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Trig addition formulae
sin π΄Β±π΅ = sin π΄ cos π΅ Β± sin π΅ cos π΄ Proof Proof cos π΄Β±π΅ = cos π΄ cos π΅ β sin π΅ sin π΄ Proof tan π΄Β±π΅ = tan π΄ Β± π‘ππ π΅ 1β tan π΄ tan π΅
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sin (A+B)
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Geometric proof of π¬π’π§ π¨+π© = π¬π’π§ π¨ ππ¨π¬ π© + π¬π’π§ π© ππ¨π¬ π¨
Take a general triangle, as shown below β¦ A B a b h c1 c2 c
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Now consider the area of the triangle as a whole and as a compound area
Area of red triangle = Β½ c1 x h Area of blue triangle = Β½ c2 x h A B a b h c1 c2 c Area of whole triangle = Β½ (c1+c2) x h = Β½ ab sin(A+B)
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Area of whole triangle = Β½ (c1+c2) x h = Β½ 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)
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Therefore : Area of whole triangle
= Β½ (c1+c2) x h = Β½ ab sin(A+B) = Β½ (asinA +bsinB)h = Β½ 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)
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cos (A-B)
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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.Β O M N So the coordinates of Q are: Q ππππ΄βππππ΅ P
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(πΆππ π΄βπΆππ π΅ ) 2 +(ππππ΄βππππ΅ ) 2 Multiply out the brackets
π π 2 = (πΆππ π΄βπΆππ π΅ ) 2 +(ππππ΄βππππ΅ ) 2 Multiply out the brackets π π 2 = (πΆπ π 2 π΄β2πΆππ π΄πΆππ π΅+πΆπ π 2 π΅) + (ππ π 2 π΄β2ππππ΄ππππ΅+ππ π 2 π΅) Rearrange π π 2 = (πΆπ π 2 π΄+ππ π 2 π΄) + (πΆπ π 2 π΅+ππ π 2 π΅) β 2(πΆππ π΄πΆππ π΅+ππππ΄ππππ΅) πΆπ π 2 ΞΈ+ππ π 2 ΞΈ β‘ 1 π π 2 = 2 β 2(πΆππ π΄πΆππ π΅+ππππ΄ππππ΅)
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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 π 2 = π 2 + π 2 β 2bcCosA Sub in the values ππ 2 = β 2Cos(B - A) Group terms ππ 2 =2 β 2Cos(B - A) Cos (B β A) = Cos (A β B) eg) Cos(60) = Cos(-60) ππ 2 =2 β 2Cos(A - B)
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Cos(A - B) = CosACosB + SinASinB
π π 2 = 2 β 2(πΆππ π΄πΆππ π΅+ππππ΄ππππ΅) ππ 2 =2 β 2Cos(A - B) 2 β 2(πΆππ π΄πΆππ π΅+ππππ΄ππππ΅) = 2 β 2Cos(A - B) Subtract 2 from both sides β 2(πΆππ π΄πΆππ π΅+ππππ΄ππππ΅) = β 2Cos(A - B) Divide by -2 πΆππ π΄πΆππ π΅+ππππ΄ππππ΅ = Cos(A - B) Cos(A - B) = CosACosB + SinASinB Cos(A + B) = CosACosB - SinASinB
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tan (A+B) You may be asked to prove either of the Tan identities using the Sin and Cos ones!
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Tan (A+B) β‘ πππ(π΄+π΅) πΆππ (π΄+π΅)
Rewrite Tan (A+B) β‘ ππππ΄πΆππ π΅+πΆππ π΄ππππ΅ πΆππ π΄πΆππ π΅βππππ΄ππππ΅ Divide top and bottom by CosACosB Tan (A+B) β‘ ππππ΄πΆππ π΅ πΆππ π΄πΆππ π΅ + πΆππ π΄ππππ΅ πΆππ π΄πΆππ π΅ πΆππ π΄πΆππ π΅ πΆππ π΄πΆππ π΅ β ππππ΄ππππ΅ πΆππ π΄πΆππ π΅ Simplify each Fraction TanA + TanB Tan (A+B) β‘ 1 - TanATanB
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