1. Double angle
formulae

2. Trigonometry: Addition formulas
KUS objectives
BAT derive the double angle formulas
BAT use the addition formulae to solve ‘show that’ problems
BAT use the addition formulae to solve equations
Starter:

3. WB6 Trig Double Angle formulas: Find a formula for Sin2A using
Sin(A + B) ≡ SinACosB + CosASinB
Sin(A + A) ≡ SinACosA + CosASinA
Sin2A ≡ 2SinACosA
1 2 Sin2A ≡ SinACosA
Sin4A ≡ 2Sin2ACos2A
Sin2A ≡ 2SinACosA
÷ 2
2A  4A
x 3
2A = 60
3Sin2A ≡ 6SinACosA
Sin60 ≡ 2Sin30Cos30

4. Replace Cos2A with (1 – Sin2A) Replace Sin2A with (1 – Cos2A)
Trig Double Angle formulas: Find a formula for Cos 2A using
Cos(A + B) ≡ CosACosB - SinASinB
Cos(A + A) ≡ CosACosA - SinASinA
Cos2A ≡ Co 𝑠 2 𝐴 −𝑆𝑖 𝑛 2 𝐴
Cos2A ≡ Co 𝑠 2 𝐴 −𝑆𝑖 𝑛 2 𝐴
Replace Cos2A with (1 – Sin2A)
Replace Sin2A with (1 – Cos2A)
Cos2A ≡ (1−𝑆𝑖 𝑛 2 𝐴)−𝑆𝑖 𝑛 2 𝐴
Cos2A ≡ Co 𝑠 2 𝐴 −(1 - Co 𝑠 2 𝐴)
Cos2A ≡ 1 − 2𝑆𝑖 𝑛 2 𝐴
Cos2A ≡ 2Co 𝑠 2 𝐴 −1
Cos2A ≡ Co 𝑠 2 𝐴 −𝑆𝑖 𝑛 2 𝐴
÷ 2
2A  4A
x 3
2A = 60
Each has three
possible answers

5. WB8 Trig Double Angle formulas: Find a formula for Tan 2A using
Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵
Tan (A + A) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐴 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐴
Tan 2A ≡ 2𝑇𝑎𝑛𝐴 1−𝑇𝑎 𝑛 2 𝐴
1 2 Tan 2A ≡ 𝑇𝑎𝑛𝐴 1−𝑇𝑎 𝑛 2 𝐴
Tan 60 ≡ 2𝑇𝑎𝑛30 1−𝑇𝑎 𝑛 2 30
Tan 2A ≡ 2𝑇𝑎𝑛𝐴 1−𝑇𝑎 𝑛 2 𝐴
÷ 2
2A = 60
x 2
2A = A
Tan A ≡ 2𝑇𝑎𝑛 𝐴 2 1−𝑇𝑎 𝑛 2 𝐴 2
2Tan 2A ≡ 4𝑇𝑎𝑛𝐴 1−𝑇𝑎 𝑛 2 𝐴

6. 𝑆𝑖𝑛2𝜃≡2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑆𝑖𝑛𝜃≡2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠𝜃 =𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
WB9:
Rewrite the following as a single Trigonometric function: 2 sin 𝜃 2 cos 𝜃 2 cos 𝜃
𝑆𝑖𝑛2𝜃≡2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
2θ  θ
𝑆𝑖𝑛𝜃≡2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2
2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠𝜃
Replace the first part
=𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
Rewrite
= 1 2 𝑠𝑖𝑛2𝜃

7. 𝐶𝑜𝑠2𝜃≡2𝑐𝑜 𝑠 2 𝜃 −1 𝐶𝑜𝑠4𝜃≡2𝑐𝑜 𝑠 2 2𝜃 −1 1+𝑐𝑜𝑠4𝜃 = 1+(2𝑐𝑜 𝑠 2 2𝜃 −1)
WB10: Show that 1+ cos 4𝜃 can be written as 2 𝑐𝑜𝑠 2 2𝜃
𝐶𝑜𝑠2𝜃≡2𝑐𝑜 𝑠 2 𝜃 −1
Double the angle parts
𝐶𝑜𝑠4𝜃≡2𝑐𝑜 𝑠 2 2𝜃 −1
1+𝑐𝑜𝑠4𝜃
Replace cos4θ
= 1+(2𝑐𝑜 𝑠 2 2𝜃 −1)
The 1s cancel out
=2𝑐𝑜 𝑠 2 2𝜃

8. WB11:
Given that cos 𝑥 = in the range [180, 360] find the exact value of 2 sin 2𝑥
𝐶𝑜𝑠𝑥= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑆𝑖𝑛𝑥= 𝑂𝑝𝑝 𝐻𝑦𝑝 4 7
𝐶𝑜𝑠𝑥= 3 4
𝑆𝑖𝑛𝑥= x 3
Use Pythagoras’ to find the missing side (ignore negatives)
𝑠𝑖𝑛2𝑥 90
180
270
360
y = Sinθ
y = Cosθ
Cosx is positive so in the range
𝑆𝑖𝑛𝑥=−
Therefore, Sinx is negative
2 Sin2x ≡ 4SinxCosx
Sub in Sinx and Cosx
2 Sin2x = 4 × 3 4 ×−
Work out and leave in surd form
2 Sin2x = −

9. WB12: Given that cos 𝑥 = 3 4 in the range [180, 360] find the exact value of tan 2𝑥
𝐶𝑜𝑠𝑥= 𝐴𝑑𝑗 𝐻𝑦𝑝
𝑇𝑎𝑛𝑥= 𝑂𝑝𝑝 𝐴𝑑𝑗 4 7
𝐶𝑜𝑠𝑥= 3 4
𝑇𝑎𝑛𝑥= x 3
Use Pythagoras’ to find the missing side (ignore negatives) 90
180
270
360
y = Cosθ
y = Tanθ
Cosx is positive so in the range
𝑇𝑎𝑛𝑥=−
Therefore, Tanx is negative
Tan 2x ≡ 2𝑇𝑎𝑛𝑥 1−𝑇𝑎 𝑛 2 𝑥
Sub in Tanx
Tan 2x = 2×− − − ×−
Work out and leave in surd form
𝑇𝑎𝑛2𝑥=−3 7

10. WB13: Show that tan 2𝜃 = 2 cot 𝜃 − tan 𝜃
𝑡𝑎𝑛2𝜃≡ 2𝑡𝑎𝑛𝜃 1−𝑡𝑎 𝑛 2 𝜃
Divide each part by tanθ
𝑡𝑎𝑛2𝜃≡ 2𝑡𝑎𝑛𝜃 𝑡𝑎𝑛𝜃 1 𝑡𝑎𝑛𝜃 − 𝑡𝑎 𝑛 2 𝜃 𝑡𝑎𝑛𝜃
Rewrite each part
Cancel terms
𝑡𝑎𝑛2𝜃≡ 2
𝑐𝑜𝑡𝜃
− 𝑡𝑎𝑛𝜃

11. Given that x =3 sin 𝜃 and 𝑦=3 −4 cos 2𝜃
WB14:
Given that x =3 sin 𝜃 and 𝑦=3 −4 cos 2𝜃
Eliminate  and express y in terms of x
𝑐𝑜𝑠2𝜃=1−2𝑠𝑖 𝑛 2 𝜃
𝑥=3𝑠𝑖𝑛𝜃
Divide by 3
𝑥 3 =𝑠𝑖𝑛𝜃
Replace Cos2θ and Sinθ
3−𝑦 4 =
1−2 𝑥 3 2
𝑦=3−4𝑐𝑜𝑠2𝜃
Multiply by 4
Subtract 3,
divide by 4
Multiply by -1
4−8 𝑥 3 2
3−𝑦 4 =𝑐𝑜𝑠2𝜃
3−𝑦 =
Subtract 3
1−8 𝑥 3 2
−𝑦 =
Multiply by -1
8 𝑥 −1
𝑦 =

12. WB15: Solve the equation 3 cos 2𝑥 −𝑐𝑜𝑠 𝑥 + 2=0 in the range 0≤𝑥≤360
3(2𝑐𝑜 𝑠 2 𝑥−1)−𝑐𝑜𝑠𝑥+2=0
𝐶𝑜𝑠2𝜃≡2𝑐𝑜 𝑠 2 𝜃 −1
6𝑐𝑜 𝑠 2 𝑥−𝑐𝑜𝑠𝑥−1=0
(3𝑐𝑜𝑠𝑥+1)(2𝑐𝑜𝑠𝑥−1)=0 90
180
270
360
y = Cosθ
1 2
− 1 3
𝑐𝑜𝑠𝑥=− or 𝑐𝑜𝑠𝑥= 1 2
𝑥=60°, 109.5°, 250.5°, 300°

13. 2𝑐𝑜 𝑠 2 𝜃 −1=7 cos 𝑥 +3 𝐶𝑜𝑠2𝜃≡2𝑐𝑜 𝑠 2 𝜃 −1 2𝑐𝑜 𝑠 2 𝜃 −7 cos 𝑥 −4=0
WB16:
Solve the equation cos2x = 7cosx for 0≤𝑥≤2π
2𝑐𝑜 𝑠 2 𝜃 −1=7 cos 𝑥 +3
𝐶𝑜𝑠2𝜃≡2𝑐𝑜 𝑠 2 𝜃 −1
2𝑐𝑜 𝑠 2 𝜃 −7 cos 𝑥 −4=0
( 2cos 𝑥 +1)( cos 𝑥 −4)=0
cos 𝑥 =− 1 2 , 4
x= 2𝜋 3 , 4𝜋 3

14. 2 sin 𝑥 cos 𝑥 = sin 𝑥 cos 𝑥 𝑠𝑖𝑛2𝐴≡2 sin 𝐴 𝑐𝑜𝑠 𝐴 2 sin 𝑥 𝑐𝑜𝑠 2 𝑥= sin 𝑥
WB17:
Solve the equation sin2x = tanx for 0≤𝑥≤2π
2 sin 𝑥 cos 𝑥 = sin 𝑥 cos 𝑥
𝑠𝑖𝑛2𝐴≡2 sin 𝐴 𝑐𝑜𝑠 𝐴
2 sin 𝑥 𝑐𝑜𝑠 2 𝑥= sin 𝑥
sin 𝑥 (2 𝑐𝑜𝑠 2 𝑥−1)=0
sin 𝑥 =0 or 𝑐𝑜𝑠 2 𝑥= 1 2
sin 𝑥 =0 or cos 𝑥 =
x=0,𝜋, 2𝜋
x= 𝜋 4 , 3𝜋 4 , 5𝜋 4 , 7𝜋 4

15. self-assess using: R / A / G ‘I am now able to ____ .
KUS objectives
BAT derive the double angle formulas
BAT use the addition formulae to solve ‘show that’ problems
BAT use the addition formulae to solve equations
self-assess using: R / A / G
‘I am now able to ____ .
To improve I need to be able to ____’