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TRIG IDENTITIES (II)

TRIG IDENTITIES (II). The following relationships are always true for two angles A and B. (1)(a) sin(A + B) = sinAcosB + cosAsinB. Supplied on a formula sheet !!. (1)(b) sin(A - B) = sinAcosB - cosAsinB. (2)(a) cos(A + B) = cosAcosB - sinAsinB. (2)(b) cos(A - B) = cosAcosB + sinAsinB.

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TRIG IDENTITIES (II)

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  1. TRIG IDENTITIES (II) The following relationships are always true for two angles A and B. (1)(a) sin(A + B) = sinAcosB + cosAsinB Supplied on a formula sheet !! (1)(b) sin(A - B) = sinAcosB - cosAsinB (2)(a) cos(A + B) = cosAcosB - sinAsinB (2)(b) cos(A - B) = cosAcosB + sinAsinB Quite tricky to prove but some of following examples should show that they do work!!

  2. Consider (cosA, sinA) Q Q’ A+B A (1,0) (1,0) O O -B (cosB, -sinB) P’ P Rotate ∆OPQ about O through angle B

  3. PQ2 = (cos A – cos B)2 + (sin A + sin B)2 = cos2 A + sin2 A + cos2 B +sin2 B – 2cosAcosB + 2sinAsinB = 2 – 2(cosAcosB – sinAsinB)

  4. P′ is (1,0) and Q′ is (cos(A+B), sin(A+B)) (P′Q′)2 = (1 – cos(A+B))2 + (sin(A+B))2 = 1 + cos2(A+B) + sin2(A+B) – 2cos(A+B) = 2 - 2cos(A+B) BUT PQ2 = (P′Q′)2 So it follows that cos (A+B) = cosAcosB – sinAsinB AND cos (A+(-B))= cos(A-B) = cosAcosB-sinAsinB

  5. Examples 1 (1) Expand cos(U – V). (use formula (2)(b) ) cos(U – V) = cosUcosV + sinUsinV (2) Simplify sinf°cosg° - cosf°sing° (use formula (1)(b) ) sinf°cosg° - cosf°sing° = sin(f – g)° (3) Simplify cos8sin + sin8cos (use formula (1)(a) ) cos8sin + sin8cos = sin(8 + ) = sin9

  6. Example 2 By taking A = 60° and B = 30°, prove the identity for cos(A – B). ***************** NB: cos(A – B) = cosAcosB + sinAsinB If A = 60° and B = 30° then LHS = cos(60 – 30 )° = cos30° = 3/2 RHS = cos60°cos30° + sin60°sin30° = ( ½ X 3/2 ) + (3/2 X ½) = 3/4 + 3/4 = 3/2 Hence LHS = RHS !!

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