The opposite and adjacent sides Here is a right-angled triangle with hypotenuse h and acute angle θ. θ h Write an expression for the length of the opposite side in terms of h and θ. Teacher notes The following slides will show that the length of the opposite side in a right-angled triangle can be written as h sin θ and that the adjacent side in a right-angle triangle can be written as h cos θ . From this it follows that when the hypotenuse is of unit length, the opposite side in a right-angled triangle can be written as sin θ and that the adjacent side can be written as cos θ . The values of sin θ and cos θ are then examined for right-angled triangles drawn on a coordinate grid. Write an expression for the length of the adjacent side in terms of h and θ.

The opposite and adjacent sides Here is a right-angled triangle with hypotenuse h and acute angle θ. θ h opp adj a) sin θ = b) cos θ = hyp hyp opp = hyp × sin θ adj = hyp × cos θ opp = h sin θ adj = h cos θ

The opposite and adjacent sides For any right-angled triangle with hypotenuse h and acute angle θ the opposite and adjacent sides can be labelled as follows: h h sin θ θ h cos θ Teacher notes Establish that tan θ = sin θ / cos θ for all values of θ . opposite h sin θ tan θ = h cos θ adjacent sin θ tan θ = cos θ

The sine of any angle Teacher notes Explain that we can find the sine of any angle by considering the movement of the point P which is fixed at 1 unit from the origin on a coordinate grid. Sin θ is given by the y-coordinate of the point P. This is shown by the length of the bold red line. Explain that moving the point P in an anticlockwise direction increases the angle between OP and the x-axis. In maths, an anti-clockwise rotation is a positive rotation. Slowly, drag the point P through 360°, starting at 0° and ending at 360°. Observe the length of the line representing the y-coordinate of the point P as it is rotated. Point out that when the line is in the first and the second quadrant, that is, when θ is between 0° and 180°, it is above the x-axis and therefore positive. In other words, the sin of angles between 0° and 180°, is positive. When the line representing sin θ is in the third and the fourth quadrant, that is, when θ is between 180° and 360°, it is below the x-axis and therefore negative. In other words, the sine of angles between 180° and 360° is negative. In this activity, P can be moved through any angle between –360° and 720°. Demonstrate the sine of these angles if required. Explain that P can be moved through any positive or negative angle in this way. Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants) demonstrate the relationships between pairs of angles that have the same sine. For example, show that sin 32° = sin 148°. Conclude that sin θ = sin (180° – θ). Focus next on angles between 180° and 270° (that is, angles in the third quadrant). By looking at the sine of the associated acute angle show that sin θ = – sin (θ – 180°). In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that sin θ = – sin (360° – θ) for angles between 270° and 360° and sin –θ = – sin θ.

Sine of angles Teacher notes Stress that the sine of any obtuse angle θ is equal to the sine of its supplementary angle (180° – θ). In the second quadrant θ can also be between –180° and –270°. In general, in the second quadrant (90 + 360n)° < θ < (180 + 360n)° for some integer n. Ask pupils to verify, using their calculators, that sin 130° = sin 50°. In the third quadrant θ can also be between –90° and –180°. In general, in the third quadrant (180 + 360n)° < θ < (270 + 360n)° for some integer n. In general, in the fourth quadrant (270 + 360n)° < θ < 360n° for some integer n. Ask pupils to verify these answers using a calculator.

The cosine of any angle Teacher notes Explain that cos θ is given by the x-coordinate of the point P. This is shown by the length of the bold orange line. Slowly, drag the point P through 360°, starting at 0° and ending at 360°. Observe the length of the line representing the x-coordinate of the point P as it is rotated. Point out that when the line is in the first and the fourth quadrants, that is, when θ is between 0° and 90° or between 270° and 360° (also between 0° and –90°), it is to the right of the y-axis and therefore positive. When the line representing cos θ is in the second and third quadrants, that is, when θ is between 90° and 270°, it is to the left the y-axis and therefore negative. In other words, the cosine of angles between 90° and 270°, is negative. In this activity, P can be moved through any angle between –360° and 720°. Demonstrate the cosine of these angles if required. Explain that P can be moved through any positive or negative angle in this way. Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants) demonstrate the relationships between pairs of angles that have the same cosine, but are of opposite sign. For example, show that cos 148° = –cos 32°. Conclude that cos θ = –cos (180° – θ). Focus next on angles between 180° and 270° (that is, angles in the third quadrant) . Show that cos θ = – cos (θ – 180°). In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that cos θ = cos (360° – θ) for angles between 270° and 360° and cos θ = cos –θ for angles between 0° and –90°.

Cosine of angles Teacher notes Stress that the cosine of any obtuse angle θ is equal to the negative cosine of its supplementary angle (180° – θ). In the second quadrant θ can also be between –180° and –270°. In general, in the second quadrant (90 + 360n)° < θ < (180 + 360n)° for some integer n. Ask pupils to verify, using their calculators, that cos 100° = –cos 80°. In the third quadrant θ can also be between –90° and –180°. In general, in the third quadrant (180 + 360n)° < θ < (270 + 360n)° for some integer n. Remind pupils that the cosine of any angle the fourth quadrant is always positive. In general, in the fourth quadrant (270 + 360n)° < θ < 360n° for some integer n. Ask pupils to verify these answers using a calculator.

The tangent of any angle Teacher notes Remind pupils that tan θ is given by sin θ/ cos θ. Tan θ is therefore given by the y-coordinate of the point P divided by the x-coordinate of the point P. Slowly, drag the point P through 360°, starting at 0° and ending at 360°. Observe the change in the value of tan θ as the point P is rotated. Point out that when P is in the first and the third quadrants, that is, when θ is between 0° and 90° or between 180° and 270°, the sin and cosine of the required angle is of the same sign. The tangent is therefore positive in these quadrants. The tangent of 90° and 270° is undefined when cos θ = 0 (because we cannot divide by 0). When the point P is in the second and fourth quadrants, that is, when θ is between 90° and 180° and between 270° and 360° , the sin and cosine of the required angle are of different signs. The tangent is therefore negative in these quadrants. Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants) demonstrate the relationships between pairs of angles that have the same tangent, but are of opposite sign. For example, show that tan 127° = –tan 53°. Conclude that tan θ = –tan (180° – θ). Focus next on angles between 180° and 270° (that is, angles in the third quadrant) . Show that tan θ = tan (θ – 180°). In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that tan θ = –tan (360° – θ) for angles between 270° and 360° and tan –θ = –tan θ for angles between 0° and –90°.

The tangent of any angle Teacher notes This demonstration shows more clearly how the value of tan θ varies as θ varies. In particular, it can be shown that tan 90° and tan 270° are undefined. This is because the tangent is parallel to the x-axis at these points. The circle has radius 1. Explain to pupils that the length of the tangent from P to the x-axis gives us the value of tan θ. Remind pupils that the tangent of a circle forms a right angle with the radius. We therefore have a right-angled triangle with opposite side of length tan θ and adjacent side of length 1. Opposite/adjacent = tan θ as required. Slowly move the point P through 0° < θ < 360° and observe how the value of tan θ varies. Draw pupils attention to the fact that when θ is close to 90° and 270° it get very large very quickly. Establish that at these points the tangent at P is parallel to the x-axis. Since the tangent will never meet the x-axis at 90° or 270° tan θ is undefined at these angles. The slope of the tangent defines whether it is positive or negative. A positive gradient gives a negative value and a negative gradient gives a positive value.

Tangent of angles Teacher notes Stress that the tangent of any obtuse angle θ is equal to the negative tangent of its supplementary angle (180° – θ). In the second quadrant θ can also be between –180° and –270°. In general, in the second quadrant (90 + 360n)° < θ < (180 + 360n)° for some integer n. Ask pupils to verify, using their calculators, that cos 116° = –tan 2.05°. Stress that the tangent of any angle in the third quadrant is positive. In the third quadrant θ can also be between –90° and –180°. In general, in the third quadrant (180 + 360n)° < θ < (270 + 360n)° for some integer n. In general, in the fourth quadrant (270 + 360n)° < θ < 360n° for some integer n. Ask pupils to verify these answers using a calculator.

sin, cos and tan of angles. 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) In the third quadrant: sin θ = –sin (θ – 180°) cos θ = –cos (θ – 180°) tan θ = tan (θ – 180°) Teacher notes This slide summarizes the results established for the sin, cos and tan of angles between 0° and 360°. In the fourth quadrant: sin θ = –sin (360° – θ) cos θ = cos (360° – θ) tan θ = –tan(180° – θ)

Remember CAST We can use CAST to remember in which quadrant each of the three ratios are positive. 2nd quadrant 1st quadrant Sine is positive All are positive 3rd quadrant 4th quadrant Teacher notes Introduce pupils to CAST to help them remember in which quadrant each of the thee ratios are positive. Tangent is positive Cosine is positive

Positive or negative? Teacher notes Start by establishing which quadrant the given angle is in. Ask pupils to use this to decide whether the sin, cos or tan of the given angle will be positive or negative. Photo credit: © Chernetskiy 2010, Shutterstock.com

The second quadrant What are the trigonometric ratios for 150°? We use the 30° angle shown in the diagram and CAST. S A 150° 30° 180° 0° In the second quadrant only sine is positive. T C Teacher notes It is important to show that for trigonometric ratios the angles are always measured from the horizontal 0-180 line. Also note that each angle can be positive (anticlockwise) and negative (clockwise). sin150 = sin –210 = sin30 cos150 = cos –210 = –cos30 tan150 = tan –210 = –tan30

The third quadrant What are the trigonometric ratios for –110°? –110° is also +250°. We use the 70° angle shown in the diagram and CAST. S A 180° 0° In the third quadrant only tangent is positive. 70° 110° T C Teacher notes It is important to show that for trigonometric ratios the angles are always measured from the horizontal 0-180 line. Also note that each angle can be positive (anticlockwise) and negative (clockwise). sin –110 = sin250 = – sin70 cos –110 = cos250 = –cos70 tan –110 = tan250 = tan70

The fourth quadrant What are the trigonometric ratios for 295°? 295° is also –65°. We use the 65° angle shown in the diagram and CAST. S A 295° 180° 0° In the fourth quadrant only cosine is positive. 65° T C Teacher notes It is important to show that for trigonometric ratios the angles are always measured from the horizontal 0-180 line. Also note that each angle can be positive (anticlockwise) and negative (clockwise). sin295 = sin –65 = –sin65 cos295 = cos –65 = cos65 tan295 = tan –65 = –tan65

Equivalents Photo credit: © Chernetskiy 2010, Shutterstock.com

Solving equations in θ Teacher notes Pupils can use their calculators to find a value for θ between 0° and 90°. Pupils should then be encouraged to recall in which quadrant the given ratio is positive (or negative). They can then use a sketch of the four quadrants to find the other three solutions in the given range. These equations can also be solved using sine, cosine or tangent graphs. Photo credit: © Chernetskiy 2010, Shutterstock.com

sin, cos and tan of 30° 2 60° 30° 1 2 60° If we cut this triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. 3 Using Pythagoras’ theorem, The height of the triangle =  2² – 1² = 3 This triangle can also be used to write exact values for sin, cos and tan 30°: 1 3 1 sin 30° = cos 30° = tan 30° = 2 2 3

sin, cos and tan of 45° A right-angled isosceles triangle has two acute angles of 45°. Suppose the equal sides are of 1 unit length. 45° 2 1 Using Pythagoras’ theorem, The hypotenuse =  1² + 1² 45° = 2 1 Teacher notes Make sure that pupils can see that these ratios would be the same for any right-angled isosceles triangle. If the equal sides were of a different length, for example 3, the hypotenuse would be of length 32. In each ratio the 3’s would cancel and so simplify to those shown above. If required, verify using a scientific calculator that sin and cos of 45° = 0.707106781… = 1/2 Stress that 1/2 is an exact answer. sin and cos of 45° cannot be written exactly as a decimal. This triangle can also be used to write exact values for sin, cos and tan 45°: 1 1 sin 45° = cos 45° = tan 45° = 1 2 2

sin, cos and tan of 60° 2 60° 30° 1 2 60° If we cut this triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. 3 Using Pythagoras’ theorem, The height of the triangle =  2² – 1² = 3 This triangle can also be used to write exact values for sin, cos and tan 60°: 3 1 sin 60° = cos 60° = tan 60° = 3 2 2

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° 45° 60° 1 2 1 3 sin 2 2 3 1 1 2 cos 2 2 1 tan 1 3 3 Teacher notes Ask pupils to give the exact values of cos 150° and tan 150°. Ask pupils for sin, cos and tan of other angles associated with 30° (for example, 210°, 330°, 390°, – 30° or –150°). Use CAST to remind pupils in which quadrant each ratio is positive. See slide 20. Ask pupils to give the exact values of sin 135° and tan 135°. Ask pupils for sin, cos and tan of other angles associated with 45° (for example, 225°, 315°, 405°, – 45° or –135°). Ask pupils to give the exact values of sin 120° and cos 120°. Ask pupils for sin, cos and tan of other angles associated with 60° (for example, 240°, 300°, 420°, – 60° or –120°). Use this table to write the exact value of: 1 –1 sin 150° = cos 135° = tan 120° = –3 2 2

Matching Teacher notes Ask pupils to complete this exercise individually before revealing the answers. Suggest to pupils that they write down which quadrant each angle falls into before writing down their solutions.