Trig addition formulae sin 𝐴±𝐵 = sin 𝐴 cos 𝐵 ± sin 𝐵 cos 𝐴 cos 𝐴±𝐵 = cos 𝐴 cos 𝐵 ∓ sin 𝐵 sin 𝐴 tan 𝐴±𝐵 = tan 𝐴 ± 𝑡𝑎𝑛 𝐵 1∓ tan 𝐴 tan 𝐵 Proofs

Trigonometry: Addition formulas KUS objectives BAT understand where the addition formula come from and derive the double angle formulas BAT use the addition formulae to solve ‘show that’ problems BAT use the addition formulae to solve equations Starter:

WB 1a: Show that 𝑏) co𝑠 75 = 6 − 2 4 𝑎) sin 15 = 6 − 2 4 𝑎) 𝑆𝑖𝑛15=𝑆𝑖𝑛(45−30) Sin(A - B) ≡ SinACosB - CosASinB Sin(45 - 30) = Sin45Cos30 – Cos45Sin30 𝑆𝑖𝑛 45 − 30 = 2 2 × 3 2 − 2 2 × 1 2 = 6 4 − 2 4 = 6 − 2 4 QED

WB 1b: Show that 𝑏) co𝑠 75 = 6 − 2 4 𝑎) sin 15 = 6 − 2 4 b) cos 75 =𝑐𝑜𝑠(45+30) 𝑐𝑜𝑠 45+ 30 = cos 45 cos 30 − sin 45 sin 30 𝑐𝑜𝑠 45+ 30 = 2 2 × 3 2 − 2 2 × 1 2 = 6 4 − 2 4 = 6 − 2 4 QED

WB 2: a) Use the identity sin⁡(𝐴+𝐵)=sin⁡𝐴 cos⁡𝐵+sin⁡𝐵 cos⁡𝐴 to show that 2 sin 𝑥+ 𝜋 3 = sin 𝑥 + 3 cos 𝑥 b) Use the identity 𝑐𝑜𝑠⁡(𝐴−𝐵)=𝑐𝑜𝑠⁡𝐴 cos⁡𝐵+ sin 𝐴 sin 𝐵 to show that sin 𝑥 −2 𝑐𝑜𝑠 𝑥− 5𝜋 6 = 3 cos 𝑥 = sin 𝑥 1 2 + cos 𝑥 3 2 𝑎) sin 𝑥+ 𝜋 3 = sin 𝑥 cos 𝜋 3 + sin 𝜋 3 cos 𝑥 So 2 sin 𝑥+ 𝜋 3 = sin 𝑥 + 3 cos 𝑥 = cos 𝑥 − 3 2 + sin 𝑥 1 2 𝑏) cos 𝑥− 𝜋 6 = cos 𝑥 cos 5𝜋 6 + sin 𝑥 cos 5𝜋 6 So sin x −2 cos 𝑥− 𝜋 6 = sin 𝑥 − − 3 cos 𝑥 + sin 𝑥 = 3 cos 𝑥

WB 3: Given that sin 𝐴 =− 3 5 in the range 180<𝐴<270 and cos 𝐵 =− 12 13 where B is Obtuse . Find the value of tan 𝐴+𝐵 𝑆𝑖𝑛𝐴= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝑇𝑎𝑛𝐴= 𝑂𝑝𝑝 𝐴𝑑𝑗 5 A 𝑆𝑖𝑛𝐴=− 3 5 3 𝑇𝑎𝑛𝐴= 3 4 90 180 270 360 y = Tanθ 4 Use Pythagoras’ to find the missing side (ignore negatives) Tan is positive in the range 180˚ - 270˚ 𝑇𝑎𝑛𝐵= 𝑂𝑝𝑝 𝐴𝑑𝑗 B 𝐶𝑜𝑠𝐵= 𝐴𝑑𝑗 𝐻𝑦𝑝 13 5 𝑇𝑎𝑛𝐵= 5 12 90 180 270 360 y = Tanθ 𝐶𝑜𝑠𝐵=− 12 13 12 𝑇𝑎𝑛𝐵=− 5 12 Use Pythagoras’ to find the missing side (ignore negatives) Tan is negative in the range 90˚ - 180˚ Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵

WB 3 (cont): Given that sin 𝐴 =− 3 5 in the range 180<𝐴<270 and cos 𝐵 =− 12 13 where B is Obtuse . Find the value of tan 𝐴+𝐵 Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 Substitute in TanA and TanB Tan (A + B) ≡ 3 4 + − 5 12 1− 3 4 ×− 5 12 Work out the Numerator and Denominator Tan (A + B) ≡ 1 3 63 48 Leave, Change and Flip Tan (A + B) ≡ 1 3 × 48 63 Simplify Tan (A + B) ≡ 16 63 Although you could just type the whole thing into your calculator, you still need to show the stages for the workings marks…

Given that 2 sin 𝑥+𝑦 =3 cos 𝑥−𝑦 Express tan 𝑥 in terms of tan 𝑦 WB 4: Given that 2 sin 𝑥+𝑦 =3 cos 𝑥−𝑦 Express tan 𝑥 in terms of tan 𝑦 2 𝑠𝑖𝑛 𝑥+𝑦 =3𝑐𝑜𝑠⁡(𝑥−𝑦) Rewrite the sin and cos parts 2 (𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦+𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦) = 3 (𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦+𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦) Multiply out the brackets 2 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦+2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 = 3 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦+3𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 Divide all by cosxcosy 2 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦+2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 = 3 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦+3𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 Simplify 2 𝑡𝑎𝑛𝑥+2𝑡𝑎𝑛𝑦 = 3 +3𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦 Subtract 3tanxtany Subtract 2tany 2 𝑡𝑎𝑛𝑥−3𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦 = 3 −2𝑡𝑎𝑛𝑦 Factorise the left side 𝑡𝑎𝑛𝑥(2−3𝑡𝑎𝑛𝑦) = 3 −2𝑡𝑎𝑛𝑦 𝑡𝑎𝑛𝑥 = 3 −2𝑡𝑎𝑛𝑦 Divide by (2 – 3tany) 2 −3𝑡𝑎𝑛𝑦

WB 5a: Solve each of the following equations for θ in the interval 0 ≤ θ ≤ 2π a) 𝑠𝑖𝑛 𝑥− 𝜋 3 =𝑐𝑜𝑠⁡𝑥 b) tan 3𝜃 + tan 𝜋 4 1− tan 3𝜃 tan 𝜋 4 = 3 2 𝑎) sin 𝑥− 𝜋 3 = sin 𝑥 cos 𝜋 3 − sin 𝜋 3 cos 𝑥 = 1 2 sin 𝑥 − 3 2 cos 𝑥 So 1 2 sin 𝑥 − 3 2 cos 𝑥 =cos x Rearrange to 1 2 sin 𝑥 = 1+ 3 2 cos x Divide through by ½ cos x tan 𝑥 =2 1+ 3 2 =2+ 3 Solve to give 𝑥= 5𝜋 12 , 17𝜋 12

WB 5b: Solve each of the following equations for θ in the interval 0 ≤ θ ≤ 2π a) 𝑠𝑖𝑛 𝑥− 𝜋 3 =𝑐𝑜𝑠⁡𝑥 b) tan 3𝑥 + tan 𝜋 4 1− tan 3𝑥 tan 𝜋 4 = 3 2 b) 𝐿𝐻𝑆= tan 3𝑥+ 𝜋 4 So tan 3𝑥+ 𝜋 4 = 3 2 Solve to give 3𝑥+ 𝜋 4 =0.983, 4.124 , 7.266, 10.408, … Solve to 𝑥=0.066, 1.113, 2.160,

self-assess using: R / A / G ‘I am now able to ____ . KUS objectives BAT understand where the addition formula come from and 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 ____’