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Chapter 7 More fun with Trigonometry

Chapter 7 More fun with Trigonometry. Compound angle formulae and memory cards. The addition formulae. In trigonometry we often have to solve problems involving compound angles. For example,.

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Chapter 7 More fun with Trigonometry

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  1. Chapter 7 More fun with Trigonometry Compound angle formulae and memory cards.

  2. The addition formulae In trigonometry we often have to solve problems involving compound angles. For example, It is important to note that trigonometric functions are not distributive over addition. In other words, In fact, for two angles A and B, This is known as one of the addition formulae (or compound angle formulae) and it should be learnt.  Cos ( A – B ) = Cos A Cos B + Sin A Sin B

  3. Q P B A O N Compound Angles Consider a unit circle, radius = 1 Angle PON = A and Angle QON = B Coordinates of P are ( Cos A, Sin A ) Coordinates of Q are ( Cos B, Sin B ) By Pythagoras PQ2 = ( Cos B – Cos A )2 +( Sin B – Sin A )2 Distance formula = Cos2 B – 2Cos A Cos B + Cos2 A + Sin2 B – 2Sin A Sin B + Sin2 A = Cos2 A + Cos2 B + Sin2 A + Sin2 B – 2 ( Cos A Cos B + Sin A Sin B ) = 2 – 2 ( Cos A Cos B + Sin A Sin B ) By Cosine Rule PQ2 = 12 + 12 – 2Cos ( A – B )** = 2 – 2Cos ( A – B ) **Remember Cos ( -A ) = Cos A  Cos ( B – A ) = Cos ( A - B )  Cos ( A – B ) = Cos A Cos B + Sin A Sin B

  4. Using Cos ( A – B ) = Cos A Cos B + Sin A Sin B We replace B with –B to obtain : Cos ( A + B ) = Cos A Cos (-B) + Sin A Sin (-B) Remember Cos (-B) = Cos B and Sin (-B) = -Sin B  Cos ( A + B ) = Cos A Cos B - Sin A Sin B

  5. Using Cos ( A – B ) = Cos A Cos B + Sin A Sin B We replace A with ( π/2 – A ) to obtain : Cos [( π/2 – A )– B ] = Cos ( π/2 – A ) Cos B + Sin ( π/2 – A ) Sin B Cos [π/2 – ( A + B) ] = Cos ( π/2 – A ) Cos B + Sin ( π/2 – A ) Sin B Now Cos ( π/2 – A ) = Sin A and Sin ( π/2 – A ) = Cos A Sin ( A + B) = Sin A Cos B + Cos A Sin B Replace B with –B we obtain : Sin ( A - B) = Sin A Cos ( -B) + Cos A Sin ( -B) Which gives : Sin ( A - B) = Sin A Cos B - Cos A Sin B

  6. tan ( A + B ) = Sin ( A + B ) = Cos ( A + B ) Sin A Cos B + Cos A Sin B Cos A Cos B - Sin A Sin B Divide throughout by Cos A Cos B Sin A Cos B + Cos A Sin B Cos A Cos B_ Cos A Cos B Cos A Cos B – Sin A Sin B Cos A Cos B Cos A Cos B / / / / tan ( A + B ) = Sin ( A + B ) = Cos ( A + B ) / / / / tan A + tan B_ 1- tan A tan B tan ( A + B ) = Similarly tan A - tan B_ 1+ tan A tan B tan ( A - B ) =

  7. Example 1 • Using • Show that • sin 15 = √6-√2 • 4 • Rewrite as • sin (45-30)=sin45 cos30 –cos45 sin 30 Sin ( A - B) = Sin A Cos B - Cos A Sin B

  8. Example 2 • Given that sin A = -3/5 and 180<A<270 and that cos B = -12/13 and B is obtuse, find • A) cos (A-B) • B) tan (A+B) • We have sin A and cos B we also need sin B and cos A to use the identity in red.  Cos ( A – B ) = Cos A Cos B + Sin A Sin B

  9. Sin B and cos A • Sin A = -3/5 so using sin2A+cos2A = 1 • cos2A = 1-9/25 • = 16/25 • cos A = - 4/5 (since A is in 3rd quad negative) • Similarly for sin B we have cos B = -12/13 so using sin2B+cos2B = 1 • Sin2 B = 1-(-12/13)2 = 25/169 • Sin B = 5/13 (B is in 2nd quad) Or better draw a RAT and use Pythagoras. Or better draw a RAT and use Pythagoras.

  10. Finally putting this together  Cos ( A – B ) = Cos A Cos B + Sin A Sin B • Cos (A-B) = (-4/5)(-12/13)+(-3/5)(5/13) • = 33/65 • Now for tan (A+B) = tan A + tan B_ 1- tan A tan B

  11. Example 2 • Given that 2sin(x+y) = 3cos(x-y) • Express tan x in terms of tan y. • Step 1 expand both sides and then divide both sides by cosx cosy (to try to obtain the tan x and tany!

  12. Volunteers

  13. Ex 7A Identities • Prove the following: • A) sin (A+60) +sin(A-60)  sin A

  14. Ex 7A Helen

  15. Ex 7A Arthur

  16. Double angle formulae let A = B Cos ( A + B ) = Cos A Cos B - Sin A Sin B

  17. Double angle formulae A = B Sin ( A + B) = Sin A Cos B + Cos A Sin B

  18. Double angle formulae A = B • tan ( A + B ) =tan A + tan B 1- tan A tan B

  19. More Examples • See mathsnet.net

  20. More Double angle practice • Rewrite the following as a single trig ratio • 2sin x/2 cos x/2 cos x

  21. Double angle practice • Rewrite the following as a single trig ratio • 1+cos4x

  22. Double angle practice • Given that cos x = ¾ and that -180<x<360 find the exact value of: • Sin 2x and tan 2x.

  23. Ex 7B Q9

  24. Ex 7B Q10

  25. Ex 7B Q12

  26. Starter • From the top of your head can you write down the addition and double angle formulae for Sine, Cosine and Tangent?

  27. Proving Identities with double angle • By expanding sin (2A+A) • show that sin 3A  3sinA-4Sin3A

  28. Proving Identities with double angle • Similarly can you prove the expansions for • Cos 3A and tan3A?

  29. Eliminating the parameter • Given that x = 3sin θ and y = 3-4 cos 2 θ eliminate θ and express y in terms of x. • Let sin θ = x/3 and cos 2 θ = (3-y)/4

  30. Solving example • Solve 3 cos 2x – cosx +2 = 0 for 0 ≤ x ≤ 360

  31. Q P B A O N Compound Angles Cos (A –B) Consider a unit circle, radius = 1 Angle PON = A and Angle QON = B Coordinates of P are ( Cos A, Sin A ) Coordinates of Q are ( Cos B, Sin B ) By Pythagoras PQ2 = Distance formula = Multiplying brackets out = = By Cosine Rule PQ2 = = **Remember Cos ( -A ) = Cos A  Cos ( B – A ) = Cos ( A - B ) So Cos (A-B) =

  32. Cos (A + B) = Using Cos ( A – B ) = Cos A Cos B + Sin A Sin B We replace B with –B to obtain : Cos ( A + B ) = Remember Cos (-B) = Cos B and Sin (-B) = -Sin B  Cos ( A + B ) =

  33. Double angle formulae let A = B Cos ( A + B ) = Cos A Cos B - Sin A Sin B

  34. Double angle formulae A = B Sin ( A + B) = Sin A Cos B + Cos A Sin B

  35. Double angle formulae A = B • tan ( A + B ) =tan A + tan B 1- tan A tan B

  36. How can we apply our knowledge? • Using Autograph investigate what • y = 3sin x + 4cos x looks like. • Can you think of one trig function that this curve fits? Try your guess and see whether they fit over your original curve. • Try to get as close as possible.

  37. How about we find the equation more formally? • 3 sin x + 4 cos X can be expressed in the form R sin (x+θ) where R>0 and 0 <θ < 90. Find the values of R and θ to 1 d.p. • Solution steps: • Use Rsin(x+ θ) = R sinxcos θ+ R cosxsin θ • R sinxcosθ+R cosxsinθ = 3sinx + 4cosx • So Rcosθ = 3 eqn(1) and Rsinθ= 4 eqn(2) • Dividing to get expression tan θ • Squaring and adding the eqns to find R or can you think of a more efficient way?

  38. A brave volunteer

  39. Another formal approach but with a sketch! • Show that sin x -√3cos x can be expressed in the form Rsin (x- θ) where R >0 and 0 <θ < /2 and hence sketch the graph of y = sin x -√3cos x • The solution steps • Use Rsinxcosθ-Rcosxsinθ = sinx-√3cos x • This gives Rcos θ=1 and Rsin θ= √3 • Dividing to get tan θ and substituting your θ to get R or squaring and adding.

  40. Helen

  41. In pairs • You will be given a question to present to the class

  42. Question 1 • Given the 5sinx +12 cosx = Rsin(x+ θ), find the value of R and tan θ

  43. Question 2 • Given the √3sinx +√6 cosx = 3cos(x- θ), where θ is acute, find θ

  44. Question 3 • Given the 2sinx -√5 cosx = -3cos(x+θ), where θ is acute, find θ

  45. Question 4 • Show that • A) cos θ+sin θ=√2 sin (θ+/4) • B) √3sin2 θ-cos2 θ=2sin(2θ-/6)

  46. Question 5 • Prove that • cos 2 θ-√3sin2 θ 2cos(2θ+/3) •  -2sin(2θ-/6)

  47. Question 6 • Show that cos θ-√3sin θ can be written in the form of Rcos (θ+x) with R>0 and x is acute. • Hence sketch the graph of y = cos θ- √3sin θ for 0<x<2 giving the points of the intersection with the axes.

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