S5_Unit_2_Outcome_3 - Higher Unit 2 Higher Outcome 3...

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www.mathsrevision.com Higher Outcome 3 Higher Unit 2 Higher Unit 2 www.mathsrevision.com www.mathsrevision.com Trigonometry identities of the form sin(A+B) Double Angle formulae Trigonometric Equations Exam Type Questions More Trigonometric Equations
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www.mathsrevision.com Higher Outcome 3 Trig Identities The following relationships are always true for two angles  A  and B. 1a. sin(A + B) = sinAcosB + cosAsinB 1b. sin(A - B)  = sinAcosB -  cosAsinB 2a. cos(A + B) = cosAcosB – sinAsinB 2b. cos(A - B) = cosAcosB + sinAsinB Quite tricky to prove but some of following examples should show  that they do work!! Supplied on a  formula sheet !!
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www.mathsrevision.com Higher Outcome 3 Examples 1 (1) Expand  cos(U – V). (use formula 2b ) cos(U – V) =  cosUcosV + sinUsinV (2)  Simplify    sinf°cosg° - cosf°sing° (use formula 1b ) sinf°cosg° - cosf°sing°  =   sin(f – g)° (3)  Simplify   cos8    θ sin θ  + sin8    θ cos  θ (use formula 1a ) cos8    θ sin  θ  + sin8    θ cos  θ   =  sin(8  θ  +  θ ) =  sin9  θ Trig Identities
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www.mathsrevision.com Higher Outcome 3 =  cos30° Example 2 By taking A = 60°  and B = 30°,   prove the identity for  cos(A – B). NB: cos(A – B)  =  cosAcosB  +  sinAsinB LHS = cos(60 – 30 )° =   3 / 2 RHS = cos60°cos30° +  sin60°sin30° = ( ½   X   3 / 2  ) + ( 3 / 2   X  ½) =   3 /  +   3 / 4 =   3 / 2 Hence  LHS  =  RHS !! Trig Identities
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www.mathsrevision.com Higher Outcome 3 Example 3 Prove that  sin15°  =  ¼( 6 -  2) sin15° = sin(45 – 30)° =   ( 1 / 2    X    3 / 2  ) - ( 1 / 2   X  ½) =    ( 3 / 2   -    1 / 2 2 )   =  sin45°cos30°  - cos45°sin30° =     ( 3  -  1)   2 2       X      2               2 =     ( 6  -  2)     4 =  ¼( 6 -  2) Trig Identities
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www.mathsrevision.com Higher Outcome 3 Example 4 α β 41 40 4 3 Show that    cos( α  -  β )  =   187 / 205 Triangle1 If missing side = x x Then   x 2  = 41 2  – 40 2  = 81 So  x = 9 sin α  =  9 / 41    and   cos α  =  40 / 41 Triangle2 If missing side = y y Then   y 2  = 4 2  + 3 2  = 25 So  y = 5 sin  β  =  3 / 5    and   cos β  =  4 / 5 Trig Identities NAB type  Question
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www.mathsrevision.com Higher Outcome 3 Trig Identities cos( α  -  β )  = cos α cos β   + sin α sin β   =   ( 40 / 41   X   4 / 5 )  +  ( 9 / 41   X    3 / 5  )  =     160 / 205   +   27 / 205 =     187 / 205 Remember this is a NAB type Question sin α  =  9 / 41    and   cos α  =  40 / 41 sin  β  =  3 / 5    and   cos β  =  4 / 5
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This note was uploaded on 02/13/2012 for the course MAT 205 math 205 taught by Professor Google during the Spring '10 term at University of Phoenix.

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S5_Unit_2_Outcome_3 - Higher Unit 2 Higher Outcome 3...

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