When working with parallel lines, remember alternate angles are equal: x = y so, for example, sin x = sin y corresponding angles are equal: : x = w so, for example, sin x = sin w co-interior angles are supplementary: z = 180 – y so, for example, sin z = sin (180 – y) = sin y Look out for compound angles. PQT = PQR – TQS So sin PQT = sin (PQR – TQS) Test Yourself? P ab x y z w P QRS T A C D ADC = ADB + BDC So sin ADC = sin (ADB + BDC) B

reveal The diagram shows the graph of y = a sin(bx) + c ; 0 ≤ x ≤ 2π (a)Find the values of a, b and c (b)Solve for x in the interval 0 ≤ x ≤ 2π when y = 1·5

The diagram shows the graph of y = a sin(bx) + c ; 0 ≤ x ≤ 2π (a)Find the values of a, b and c (b)Solve for x in the interval 0 ≤ x ≤ 2π when y = 1·5 (a) a = 3 (amplitude) b = 2 (waves per 2π) c = 1 (vertical-translation) (b) sin 2x = 0·5 ÷ 3 = 0·167 (3 s.f.) So2x = sin –1 (0·167), or π – sin –1 (0·167), or sin –1 (0·167) + 2π, (π – sin –1 (0·167)) + 2π So2x = 0·167, π – 0·167, 0· π, and (π – 0·167) + 2π, … = 0·167, 2·97, 6·45, 9·26, … Sox = 0·084, 1·49, 3·23, 4·63 1 st Solution from calculator: sin –1 (…) 2 nd Solution from symmetry: π – sin –1 (…) More solutions from periodicity of function: If x is a solution then so is 2π + x

Q R SP T PQRS is a parallelogram with a base of 11 cm. Its altitude, ST, is 12 cm. QR is extended to meet the altitude at T. RT = 5 cm. (a)State the length of RS. (b)Find the exact value of sin PQR. (c)State the exact value of sin SQR. (d)Hence find the exact value of sin PQS. reveal

Q R SP T PQRS is a parallelogram with a base of 11 cm. Its altitude, ST, is 12 cm. QR is extended to meet the altitude at T. RT = 5 cm. (a)State the length of RS and of QS. (b)Find the exact value of sin PQR. (c)State the exact value of sin SQR. (d)Hence find the exact value of sin PQS. (a) By Pythagoras, SR = 13 and QS = 25. (b)PQR = SRT (corresponding angles) So sin PQR = sin SRT = 12 / 13 (c) SQR = SQT (corresponding angles) So sin SQR = sin SQT = 12 / 25 = 3 / 4 (d)Sin PQS = sin(PQR – SQR) = sin PQR.cos SQR – cosPQR.sin SQR = 12 / / 25 – 5 / / 25 = 132 / 325

Find the exact value of (a) sin 15˚ (b)cos 75˚ (c) tan 105˚ reveal

Find the exact value of (a) sin 15˚ (b)cos 75˚ (c) tan 105˚ 1 1 √2 45˚ 2 1 √3 60˚ 30˚ (a)sin 15˚ = sin (60 – 45)˚ = sin 60˚cos 45˚ – cos 60˚sin 45˚ (b)cos 75˚ = cos ( )˚ = cos 45˚ cos 30˚ – sin45˚ sin30˚ (c)tan 105˚ = tan ( )˚

reveal BC A x˚x˚ Triangle ABC is isosceles with AB = AC Angle ABC = x˚. sin x˚ = 0·8 exactly. Calculate the exact value of (a)cos ACB (b)cos BAC

BC A x˚x˚ Triangle ABC is isosceles with AB = AC Angle ABC = x˚. sin x˚ = 0·8 exactly. Calculate the exact value of (a)cos ACB (b)cos BAC (a)Angle ACB = angle ABC = x˚ We know sin x˚ = 0·8. We can draw a right angle triangle, hypotenuse 1 and opposite side 0·8 and use Pythagoras’ Theorem to deduce that the third side is 0·6. cos ACB = cos x˚ = 0·6.[from triangle] (b)cos BAC = cos (180 – 2x) = – cos 2x˚ = – [cos 2 x˚ – sin 2 x˚] = – [0·6 2 – 0·8 2 ] = 0·28 exactly 0·8 1 0·6 x˚x˚

reveal 1a Expand sin(30 + x)˚ b Hence solve the equation cos x˚ + √3 sin x˚ = 1, 0 ≤ x ≤ 360 2Solve sin 2x – cos x = 0,0 ≤ x ≤ 2π 2 sin x cos x – cos x = 0  cos x(2 sin x – 1) = 0  cos x = 0 or sin x = ½  x = π / 2 or 2π – π / 2 or π / 6 or 2π – π / 6  x = π / 2, 3π / 2, π / 6, 11π / 6 3Solve 3 cos 2x˚ – 14 cos x˚ + 7 = 0, 0 ≤ x ≤ 360 3[2cos 2 x˚ – 1] – 14 cos x˚ + 7 = 0  6cos 2 x˚ – 14 cos x˚ + 4 = 0  cos x˚ = [14 ± √(14 2 – 4.6.4)] ÷ 12 = 24 / 12 or 4 / 12  cos x˚ = 2 (no solutions) or cos x˚ = 1 / 3  x = 70·5 or 360 – 70·5 = 289·5(to 1 d.p.)

1a Expand sin(30 + x)˚ b Hence solve the equation cos x˚ + √3 sin x˚ = 1, 0 ≤ x ≤ 360 2Solve sin 2x – cos x = 0,0 ≤ x ≤ 2π 2 sin x cos x – cos x = 0 cos x(2 sin x – 1) = 0 cos x = 0 or sin x = ½ x = π / 2 or 2π – π / 2 or π / 6 or 2π – π / 6 x = π / 2, 3π / 2, π / 6, 11π / 6 3Solve 3 cos 2x˚ – 14 cos x˚ + 7 = 0, 0 ≤ x ≤ 360 3[2cos 2 x˚ – 1] – 14 cos x˚ + 7 = 0 6cos 2 x˚ – 14 cos x˚ + 4 = 0 cos x˚ = [14 ± √(14 2 – 4.6.4)] ÷ 12 = 24 / 12 or 4 / 12 cos x˚ = 2 (no solutions) or cos x˚ = 1 / 3 x = 70·5 or 360 – 70·5 = 289·5(to 1 d.p.) a sin(30 + x)˚ = sin 30˚ cos x˚ + cos 30˚ sin x˚ = 1 / 2 cos x˚ + √3 / 2 cos x˚ b Given cos x˚ + √3 sin x˚ = 1 then 1 / 2 cos x˚ + √3 / 2 cos x˚ = 1 / 2 sin(30 + x)˚ = 1 / x = 30 or 180 – 30 = 150 or or or … In the desired domain, x = 30 or 150.