1. Starter

2. Identify Trig Ratios with Obtuse Angles
We are Learning to……
Identify Trig Ratios with Obtuse Angles

3. 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.
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.

4. 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
b) cos θ =
adj
hyp
opp = hyp × sin θ
adj = hyp × cos θ
opp = h sin θ
adj = h cos θ

5. 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: h
h sin θ θ
h cos θ
Establish that tan θ = sin θ / cos θ for all values of θ .
opposite
tan θ =
h sin θ
h cos θ
We can write,
adjacent
tan θ =
sin θ
cos θ

6. The sine of any angle
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 show 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.

7. 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)
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 ( n)° < θ < ( n)° for any integer n.
Ask pupils to verify, using their calculators, that sin 130° = sin 50°.
For example,
sin 130° =
sin (180° – 130°)
= sin 50°
= (to 3 sig. figs)

8. Sine of angles in the third quadrant
sin θ = –sin (θ – 180°)
For example,
sin 220° =
– sin (220° – 180°)
= – sin 40°
In the third quadrant θ can also be between –90° and –180°.
In general, in the third quadrant ( n)° < θ < ( n)° for any integer n.
= – (to 3 sig. figs)
Verify, using a scientific calculator, that sin 220° = –sin 40°

9. Sine of angles in the fourth quadrant
In the fourth quadrant, 270° < θ < 360° or 0° > 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)
In general, in the fourth quadrant ( n)° < θ < 360n° for any integer n.
Ask pupils to verify these answers using a calculator.
sin –35° =
–sin 35°
= –0.574 (to 3 sig. figs)

10. The cosine of any angle
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°.

11. 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)
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 ( n)° < θ < ( n)° for any integer n.
Ask pupils to verify, using their calculators, that cos 100° = –cos 80°.
For example,
cos 100° =
–cos (180° – 100°)
= –cos 80°
= –0.174 (to 3 sig. figs)

12. Cosine of angles in the third quadrant
cos θ = –cos (θ – 180°)
For example,
cos 250° =
–cos (250° – 180°)
= –cos 70°
In the third quadrant θ can also be between –90° and –180°.
In general, in the third quadrant ( n)° < θ < ( n)° for any integer n.
= –0.342 (to 3 sig. figs.)
Verify, using a scientific calculator, that cos 250° = –cos 70°

13. Sine of angles in the fourth quadrant
In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90°
cos θ = cos(360° – θ) or cos –θ = cos θ
For example,
cos 317° =
cos (360° – 317°)
= cos 43°
= (to 3 sig. figs.)
Remind pupils that the cosine of any angle the fourth quadrant is always positive.
In general, in the fourth quadrant ( n)° < θ < 360n° for any integer n.
Ask pupils to verify these answers using a calculator.
cos –28° =
cos 28°
= (to 3 sig. figs.)

14. The tangent of any angle
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°.

15. The tangent of any angle
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.

16. 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)
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 ( n)° < θ < ( n)° for any integer n.
Ask pupils to verify, using their calculators, that cos 116° = –tan 2.05°.
For example,
tan 116° =
–tan (180° – 116°)
= –tan 64°
= –2.05 (to 3 sig. figs)

17. Tangent of angles in the third quadrant
tan θ = tan (θ – 180°)
For example,
tan 236° =
tan (236° – 180°)
= tan 56°
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 ( n)° < θ < ( n)° for any integer n.
= 1.48 (to 3 sig. figs)
Verify, using a scientific calculator, that tan 236° = tan 56°

18. Tangent of angles in the fourth quadrant
In the fourth quadrant, 270° < θ < 360° or 0° > 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)
In general, in the fourth quadrant ( n)° < θ < 360n° for any integer n.
Ask pupils to verify these answers using a calculator.
tan –16° =
–tan 16°
= –0.287 (to 3 sig. figs)

19. 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)
In the third quadrant:
sin θ = –sin (θ – 180°)
cos θ = –cos (θ – 180°)
tan θ = tan (θ – 180°)
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)

20. 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 S
All are positive A
3rd quadrant
4th quadrant
Introduce pupils to CAST to help them remember in which quadrant each of the thee ratios are positive.
Tangent is positive T
Cosine is positive C

21. Positive or negative?
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.

22. McGraw-Hill Page 93 #s 1 – 6 BLM 2.6 #s 1, 2, 4 & 6
To succeed at this lesson today you need to…
1. Remember that the quadrants turn counter-clockwise
2. Remember the word CAST
3. Know what a terminal arm is
McGraw-Hill Page 93 #s 1 – 6 BLM 2.6 #s 1, 2, 4 & 6

23. Homework
McGraw-Hill 12 Page 94 #s 7, 8 & 10
BLM 2.6 #s 8 – 13

24. Find the equivalent ratio
For each example find at least three equivalent ratios.

25. Solving equations in θ
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.