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

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

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 ?

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

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 ?

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

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 ?

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

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

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

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.

Test Your Understanding Solve cos 𝑥 =0.9 in the range 0≤𝑥<360 ? 𝑥= cos −1 0.9 =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 =−0.68 → 𝜃=−42.84° 180−−42.84=𝟐𝟐𝟐.𝟖𝟒° −42.84+360=𝟑𝟏𝟕.𝟏𝟔° ?

Exercise 1 Solve the following in the range 0≤𝑥≤360 sin 𝑥 =0.5 → 𝒙=𝟑𝟎°, 𝟏𝟓𝟎° cos 𝑥 = 3 2 → 𝒙=𝟑𝟎°, 𝟑𝟑𝟎° tan 𝑥 = 3 → 𝒙=𝟔𝟎°, 𝟐𝟒𝟎° sin 𝑥 =0.1 → 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 =−0.5 → 𝟐𝟏𝟎°, 𝟑𝟑𝟎° cos 𝜃 =−0.5 → 𝟏𝟐𝟎°, 𝟐𝟒𝟎° tan 𝜃 =−1 → 𝟏𝟑𝟓°, 𝟑𝟏𝟓° sin 𝜃 =−0.4 → 𝟐𝟎𝟑.𝟓𝟖°, 𝟑𝟑𝟔.𝟒𝟐° cos 𝜃 =−0.7 → 𝟏𝟑𝟒.𝟒°, 𝟐𝟐𝟓.𝟔° tan 𝜃 =−0.2 → 𝟏𝟔𝟖.𝟔𝟗°, 𝟑𝟒𝟖.𝟔𝟗° 1 a ? b ? c ? d ? e ? f ? g ? 2 a ? b ? c ? d ? e ? f ?

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 2 . It does NOT mean the sin is being squared – this does not make sense as sin is not a quantity that we can square!

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°, 243.43° ? ? 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.

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°, 206.57° ? 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

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, 108.43°, 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

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

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 =− 1 2 𝑜𝑟 sin 𝜃=1 𝜃=210°, 330°, 90° ?

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 =0 → 𝜽=𝟗𝟎, 𝟐𝟕𝟎, 𝟎 tan 2 𝜃 −3 tan 𝜃 =0 → 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 = 3 4 → 𝜽=𝟔𝟎°, 𝟏𝟐𝟎°, 𝟐𝟒𝟎°, 𝟑𝟎𝟎° cos 2 𝜃 = 3 4 →𝜽=𝟑𝟎°, 𝟏𝟓𝟎°, 𝟐𝟏𝟎°, 𝟑𝟑𝟎° tan 2 𝜃 =3 →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=0 → 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

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

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…

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. ? ? ?

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 ?

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 2 + sin 𝜃 +2 cos 𝜃 2 ≡5 1 ? 2 ? ? ? 3

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 = 5 2 and that 𝜃 is acute, find the value of: sin 𝜃 = 𝟓 𝟑 cos 𝜃= 𝟐 𝟑 Given that cos 𝜃 = 5 13 and that 𝜃 is acute, find the value of: sin 𝜃 = 𝟏𝟐 𝟏𝟑 tan 𝜃= 𝟏𝟐 𝟓 1 2 ? ? ? ?