C2: Trigonometrical Equations

Learning Objective: to be able to solve simple trigonometrical equations in a given range

Starter:

1. Calculate the area of the shaded segment

15 cmπ/4

O

A

B

θ

OPPOSITE

HY

PO

TE

NU

SE

A D J A C E N T

The three trigonometric ratios

Sin θ =Opposite

Hypotenuse S O HS O H

Cos θ =Adjacent

Hypotenuse C A HC A H

Tan θ =OppositeAdjacent T O AT O A

Remember: S O HS O H C A HC A H T O AT O A

The three trigonometric ratios, sine, cosine and tangent, can be defined using the ratios of the sides of a right-angled triangle as follows:

x

y

O

P(x, y)

r

The sine, cosine and tangent of any angleThese definitions are limited because a right-angled triangle cannot contain any angles greater than 90°.

To extend the three trigonometric ratios to include angles greater than 90° and less than 0° we consider the rotation of a straight line OP of fixed length r about the origin O of a coordinate grid.

Angles are then measured anticlockwise from the positive x-axis.

For any angle θ there is an associated acute angle α between the line OP and the x-axis.

α

θ

The graph of y = sin θ

The graph of y = cos θ

The graph of y = tan θ

3rd quadrant

2nd quadrant 1st quadrant

4th quadrant

Tangent is positiveTT

Sine is positiveSS All are positiveAA

Remember CASTWe can use CAST to remember in which quadrant each of the three ratios is positive.

Cosine is positiveCC

Task 1:Write each of the following as trigonometric ratios of

positive acute angles:• sin 260°• cos 140°• tan 185°• tan 355°• cos 137°• sin 414°• sin (-194)°• cos (-336)°• tan 396°• tan 148°

Sin, cos and tan of 45°A right-angled isosceles triangle has two acute angles of 45°.

45°

45°Suppose the equal sides are of unit length.

1

1

Using Pythagoras’ theorem:

We can use this triangle to write exact values for sin, cos and tan 45°:

cos 45° = tan 45° = 1

2

2

The hypotenuse 2 21 1

1

2sin 45° =

1

2

2 2

2

60° 60°

60°

2

60°

30°

1

3

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°:

If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.

Using Pythagoras’ theorem:

The height of the triangle 2 22 1

3

sin 30° =1

2cos 30° =

3

2tan 30° =

1

3

Sin, cos and tan of 60°Suppose we have an equilateral triangle of side length 2.

We can also use this triangle to write exact values for sin, cos and tan 60°:

3

sin 60° =3

2cos 60° =

1

2tan 60° = 3

2 2

2

60° 60°

60°

2

60°

30°

1

3

If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.

Using Pythagoras’ theorem:

The height of the triangle 2 22 1

Sin, cos and tan of 30°, 45° and 60°Write the following ratios exactly:

1) cos 300° =

3) tan 240° =

5) cos –30° =

7) sin 210° =

2) tan 315° =

4) sin –330° =

6) tan –135° =

8) cos 315° =

12

12

12

32

1

2

3

–1

1

Task 2 :

Write down the value of the following leaving your answers in terms of surds where appropriate:

1. sin 120°2. cos 150°3. tan 225°4. cos 300°5. sin (-30)°6. cos (-120)°7. sin 240°8. sin 420°9. cos 315°

Equations of the form sin θ = kEquations of the form sin θ = k, where –1 ≤ k ≤ 1, have an infinite number of solutions.

If we use a calculator to find sin–1 k the calculator will give a value for θ between –90° and 90°.

This is called the principal solution of sin θ = k.

Other solutions in a given range can be found by considering the unit circle.

For example:

There is one and only one solution in this range.

Solve sin θ = 0.7 for –360° < θ < 360°.

sin-1 0.7 = 44.4° (to 1 d.p.)

44.4°

Equations of the form sin θ = kWe solve sin θ = 0.7 for –360° < θ < 360° by considering angles in the first and second quadrants of a unit circle where the sine ratio is positive.Start by sketching the principal solution 44.4° in the first quadrant.

Next, sketch the associated acute angle in the second quadrant.

135.6°–224.4° –315.6°

Moving anticlockwise from thex-axis gives the second solution:

Moving clockwise from thex-axis gives the third and fourth solutions:

180° – 44.4° = 135.6°

–(180° + 44.4°) = –224.4°–(360° – 44.4°) = –315.6°

Examples:

1. Solve for 0 ≤ x ≤ 360°, cos x = ½

2. Solve for 0 ≤ x ≤ 360°, sin x = - 0.685

Task 3: Solve for 0 ≤ x ≤ 360°

1. Sin x = 0.6

2. Cos x = 0.8

3. Tan x = 0.4

4. Sin x = -0.8

5. Cos x = -0.6

6. Tan x = -0.5

Task 4: Solve for 0 ≤ x ≤ 2π

1. sin x = 1/2

2. cos x = 1/ √2

3. tan x = - √3

4. sin x = √3 / 2

5. cos x = 1/2

6. cos x = - 1 / √2

Examples:

Solve for -180° ≤ x ≤ 180°,

tan 2x = 1.424

Solve for 0 ≤ x ≤ 360°,

sin (x + 30°) = 0.781

Task 5: Solve for 0 ≤ x ≤ 360°

1. sin 3x = 0.72. sin (x / 3) = 2/3 3. tan 4x = 1/34. cos 2x = 0.635. cos (x + 72°) = 0.515

Solve for 0 ≤ x ≤ 2π1. tan 2x = 12. sin (x / 3) = ½3. sin (x + π/6) = √3 / 2