1. GCSE Trigonometry Parts 3 and 4 – Trigonometric Graphs and Equations
Dr J Frost
Objectives: (from the IGCSE FM specification)
Last modified: 28th December 2016

2. Sin Graph
What does it look like? ?
-360
-270
-180
-90 90
180
270
360

3. Sin Graph ? ? ? sin(150) = 0.5 sin(-30) = -0.5 sin(210) = -0.5
What do the following graphs look like?
-360
-270
-180
-90 90
180
270
360
Suppose we know that sin(30) = 0.5. By thinking about symmetry in the graph, how could we work out:
sin(150) = 0.5 ?
sin(-30) = -0.5 ?
sin(210) = -0.5 ?

4. Cos Graph ? What do the following graphs look like? -360 -270 -180 -90

5. Cos Graph ? ? ? cos(120) = -0.5 cos(-60) = 0.5 cos(240) = -0.5
What does it look like?
-360
-270
-180
-90 90
180
270
360
Suppose we know that cos(60) = 0.5. By thinking about symmetry in the graph, how could we work out:
cos(120) = -0.5 ?
cos(-60) = 0.5 ?
cos(240) = -0.5 ?

6. Tan Graph
What does it look like? ?
-360
-270
-180
-90 90
180
270
360

7. Tan Graph ? ? tan(-30) = -1/√3 tan(150) = -1/√3
What does it look like?
-360
-270
-180
-90 90
180
270
360
Suppose we know that tan(30) = 1/√3. By thinking about symmetry in the graph, how could we work out:
tan(-30) = -1/√3 ?
tan(150) = -1/√3 ?

8. Solving Trig Equations
Solve sin 𝑥 =0.6 in the range 0≤𝑥<360
𝒙= 𝐬𝐢𝐧 −𝟏 𝟎.𝟔 =𝟑𝟔.𝟖𝟕°, 𝟏𝟒𝟑.𝟏𝟑° ? ?
Angle Law #1:
𝐬𝐢𝐧 𝒙 =𝐬𝐢𝐧⁡(𝟏𝟖𝟎−𝒙)
0.6
𝟑𝟔.𝟖𝟕° ?
-360
-270
-180
-90 90
180
270
360

9. Solving Trig Equations
Solve 3cos 𝑥 =2 in the range 0≤𝑥<360 cos 𝑥 = 2 3
𝒙=𝒄𝒐 𝒔 −𝟏 𝟐 𝟑 =𝟒𝟖.𝟏𝟗°, 𝟑𝟏𝟏.𝟖𝟏°
Angle Law #2:
𝒄𝒐𝒔 𝒙 =𝒄𝒐𝒔⁡(𝟑𝟔𝟎−𝒙) ? ? ?
𝟐 𝟑
𝟒𝟖.𝟏𝟗° ?
-360
-270
-180
-90 90
180
270
360

10. Solving Trig Equations
Solve sin 𝑥 =−0.3 in the range 0≤𝑥<360
𝒙= 𝐬𝐢𝐧 −𝟏 −𝟎.𝟑 =−𝟏𝟕.𝟒𝟔°, 𝟏𝟗𝟕.𝟒𝟔°, 𝟑𝟒𝟐.𝟓𝟒 ? ?
Angle Law #3:
Sin and cos repeat every 360° ?
−𝟏𝟕.𝟒𝟔° ?
-360
-270
-180
-90 90
180
270
360
-0.3

11. sin and cos repeat every 360°
Laws of Trigonometric Functions ? !
sin 𝑥 = sin 180−𝑥
cos 𝑥 = cos 360−𝑥 ?
sin and cos repeat every 360° ?
tan repeats every 180° ?
Solutions generally come in pairs (e.g. 30° and 150° for sin) for each ‘cycle’ of sin/cos. We can therefore get the next pair of solutions by going to the next cycle.

12. Test Your Understanding
Solve cos 𝑥 =0.9 in the range 0≤𝑥<360 ?
𝑥= cos − =𝟐𝟓.𝟖𝟒° 360−25.84°=𝟑𝟑𝟒.𝟏𝟔°
Solve tan 𝑥 =1 in the range 0≤𝑥<360
𝑥= tan −1 1 =𝟒𝟓° 45°+180°=𝟐𝟐𝟓° ?
Set 4 Paper 2 Q14
sin 𝜃 =− → 𝜃=−42.84°
180−−42.84=𝟐𝟐𝟐.𝟖𝟒° − =𝟑𝟏𝟕.𝟏𝟔° ?

13. Exercise 1
Solve the following in the range 0≤𝑥≤360
sin 𝑥 = → 𝒙=𝟑𝟎°, 𝟏𝟓𝟎° cos 𝑥 = → 𝒙=𝟑𝟎°, 𝟑𝟑𝟎°
tan 𝑥 = → 𝒙=𝟔𝟎°, 𝟐𝟒𝟎° sin 𝑥 = → 𝒙=𝟓.𝟕𝟒°, 𝟏𝟕𝟒.𝟐𝟔°
4 cos 𝑥 =3 → 𝟒𝟏.𝟒𝟏°, 𝟑𝟏𝟖.𝟓𝟗° 6 tan 𝑥 =5 → 𝟑𝟗.𝟖𝟏°, 𝟐𝟏𝟗.𝟖𝟏° sin 𝑥 =0.4 → 𝟐𝟑.𝟓𝟖°, 𝟏𝟓𝟔.𝟒𝟐°
sin 𝜃 =− → 𝟐𝟏𝟎°, 𝟑𝟑𝟎° cos 𝜃 =− → 𝟏𝟐𝟎°, 𝟐𝟒𝟎° tan 𝜃 =− → 𝟏𝟑𝟓°, 𝟑𝟏𝟓° sin 𝜃 =− → 𝟐𝟎𝟑.𝟓𝟖°, 𝟑𝟑𝟔.𝟒𝟐° cos 𝜃 =− → 𝟏𝟑𝟒.𝟒°, 𝟐𝟐𝟓.𝟔° tan 𝜃 =− → 𝟏𝟔𝟖.𝟔𝟗°, 𝟑𝟒𝟖.𝟔𝟗° 1 a ? b ? c ? d ? e ? f ? g ? 2 a ? b ? c ? d ? e ? f ?

14. 1 sin 𝜃  cos⁡ Trigonometric Identities 𝒔𝒊 𝒏 𝟐 𝜽+𝒄𝒐 𝒔 𝟐 𝜽=𝟏 ? ? ?
Using basic trigonometry to find these two missing sides… 1
sin 𝜃 ? 
cos⁡ ? ?
Then 𝒕𝒂𝒏 𝜽= 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 𝜽 1
These two identities are all you will need for IGCSE FM. ?
Pythagoras gives you...
𝒔𝒊 𝒏 𝟐 𝜽+𝒄𝒐 𝒔 𝟐 𝜽=𝟏 2
sin 2 𝜃 is a shorthand for sin 𝜃 It does NOT mean the sin is being squared – this does not make sense as sin is not a quantity that we can square!

15. Application #1: Solving Harder Trig Equations
Solve sin 𝑥 =2 cos 𝑥 in the range 0≤𝑥<360°
The problem here is that we have two different trig functions. Is there anything we could divide by to get just one trig function? ?
sin 𝑥 cos 𝑥 = 2 cos 𝑥 cos 𝑥
tan 𝑥 =2
𝑥=63.43°, ° ? ?
sin 𝑥 = sin 180−𝑥
cos 𝑥 = cos 360−𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
tan 𝑥 = sin 𝑥 cos 𝑥
sin 2 𝑥 + cos 2 𝑥 =1
Bro Tip: In general, when you have a mixture of sin and cos, divide everything by cos.

16. Test Your Understanding
Solve 2 s𝑖𝑛 𝑥 = cos 𝑥 in the range 0≤𝑥<360°
2 tan 𝑥 =1 tan 𝑥 = 1 2 𝑥=26.57°, ° ?
Solve cos 𝑥 = sin 𝑥 in the range 0≤𝑥<360°
1= tan 𝑥 𝑥=45°, 225° ?
sin 𝑥 = sin 180−𝑥
cos 𝑥 = cos 360−𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
tan 𝑥 = sin 𝑥 cos 𝑥
sin 2 𝑥 + cos 2 𝑥 =1

17. Application #1: Solving Harder Trig Equations
June 2013 Paper 2 Q22
Solve tan 2 𝜃 +3 tan 𝜃 =0 in the range 0≤𝑥<360°
This looks a bit like a quadratic. What would be our usual strategy to solve! ?
tan 𝜃 tan 𝜃 +3 =0 tan 𝜃 =0 𝑜𝑟 tan 𝜃=−3 𝜃=0, 180, °, 288.4° ? ?
sin 𝑥 = sin 180−𝑥
cos 𝑥 = cos 360−𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
tan 𝑥 = sin 𝑥 cos 𝑥
sin 2 𝑥 + cos 2 𝑥 =1

18. sin 𝜃 2 sin 𝜃 −1 =0 sin 𝜃=0 𝑜𝑟 sin 𝜃 = 1 2 𝜃=0, 180°, 30°, 150°
More Examples
Solve 2sin 2 𝜃 − sin 𝜃 =0 in the range 0≤𝑥<360° ?
sin 𝜃 2 sin 𝜃 −1 =0 sin 𝜃=0 𝑜𝑟 sin 𝜃 = 𝜃=0, 180°, 30°, 150°
Solve cos 2 𝜃 = 1 4 in the range 0≤𝑥<360°
cos 𝜃 = 𝑜𝑟 cos 𝜃 =− 1 2 𝜃=60°, 300°, 120°, 240° ?

19. Test Your Understanding
Solve cos 2 𝜃 + cos 𝜃 =0 in the range 0≤𝑥<360° ?
cos 𝜃 cos 𝜃 +1 =0 cos 𝜃 =0 𝑜𝑟 cos 𝜃 =−1 𝜃=90°, 270°, 180°
Expand and simplify (2𝑠+1)(𝑠−1). Hence or otherwise, solve 2 sin 2 𝜃 − sin 𝜃 −1=0 for 0°≤𝜃<360°
2 sin 𝜃 +1 sin 𝜃 −1 =0 sin 𝜃 =− 𝑜𝑟 sin 𝜃=1 𝜃=210°, 330°, 90° ?

20. Exercise 2 ? ? ? ? ? ? ? ? ? ? ? 1 Solve the following in the range
0°≤𝑥<360°
sin 𝜃 =3 cos 𝜃 → 𝜽=𝟕𝟏.𝟓𝟕°, 𝟐𝟓𝟏.𝟓𝟕°
2 sin 𝜃 =3 cos 𝜃 → 𝜽=𝟓𝟔.𝟑𝟏°, 𝟐𝟑𝟔.𝟑𝟏°
Solve the following by first factorising. 0°≤𝑥<360°
cos 2 𝜃 − cos 𝜃 = → 𝜽=𝟗𝟎, 𝟐𝟕𝟎, 𝟎 tan 2 𝜃 −3 tan 𝜃 = → 𝜽=𝟎,𝟏𝟖𝟎,𝟕𝟏.𝟓𝟕, 𝟐𝟓𝟏.𝟓𝟕° sin 𝑥 cos 𝑥 + sin 𝑥 =0 → 𝜽=𝟎°, 𝟏𝟖𝟎°
Solve the following: 0°≤𝑥<360°
sin 2 𝜃 = → 𝜽=𝟔𝟎°, 𝟏𝟐𝟎°, 𝟐𝟒𝟎°, 𝟑𝟎𝟎°
cos 2 𝜃 = →𝜽=𝟑𝟎°, 𝟏𝟓𝟎°, 𝟐𝟏𝟎°, 𝟑𝟑𝟎°
tan 2 𝜃 = →𝜽=𝟔𝟎°, 𝟏𝟐𝟎°, 𝟐𝟒𝟎°, 𝟑𝟎𝟎°
By factorising these ‘quadratics’, solve in the range 0≤𝑥<360
3 cos 2 𝜃 +2 cos 𝜃 −1=0 → 𝜽=𝟕𝟎.𝟓𝟑°, 𝟏𝟖𝟎°, 𝟐𝟖𝟗.𝟒𝟕°
6 sin 2 𝜃 − sin 𝜃 −1= → 𝜽=𝟑𝟎°, 𝟏𝟓𝟎°, 𝟏𝟗𝟗.𝟒𝟕°, 𝟑𝟒𝟎.𝟓𝟑° sin 𝜃 cos 𝜃 + sin 𝜃 + cos 𝜃 =−1 → 𝐬𝐢𝐧 𝜽 +𝟏 𝐜𝐨𝐬 𝜽 +𝟏 =𝟎 → 𝜽=𝟏𝟖𝟎°, 𝟐𝟕𝟎° a ? b ? 2 ? a ? b ? c 3 a ? b ? c ? 4 a ? b ? ? N

21. Review of what we’ve done so far 
partly 

22. Application of identities #2: Proofs
Prove that 1− tan 𝜃 sin 𝜃 cos 𝜃 ≡ cos 2 𝜃
Recall that ≡ means ‘equivalent to’, and just means the LHS is always equal to the RHS for all values of 𝜃.
𝐿𝐻𝑆=1− sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
=1− sin 2 𝜃 𝑐𝑜𝑠𝜃 cos 𝜃 =1− sin 2 𝜃 = cos 2 𝜃 =𝑅𝐻𝑆 ? ? ? ?
sin 𝑥 = sin 180−𝑥
cos 𝑥 = cos 360−𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙
𝐬𝐢𝐧 𝟐 𝒙 + 𝐜𝐨𝐬 𝟐 𝒙 =𝟏
We want to use these…

23. Another Example ? ? ? ? Prove that tan 𝜃 + 1 tan 𝜃 ≡ 1 sin 𝜃 cos 𝜃
June 2012 Paper 1 Q16
Prove that tan 𝜃 + 1 tan 𝜃 ≡ 1 sin 𝜃 cos 𝜃 ?
𝐿𝐻𝑆= sin 𝜃 cos 𝜃 + cos 𝜃 sin 𝜃
= sin 2 𝜃 sin 𝜃 cos 𝜃 + cos 2 𝜃 sin 𝜃 cos 𝜃
= sin 2 𝜃 + cos 2 𝜃 sin 𝜃 cos 𝜃
= 1 sin 𝜃 cos 𝜃 =𝑅𝐻𝑆
Bro Tip: Whenever you have a fraction in a proof question, always add the fractions. ? ? ?

24. Test Your Understanding
Prove that tan 𝑥 cos 𝑥 1− cos 2 𝑥 ≡1
= sin 𝑥 cos 𝑥 cos 𝑥 sin 2 𝑥 = sin 𝑥 sin 𝑥 =1 ?
AQA Worksheet
Prove that tan 2 𝜃 ≡ 1 cos 2 𝜃 −1
tan 2 𝜃 = sin 2 𝜃 cos 2 𝜃
= 1− cos 2 𝜃 cos 2 𝜃
= 1 cos 2 𝜃 − cos 2 𝜃 cos 2 𝜃 ?

25. Exercise 3 ? ? ? ? Simplify 3 sin 𝑥 sin 𝑥 +2 −3 2 sin 𝑥 − cos 2 𝑥 1
=𝟑 𝐬𝐢𝐧 𝟐 𝒙 +𝟔 𝐬𝐢𝐧 𝒙 −𝟔 𝐬𝐢𝐧 𝒙 +𝟑 𝐜𝐨𝐬 𝟐 𝒙 =𝟑 𝐬𝐢𝐧 𝟐 𝒙 + 𝐜𝐨𝐬 𝟐 𝒙 =𝟑
Write out the following in terms of sin 𝑥 :
cos 2 𝑥 tan 2 𝑥 = 𝒄𝒐𝒔 𝟐 𝒙 𝐬𝐢𝐧 𝟐 𝒙 𝐜𝐨𝐬 𝟐 𝒙 = 𝐬𝐢𝐧 𝟐 𝒙
tan 𝑥 cos 3 𝑥 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝟑 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝟐 𝒙 = 𝐬𝐢𝐧 𝒙 𝟏− 𝐬𝐢𝐧 𝟐 𝒙
cos 𝑥 2 cos 𝑥 −3 tan 𝑥 =𝟐 𝐜𝐨𝐬 𝟐 𝒙 −𝟑 𝐬𝐢𝐧 𝒙 =𝟐 𝟏− 𝐬𝐢𝐧 𝟐 𝒙 −𝟑 𝐬𝐢𝐧 𝒙
Prove the following:
tan 𝑥 1− sin 2 𝑥 ≡ sin 𝑥
1− cos 2 𝑥 1− sin 2 𝑥 ≡ tan 2 𝑥
1+ sin 𝑥 1− sin 𝑥 ≡ cos 2 𝑥
2 sin 𝑥 cos 𝑥 tan 𝑥 ≡2−2 sin 2 𝑥
1 cos 𝜃 − cos 𝜃 ≡ sin 𝜃 tan 𝜃
2 sin 𝜃 − cos 𝜃 sin 𝜃 +2 cos 𝜃 2 ≡5 1 ? 2 ? ? ? 3

26. Represent as a triangle
Using triangles to change between sin/cos/tan
Given that sin 𝜃 = 3 5 and that 𝜃 is acute, find the exact value of:
cos 𝜃 = 𝟒 𝟓
tan 𝜃= 𝟑 𝟒 ?
Represent as a triangle 5 ? ? 3 ? 𝜃 4
Test Your Understanding
Given that tan 𝜃 = and that 𝜃 is acute, find the value of:
sin 𝜃 = 𝟓 𝟑
cos 𝜃= 𝟐 𝟑
Given that cos 𝜃 = and that 𝜃 is acute, find the value of:
sin 𝜃 = 𝟏𝟐 𝟏𝟑
tan 𝜃= 𝟏𝟐 𝟓 1 2 ? ? ? ?