Further Trigonometric Identities and their Applications

Further Trigonometric Identities and their Applications

You need to know and be able to use the addition formulae Q By GCSE Trigonometry: 1 P 1 So the coordinates of P are: Type equation here.Type equation here. here. B A O M N So the coordinates of Q are: Q 𝑆𝑖𝑛𝐴−𝑆𝑖𝑛𝐵 P 7A

You need to know and be able to use the addition formulae 𝑃 𝑄 2 = (𝐶𝑜𝑠𝐴−𝐶𝑜𝑠𝐵 ) 2 +(𝑆𝑖𝑛𝐴−𝑆𝑖𝑛𝐵 ) 2 Multiply out the brackets 𝑃 𝑄 2 = (𝐶𝑜 𝑠 2 𝐴−2𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝐶𝑜 𝑠 2 𝐵) + (𝑆𝑖 𝑛 2 𝐴−2𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵+𝑆𝑖 𝑛 2 𝐵) Rearrange 𝑃 𝑄 2 = (𝐶𝑜 𝑠 2 𝐴+𝑆𝑖 𝑛 2 𝐴) + (𝐶𝑜 𝑠 2 𝐵+𝑆𝑖 𝑛 2 𝐵) − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) 𝐶𝑜 𝑠 2 θ+𝑆𝑖 𝑛 2 θ ≡ 1 𝑃 𝑄 2 = 2 − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) 7A

You need to know and be able to use the addition formulae Q You can also work out PQ using the triangle OPQ: Q 1 P 1 1 B P A B - A 1 O M N 𝑎 2 = 𝑏 2 + 𝑐 2 − 2bcCosA Sub in the values 𝑃𝑄 2 = − 2Cos(B - A) Group terms 𝑃𝑄 2 =2 − 2Cos(B - A) Cos (B – A) = Cos (A – B) eg) Cos(60) = Cos(-60) 𝑃𝑄 2 =2 − 2Cos(A - B) 7A

You need to know and be able to use the addition formulae 𝑃 𝑄 2 = 2 − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) 𝑃𝑄 2 =2 − 2Cos(A - B) 2 − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) = 2 − 2Cos(A - B) Subtract 2 from both sides − 2(𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵) = − 2Cos(A - B) Divide by -2 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵+𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 = Cos(A - B) Cos(A - B) = CosACosB + SinASinB Cos(A + B) = CosACosB - SinASinB 7A

You need to know and be able to use the addition formulae Cos(A - B) ≡ CosACosB + SinASinB Cos(A + B) ≡ CosACosB - SinASinB Sin(A + B) ≡ SinACosB + CosASinB Sin(A - B) ≡ SinACosB - CosASinB 7A

Tan (A+B) ≡ 𝑆𝑖𝑛(𝐴+𝐵) 𝐶𝑜𝑠(𝐴+𝐵) You need to know and be able to use the addition formulae Show that: Rewrite Sin(A + B) = SinACosB + CosASinB Tan (A+B) ≡ 𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵+𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵−𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 Sin(A - B) = SinACosB - CosASinB Divide top and bottom by CosACosB Cos(A - B) = CosACosB + SinASinB Tan (A+B) ≡ 𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 − 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 Cos(A + B) = CosACosB - SinASinB Simplify each Fraction TanA + TanB Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 Tan (A+B) ≡ 1 - TanATanB Tan θ ≡ 𝑆𝑖𝑛θ 𝐶𝑜𝑠θ 7A

You need to know and be able to use the addition formulae Cos(A - B) ≡ CosACosB + SinASinB Cos(A + B) ≡ CosACosB - SinASinB Sin(A + B) ≡ SinACosB + CosASinB Sin(A - B) ≡ SinACosB - CosASinB Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 You may be asked to prove either of the Tan identities using the Sin and Cos ones! Tan (A - B) ≡ 𝑇𝑎𝑛𝐴−𝑇𝑎𝑛𝐵 1+𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 7A

This is how they look on the formula sheet

You need to know and be able to use the addition formulae Show, using the formula for Sin(A – B), that: 𝑆𝑖𝑛15=𝑆𝑖𝑛(45−30) Cos(A + B) ≡ CosACosB - SinASinB Sin(A - B) ≡ SinACosB - CosASinB Cos(A - B) ≡ CosACosB + SinASinB A=45, B=30 Sin(A + B) ≡ SinACosB + CosASinB Sin( ) ≡ Sin45Cos30 – Cos45Sin30 These can be written as surds Sin(A - B) ≡ SinACosB - CosASinB 2 2 3 2 2 2 1 2 Sin( ) ≡ × − × Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 Multiply each pair Tan (A - B) ≡ 𝑇𝑎𝑛𝐴−𝑇𝑎𝑛𝐵 1+𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 6 4 2 4 Sin( ) ≡ − Group the fractions up 6 − 2 4 Sin(15) ≡ 𝑆𝑖𝑛15= 6 − 2 4 7A

You need to know and be able to use the addition formulae Given that: Find the value of: 𝑆𝑖𝑛𝐴= 𝑂𝑝𝑝 𝐻𝑦𝑝 𝑇𝑎𝑛𝐴= 𝑂𝑝𝑝 𝐴𝑑𝑗 A 5 𝑆𝑖𝑛𝐴=− 3 5 3 𝑇𝑎𝑛𝐴= 3 4 SinA=− ˚ < A < 270˚ 4 CosB=− B=Obtuse Use Pythagoras’ to find the missing side (ignore negatives) Tan is positive in the range 180˚ - 270˚ Tan(A+B) 𝑇𝑎𝑛𝐵= 𝑂𝑝𝑝 𝐴𝑑𝑗 B 𝐶𝑜𝑠𝐵= 𝐴𝑑𝑗 𝐻𝑦𝑝 13 5 Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 𝑇𝑎𝑛𝐵= 5 12 𝐶𝑜𝑠𝐵=− 12 13 12 𝑇𝑎𝑛𝐵=− 5 12 y = Tanθ Use Pythagoras’ to find the missing side (ignore negatives) 90 180 270 360 Tan is negative in the range 90˚ - 180˚ 7A

You need to know and be able to use the addition formulae Given that: Find the value of: Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 Substitute in TanA and TanB SinA=− ˚ < A < 270˚ Tan (A + B) ≡ − − 3 4 ×− 5 12 CosB=− B=Obtuse Work out the Numerator and Denominator Tan (A + B) ≡ Tan(A+B) Leave, Change and Flip Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 Tan (A + B) ≡ 1 3 × 48 63 Simplify 𝑇𝑎𝑛𝐴= 3 4 𝑇𝑎𝑛𝐵=− 5 12 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… 7A

You need to know and be able to use the addition formulae Given that: Express Tanx in terms of Tany… 2 𝑠𝑖𝑛 𝑥+𝑦 =3𝑐𝑜𝑠⁡(𝑥−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 7A

Have a go at exercise 7A

Can you derive some formula for sin2A, cos2A, tan2A

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Sin(A + B) ≡ SinACosB + CosASinB Replace B with A Sin(A + A) ≡ SinACosA + CosASinA Simplify Sin2A ≡ 2SinACosA 1 2 Sin2A ≡ SinACosA Sin4A ≡ 2Sin2ACos2A ÷ 2 2A  4A Sin2A ≡ 2SinACosA x 3 2A = 60 3Sin2A ≡ 6SinACosA Sin60 ≡ 2Sin30Cos30 7B

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Cos(A + B) ≡ CosACosB - SinASinB Replace B with A Cos(A + A) ≡ CosACosA - SinASinA Simplify 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 7B

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Tan (A + B) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐵 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐵 Replace B with A Tan (A + A) ≡ 𝑇𝑎𝑛𝐴+𝑇𝑎𝑛𝐴 1−𝑇𝑎𝑛𝐴𝑇𝑎𝑛𝐴 Simplify 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 ÷ 2 2A = 60 Tan 2A ≡ 2𝑇𝑎𝑛𝐴 1−𝑇𝑎 𝑛 2 𝐴 x 2 Tan A ≡ 2𝑇𝑎𝑛 𝐴 2 1−𝑇𝑎 𝑛 2 𝐴 2 2Tan 2A ≡ 4𝑇𝑎𝑛𝐴 1−𝑇𝑎 𝑛 2 𝐴 2A = A 7B

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Rewrite the following as a single Trigonometric function: 𝑆𝑖𝑛2𝜃≡2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 2θ  θ 𝑆𝑖𝑛𝜃≡2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠𝜃 2𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠𝜃 Replace the first part =𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 Rewrite = 1 2 𝑠𝑖𝑛2𝜃 7B

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Show that: Can be written as: 𝐶𝑜𝑠2𝜃≡2𝑐𝑜 𝑠 2 𝜃 −1 Double the angle parts 𝐶𝑜𝑠4𝜃≡2𝑐𝑜 𝑠 2 2𝜃 −1 1+𝑐𝑜𝑠4𝜃 1+𝑐𝑜𝑠4𝜃 Replace cos4θ 2𝑐𝑜 𝑠 2 2𝜃 = 1+(2𝑐𝑜 𝑠 2 2𝜃 −1) The 1s cancel out =2𝑐𝑜 𝑠 2 2𝜃 7B

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Given that: Find the exact value of: 𝐶𝑜𝑠𝑥= 𝐴𝑑𝑗 𝐻𝑦𝑝 𝑆𝑖𝑛𝑥= 𝑂𝑝𝑝 𝐻𝑦𝑝 4 7 𝐶𝑜𝑠𝑥= 3 4 𝑆𝑖𝑛𝑥= x 3 𝑐𝑜𝑠𝑥= 3 4 180˚<𝑥<360˚ Use Pythagoras’ to find the missing side (ignore negatives) Cosx is positive so in the range 𝑠𝑖𝑛2𝑥 𝑆𝑖𝑛𝑥=− Therefore, Sinx is negative y = Cosθ 90 180 270 360 Sin2x ≡ 2SinxCosx Sub in Sinx and Cosx Sin2x = 2 × 3 4 ×− Work out and leave in surd form y = Sinθ Sin2x = − 7B

You can express sin2A, cos 2A and tan2A in terms of angle A, using the double angle formulae Given that: Find the exact value of: 𝐶𝑜𝑠𝑥= 𝐴𝑑𝑗 𝐻𝑦𝑝 𝑇𝑎𝑛𝑥= 𝑂𝑝𝑝 𝐴𝑑𝑗 4 7 𝐶𝑜𝑠𝑥= 3 4 𝑇𝑎𝑛𝑥= x 3 𝑐𝑜𝑠𝑥= 3 4 180˚<𝑥<360˚ Use Pythagoras’ to find the missing side (ignore negatives) Cosx is positive so in the range 𝑡𝑎𝑛2𝑥 𝑇𝑎𝑛𝑥=− Therefore, Tanx is negative y = Cosθ Tan 2x ≡ 2𝑇𝑎𝑛𝑥 1−𝑇𝑎 𝑛 2 𝑥 90 180 270 360 Sub in Tanx Tan 2x = 2×− − − ×− y = Tanθ Work out and leave in surd form 90 180 270 360 𝑇𝑎𝑛2𝑥=−3 7 7B

Exercise 7B

The double angle formulae allow you to solve more equations and prove more identities Prove the identity: 𝑡𝑎𝑛2𝜃≡ 2𝑡𝑎𝑛𝜃 1−𝑡𝑎 𝑛 2 𝜃 Divide each part by tanθ 𝑡𝑎𝑛2𝜃≡ 2𝑡𝑎𝑛𝜃 𝑡𝑎𝑛𝜃 1 𝑡𝑎𝑛𝜃 − 𝑡𝑎 𝑛 2 𝜃 𝑡𝑎𝑛𝜃 𝑡𝑎𝑛2𝜃≡ 2 𝑐𝑜𝑡𝜃−𝑡𝑎𝑛𝜃 Rewrite each part 2 𝑡𝑎𝑛2𝜃≡ 𝑐𝑜𝑡𝜃 − 𝑡𝑎𝑛𝜃 7C

The double angle formulae allow you to solve more equations and prove more identities By expanding: Show that: 𝑠𝑖 𝑛 𝐴+𝐵 ≡𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵+𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 Replace A and B 𝑠𝑖 𝑛 2𝐴+𝐴 ≡𝑠𝑖𝑛2𝐴𝑐𝑜𝑠𝐴+𝑐𝑜𝑠2𝐴𝑠𝑖𝑛𝐴 Replace Sin2A and Cos 2A 𝑠𝑖 𝑛 3𝐴 ≡ (2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴)𝑐𝑜𝑠𝐴 + (1−2𝑠𝑖 𝑛 2 𝐴)𝑠𝑖𝑛𝐴 𝑠𝑖𝑛⁡(2𝐴+𝐴) Multiply out 𝑠𝑖 𝑛 3𝐴 ≡ 2𝑠𝑖𝑛𝐴𝑐𝑜 𝑠 2 𝐴 + 𝑠𝑖𝑛𝐴−2𝑠𝑖 𝑛 3 𝐴 Replace cos2A 𝑠𝑖 𝑛 3𝐴 ≡3𝑠𝑖𝑛𝐴−4𝑠𝑖 𝑛 3 𝐴 𝑠𝑖 𝑛 3𝐴 ≡ 2𝑠𝑖𝑛𝐴(1− 𝑠𝑖𝑛 2 𝐴) + 𝑠𝑖𝑛𝐴−2𝑠𝑖 𝑛 3 𝐴 Multiply out 𝑠𝑖 𝑛 3𝐴 ≡ 2𝑠𝑖𝑛𝐴− 2𝑠𝑖𝑛 3 𝐴 + 𝑠𝑖𝑛𝐴−2𝑠𝑖 𝑛 3 𝐴 Group like terms 𝑠𝑖 𝑛 3𝐴 ≡ 3𝑠𝑖𝑛𝐴− 4𝑠𝑖𝑛 3 𝐴 7C

The double angle formulae allow you to solve more equations and prove more identities Given that: Eliminate θ and express y in terms of x… 𝑐𝑜𝑠2𝜃=1−2𝑠𝑖 𝑛 2 𝜃 Replace Cos2θ and Sinθ 3−𝑦 4 = 1−2 𝑥 3 2 Multiply by 4 𝑥=3𝑠𝑖𝑛𝜃 and 𝑦=3−4𝑐𝑜𝑠2𝜃 4−8 𝑥 3 2 3−𝑦 = Subtract 3 1−8 𝑥 3 2 𝑥=3𝑠𝑖𝑛𝜃 −𝑦 = Divide by 3 Multiply by -1 𝑥 3 =𝑠𝑖𝑛𝜃 8 𝑥 −1 𝑦 = 𝑦=3−4𝑐𝑜𝑠2𝜃 Subtract 3, divide by 4 Multiply by -1 3−𝑦 4 =𝑐𝑜𝑠2𝜃 7C

The double angle formulae allow you to solve more equations and prove more identities Solve the following equation in the range stated: (All trigonometrical parts must be in terms x, rather than 2x) 3𝑐𝑜𝑠2𝑥−𝑐𝑜𝑠𝑥+2=0 Replace cos2x 3(2𝑐𝑜 𝑠 2 𝑥−1)−𝑐𝑜𝑠𝑥+2=0 Multiply out the bracket 6𝑐𝑜 𝑠 2 𝑥−3−𝑐𝑜𝑠𝑥+2=0 Group terms 6𝑐𝑜 𝑠 2 𝑥−𝑐𝑜𝑠𝑥−1=0 3𝑐𝑜𝑠2𝑥−𝑐𝑜𝑠𝑥+2=0 Factorise (3𝑐𝑜𝑠𝑥+1)(2𝑐𝑜𝑠𝑥−1)=0 0°≤𝑥≤360° 𝑐𝑜𝑠𝑥=− 1 3 𝑐𝑜𝑠𝑥= 1 2 or Solve both pairs 𝑥=𝑐𝑜 𝑠 −1 − 1 3 𝑥=𝑐𝑜 𝑠 − y = Cosθ Remember to find additional answers! 1 2 𝑥=109.5° , 250.5° 𝑥=60° , 300° − 1 3 90 180 270 360 𝑥=60°, 109.5°, 250.5°, 300° 7C

Exercise 7D

𝑅𝑠𝑖𝑛(𝑥+α) =𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠α + 𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛α You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only Show that: Can be expressed in the form: So: Replace with the expression 3𝑠𝑖𝑛𝑥+4𝑐𝑜𝑠𝑥 =𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠α + 𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛α Compare each term – they must be equal! 𝑅𝑐𝑜𝑠α=3 𝑅𝑠𝑖𝑛α=4 Square and add to get R2 3𝑠𝑖𝑛𝑥+4𝑐𝑜𝑠𝑥 𝑅=5 𝑅𝑠𝑖𝑛(𝑥+α) 𝑅>0 0°<α<90° Divide sin equation by cos equation to get tan equation α=53.1° 3𝑠𝑖𝑛𝑥+4𝑐𝑜𝑠𝑥 =5sin⁡(𝑥+53.1°)

You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only Show that you can express: In the form: So: 𝑅𝑠𝑖𝑛(𝑥−α) = 𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛼−𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛼 Replace with the expression 𝑠𝑖𝑛𝑥− 3 𝑐𝑜𝑠𝑥 = 𝑅𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝛼−𝑅𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝛼 Compare each term – they must be equal! 𝑅𝑐𝑜𝑠𝛼=1 𝑅𝑠𝑖𝑛𝛼= 3 𝑠𝑖𝑛𝑥− 3 𝑐𝑜𝑠𝑥 Square and add to get R2 𝑅>0 0<α< 𝜋 2 𝑅=2 𝑅𝑠𝑖𝑛(𝑥−α) Divide sin equation by cos equation to get tan equation 𝛼= 𝜋 3 𝑠𝑖𝑛𝑥− 3 𝑐𝑜𝑠𝑥 =2sin⁡ 𝑥− 𝜋 3 7D

Sketch the graph of: 𝑠𝑖𝑛𝑥− 3 𝑐𝑜𝑠𝑥 You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only Show that you can express: In the form: So: 2sin⁡ 𝑥− 𝜋 3 = Sketch the graph of: 1 y=sin⁡𝑥 Start out with sinx 𝑠𝑖𝑛𝑥− 3 𝑐𝑜𝑠𝑥 -1 π/2 π 3π/2 2π 𝑅>0 0<α< 𝜋 2 𝑅𝑠𝑖𝑛(𝑥−α) y=sin⁡ 𝑥− 𝜋 3 1 Translate π/3 units right 𝑠𝑖𝑛𝑥− 3 𝑐𝑜𝑠𝑥 -1 π/3 π/2 π 4π/3 3π/2 2π =2sin⁡ 𝑥− 𝜋 3 2 y=2sin⁡ 𝑥− 𝜋 3 Vertical stretch, scale factor 2 1 2sin⁡ − 𝜋 3 -1 π/3 π/2 π 4π/3 3π/2 2π =− 3 At the y-intercept, x = 0 -2 7D

𝑅𝑐𝑜𝑠(𝜃−𝛼) = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼 You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only Express: in the form: So: Replace with the expression 2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃 = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼 Compare each term – they must be equal! 𝑅𝑐𝑜𝑠𝛼=2 𝑅𝑠𝑖𝑛𝛼=5 2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃 Square and add to get R2 𝑅= 29 𝑅>0 0°<𝛼<90° 𝑅𝑐𝑜𝑠(𝜃−𝛼) Divide sin equation by cos equation to get tan equation 2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃 𝛼=68.2 = 29 cos(𝜃−68.2) 7D

29 cos(𝜃−68.2)=3 You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only Solve in the given range, the following equation: Divide by √29 cos(𝜃−68.2)= Inverse Cos 𝜃−68.2=𝑐𝑜 𝑠 − Remember to work out other values in the adjusted range 2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃=3 0°<𝜃<360° 𝜃−68.2=56.1 , −56.1 , 303.9 We just showed that the original equation can be rewritten… Add 68.2 (and put in order!) 𝜃=12.1 , 124.3 2𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃= 29 cos(𝜃−68.2) Hence, we can solve this equation instead! 29 cos(𝜃−68.2)=3 y = Cosθ -56.1 56.1 303.9 0°<𝜃<360° Remember to adjust the range for (θ – 68.2) -90 90 180 270 360 −68.2°<𝜃−68.2<291.2° 7D

Rcos(θ – α) chosen as it gives us the same form as the expression You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only Find the maximum value of the following expression, and the smallest positive value of θ at which it arises: 𝑅𝑐𝑜𝑠(𝜃−𝛼) = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼 Replace with the expression 12𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃 = 𝑅𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛼+𝑅𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛼 Compare each term – they must be equal! 𝑅𝑐𝑜𝑠𝛼=12 𝑅𝑠𝑖𝑛𝛼=5 Square and add to get R2 𝑅=13 12𝑐𝑜𝑠𝜃+5𝑠𝑖𝑛𝜃 =13cos⁡(𝜃−22.6) Divide sin equation by cos equation to get tan equation 13cos⁡(𝜃−22.6) Max value of cos(θ ) = 1 𝛼=22.6 13(1) Overall maximum therefore = 13 𝑀𝑎𝑥=13 Cos peaks at 0 𝜃−22.6=0 θ = 22.6 gives us 0 𝜃=22.6 7D

You can write expressions of the form acosθ + bsinθ, where a and b are constants, as a sine or cosine function only 𝑎𝑠𝑖𝑛𝜃±𝑏𝑐𝑜𝑠𝜃 𝑅𝑠𝑖𝑛 𝜃±𝛼 𝑎𝑐𝑜𝑠𝜃±𝑏𝑠𝑖𝑛𝜃 𝑅𝑐𝑜𝑠 𝜃∓𝛼 Whichever ratio is at the start, change the expression into a function of that (This makes solving problems easier) Remember to get the + or – signs the correct way round! 7D

Exercise 7D

You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 You get given all these in the formula booklet! 7E

Using the formulae for Sin(A + B) and Sin (A – B), derive the result that: You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 1) 𝑆𝑖𝑛 𝐴+𝐵 =𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵+𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 2) 𝑆𝑖𝑛 𝐴−𝐵 =𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵−𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 Add both sides together (1 + 2) 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑆𝑖𝑛 𝐴+𝐵 +𝑠𝑖𝑛⁡(𝐴−𝐵)=2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 Let (A+B) = P Let (A-B) = Q 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 1) 𝐴+𝐵=𝑃 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 1) 𝐴+𝐵=𝑃 2) 𝐴−𝐵=𝑄 2) 𝐴−𝐵=𝑄 1 + 2 1 - 2 2𝐴=𝑃+𝑄 2𝐵=𝑃−𝑄 𝑆𝑖𝑛 𝐴+𝐵 +𝑠𝑖𝑛⁡(𝐴−𝐵)=2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 Divide by 2 Divide by 2 𝐴= 𝑃+𝑄 2 𝐵= 𝑃−𝑄 2 𝑃+𝑄 2 𝑃−𝑄 2 𝑆𝑖𝑛𝑃 + 𝑆𝑖𝑛𝑄 =2𝑠𝑖𝑛 𝑐𝑜𝑠 7E

You can derive the rest in the same way if you really want to……

Show that: You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ 𝑠𝑖𝑛105−𝑠𝑖𝑛15= 1 2 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 P = Q = 15 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑠𝑖𝑛105−𝑠𝑖𝑛15=2𝑐𝑜𝑠 𝑠𝑖𝑛 105−15 2 Work out the fraction parts 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛105−𝑠𝑖𝑛15=2𝑐𝑜𝑠60𝑠𝑖𝑛45 Sub in values for Cos60 and Sin45 1 2 1 2 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑠𝑖𝑛105−𝑠𝑖𝑛15= 2 × × Work out the right hand side 1 2 𝑠𝑖𝑛105−𝑠𝑖𝑛15= 7E

Solve in the range indicated: You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ 𝑠𝑖𝑛4𝜃−𝑠𝑖𝑛3𝜃=0 0≤𝜃≤𝜋 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 P = 4θ Q = 3θ 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛4𝜃−𝑠𝑖𝑛3𝜃=2𝑐𝑜𝑠 4𝜃+3𝜃 2 𝑠𝑖𝑛 4𝜃−3𝜃 2 Work out the fractions 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑠𝑖𝑛4𝜃−𝑠𝑖𝑛3𝜃=2𝑐𝑜𝑠 7𝜃 2 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Set equal to 0 2𝑐𝑜𝑠 7𝜃 2 𝑠𝑖𝑛 𝜃 2 =0 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 Either the cos or sin part must equal 0… 𝑐𝑜𝑠 7𝜃 2 =0 0≤𝜃≤𝜋 Adjust the range Inverse cos 0≤ 7𝜃 2 ≤ 7𝜋 2 7𝜃 2 =𝑐𝑜 𝑠 −1 0 Solve, remembering to take into account the different range Once you have all the values from 0-2π, add 2π to them to obtain equivalents… 7𝜃 2 = 𝜋 2 , 3𝜋 2 , 5𝜋 2 , 7𝜋 2 y = Cosθ π/2 π 3π/2 2π Multiply by 2 and divide by 7 𝜃= 𝜋 7 , 3𝜋 7 , 5𝜋 7 , 𝜋 7E

Solve in the range indicated: You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ 𝑠𝑖𝑛4𝜃−𝑠𝑖𝑛3𝜃=0 0≤𝜃≤𝜋 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 P = 4θ Q = 3θ 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛4𝜃−𝑠𝑖𝑛3𝜃=2𝑐𝑜𝑠 4𝜃+3𝜃 2 𝑠𝑖𝑛 4𝜃−3𝜃 2 Work out the fractions 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑠𝑖𝑛4𝜃−𝑠𝑖𝑛3𝜃=2𝑐𝑜𝑠 7𝜃 2 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Set equal to 0 2𝑐𝑜𝑠 7𝜃 2 𝑠𝑖𝑛 𝜃 2 =0 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 Either the cos or sin part must equal 0… 𝑠𝑖𝑛 𝜃 2 =0 0≤𝜃≤𝜋 Adjust the range Inverse sin 0≤ 𝜃 2 ≤ 𝜋 2 𝜃 2 =𝑠𝑖 𝑛 −1 0 Solve, remembering to take into account the different range Once you have all the values from 0-2π, add 2π to them to obtain equivalents 𝜃 2 =0 y = Sinθ π/2 π 3π/2 2π Multiply by 2 𝜃=0 7E

In the numerator: You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ Prove that: 𝑠𝑖𝑛 𝑥+2𝑦 +𝑠𝑖𝑛 𝑥+𝑦 +𝑠𝑖𝑛𝑥 Ignore sin(x + y) for now as we want x+y in solution… 𝑠𝑖𝑛 𝑥+2𝑦 +𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Use the identity for adding 2 sines 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 P = x + 2y Q = x 𝑥+2𝑦+𝑥 2 𝑥+2𝑦−𝑥 2 𝑠𝑖𝑛⁡(𝑥+2𝑦)+𝑠𝑖𝑛𝑥=2𝑠𝑖𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Simplify Fractions =2 𝑠𝑖𝑛 𝑥+𝑦 𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 Bring back the sin(x + y) we ignored earlier =2 𝑠𝑖𝑛 𝑥+𝑦 𝑐𝑜𝑠𝑦+𝑠𝑖𝑛⁡(𝑥+𝑦) Factorise 𝑠𝑖𝑛 𝑥+2𝑦 +𝑠𝑖𝑛 𝑥+𝑦 +𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 𝑥+2𝑦 + 𝑐𝑜𝑠 𝑥+𝑦 +𝑐𝑜𝑠𝑥 =𝑡𝑎𝑛⁡(𝑥+𝑦) = 𝑠𝑖𝑛 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) Numerator: 𝑠𝑖𝑛 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) 7E

In the denominator: You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ Prove that: 𝑐𝑜𝑠 𝑥+2𝑦 +𝑐𝑜𝑠 𝑥+𝑦 +𝑐𝑜𝑠𝑥 Ignore cos(x + y) for now… 𝑐𝑜𝑠 𝑥+2𝑦 +𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Use the identity for adding 2 cosines 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 P = x + 2y Q = x 𝑥+2𝑦+𝑥 2 𝑥+2𝑦−𝑥 2 𝑐𝑜𝑠⁡(𝑥+2𝑦)+𝑐𝑜𝑠𝑥=2𝑐𝑜𝑠 𝑐𝑜𝑠 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Simplify Fractions =2 𝑐𝑜𝑠 𝑥+𝑦 𝑐𝑜𝑠𝑦 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 Bring back the cos(x + y) we ignored earlier =2 𝑐𝑜𝑠 𝑥+𝑦 𝑐𝑜𝑠𝑦+𝑐𝑜𝑠⁡(𝑥+𝑦) Factorise 𝑠𝑖𝑛 𝑥+2𝑦 +𝑠𝑖𝑛 𝑥+𝑦 +𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 𝑥+2𝑦 + 𝑐𝑜𝑠 𝑥+𝑦 +𝑐𝑜𝑠𝑥 =𝑡𝑎𝑛⁡(𝑥+𝑦) = 𝑐𝑜𝑠 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) Numerator: 𝑠𝑖𝑛 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) Denominator: 𝑐𝑜𝑠 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) 7E

You can express sums and differences of sines and cosines as products of sines and cosines by using the ‘factor formulae’ Prove that: 𝑠𝑖𝑛 𝑥+2𝑦 +𝑠𝑖𝑛 𝑥+𝑦 +𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 𝑥+2𝑦 + 𝑐𝑜𝑠 𝑥+𝑦 +𝑐𝑜𝑠𝑥 Replace the numerator and denominator = 𝑠𝑖𝑛(𝑥+𝑦)(2𝑐𝑜𝑠𝑦+1) 𝑐𝑜𝑠(𝑥+𝑦)(2𝑐𝑜𝑠𝑦+1) 𝑠𝑖𝑛𝑃+𝑠𝑖𝑛𝑄=2𝑠𝑖𝑛 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 Cancel out the (2cosy + 1) brackets 𝑠𝑖𝑛𝑃−𝑠𝑖𝑛𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 = 𝑠𝑖𝑛(𝑥+𝑦) 𝑐𝑜𝑠(𝑥+𝑦) Use one of the identities from C2 𝑐𝑜𝑠𝑃+𝑐𝑜𝑠𝑄=2𝑐𝑜𝑠 𝑃+𝑄 2 𝑐𝑜𝑠 𝑃−𝑄 2 =𝑡𝑎𝑛⁡(𝑥+𝑦) 𝑐𝑜𝑠𝑃−𝑐𝑜𝑠𝑄=−2𝑠𝑖𝑛 𝑃+𝑄 2 𝑠𝑖𝑛 𝑃−𝑄 2 𝑠𝑖𝑛 𝑥+2𝑦 +𝑠𝑖𝑛 𝑥+𝑦 +𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 𝑥+2𝑦 + 𝑐𝑜𝑠 𝑥+𝑦 +𝑐𝑜𝑠𝑥 =𝑡𝑎𝑛⁡(𝑥+𝑦) Numerator: 𝑠𝑖𝑛 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) Denominator: 𝑐𝑜𝑠 𝑥+𝑦 (2𝑐𝑜𝑠𝑦+1) 7E

Exercise 7E and 7F

Summary We have extended the range of techniques we have for solving trigonometrical equations We have seen how to combine functions involving sine and cosine into a single transformation of sine or cosine We have learnt several new identities

Further Trigonometric Identities and their Applications

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Further Trigonometric Identities and their Applications

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Presentation on theme: "Further Trigonometric Identities and their Applications"— Presentation transcript:

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