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

sin (A+B)

Geometric proof of 𝐬𝐢𝐧 𝑨+𝑩 = 𝐬𝐢𝐧 𝑨 𝐜𝐨𝐬 𝑩 + 𝐬𝐢𝐧 𝑩 𝐜𝐨𝐬 𝑨 Take a general triangle, as shown below … A B a b h c1 c2 c

Now consider the area of the triangle as a whole and as a compound area Area of red triangle = ½ c1 x h Area of blue triangle = ½ c2 x h A B a b h c1 c2 c Area of whole triangle = ½ (c1+c2) x h = ½ ab sin(A+B)

Area of whole triangle = ½ (c1+c2) x h = ½ ab sin(A+B) A B a b h c1 c2 c But h = a cos(A) = b cos(B) with c1 = a sin(A), c2 = b sin(B)

Therefore : Area of whole triangle = ½ (c1+c2) x h = ½ ab sin(A+B) = ½ (asinA +bsinB)h = ½ ab sin(A+B) A B a b h c1 c2 c Substituting values of h gives : absinAcosB + absinBcosA = ab sin(A+B) So finally : sinAcosB + sinBcosA = sin(A+B)

cos (A-B)

Type equation here.Type equation here.here. By GCSE Trigonometry:   1 P   1   So the coordinates of P are: B     A Type equation here.Type equation here.here.  O M N     So the coordinates of Q are:       Q     𝑆𝑖𝑛𝐴−𝑆𝑖𝑛𝐵     P    

(𝐶𝑜𝑠𝐴−𝐶𝑜𝑠𝐵 ) 2 +(𝑆𝑖𝑛𝐴−𝑆𝑖𝑛𝐵 ) 2 Multiply out the brackets 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)

You can also work out PQ using the triangle OPQ:   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 = 1 2 + 1 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)

Cos(A - B) = CosACosB + SinASinB 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

tan (A+B) You may be asked to prove either of the Tan identities using the Sin and Cos ones!

Tan (A+B) ≡ 𝑆𝑖𝑛(𝐴+𝐵) 𝐶𝑜𝑠(𝐴+𝐵) Rewrite Tan (A+B) ≡ 𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵+𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵−𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 Divide top and bottom by CosACosB Tan (A+B) ≡ 𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 + 𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 − 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐵 𝐶𝑜𝑠𝐴𝐶𝑜𝑠𝐵 Simplify each Fraction TanA + TanB Tan (A+B) ≡ 1 - TanATanB