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Expressing asinx + bcosx in the forms Rsin(x ±  ) or Rcos(x ±  ) The graph below is y = 3cosx + 4sinx. This can be considered as either a sine or a cosine.

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Presentation on theme: "Expressing asinx + bcosx in the forms Rsin(x ±  ) or Rcos(x ±  ) The graph below is y = 3cosx + 4sinx. This can be considered as either a sine or a cosine."— Presentation transcript:

1 Expressing asinx + bcosx in the forms Rsin(x ±  ) or Rcos(x ±  ) The graph below is y = 3cosx + 4sinx. This can be considered as either a sine or a cosine graph which has been translated horizontally and stretched vertically.

2 If it is considered to be a cosine curve then it has been translated horizontally and stretched vertically by a factor of The object is to find the horizontal translation and vertical stretch.

3 If it is considered to be a sine curve then it has been translated horizontally and stretched vertically by a factor of The object is to find the horizontal translation and vertical stretch.

4 If the curve is taken to be a translated cosine curve then its equation will be of the form 3cosx + 4sinx = Rcos(x -  ) Where  is the horizontal translation And R is the vertical stretch Note the question contains a PLUS in the middle and the translated equation contains a MINUS because the curve is translated in a positive x direction.

5 cosx + sinx = Rcos(x -  = R(cosxcos  + sinxsin  ) Using cos(A - B) cosx Matching up the left and right hand side then Rcos  = 3 Rsin  = 4 = (Rcos  ) sinx + (Rsin  ) 3 4

6 Rsin  = 4 Rcos  = 3 Dividing these two equations  = 53.1 o

7 R 2 sin 2  = 16 andR 2 cos 2  = 9 Adding these two equations R 2 sin 2  + R 2 cos 2  = = 25 R 2 (sin 2  + cos 2  ) = 25 R = 5 as sin 2  + cos 2  = 1 Rsin  = 4 Rcos  = 3 Squaring these two equations

8 Hence 3cosx + 4sinx = Rcos(x -  ) = 5 cos(x – 53.1) This is a cosine graph which has been translated horizontally and stretched vertically by a factor of 5 This is evident from the graph on the right

9 If the curve is taken to be a translated sine curve then its equation will be of the form 3cosx + 4sinx = Rsin(x +  ) Where  is the horizontal translation And R is the vertical stretch Note the question contains a PLUS in the middle and the translated equation contains a PLUS because the curve is translated in a negative x direction.

10 cosx + sinx = Rsin(x +  = R(sinxcos  + cosxsin  ) Using sin(A + B) sinx Matching up the left and right hand side then Rcos  = 4 Rsin  = 3 = (Rcos  ) cosx + (Rsin  ) 3 4

11 Rsin  = 3 Rcos  = 4 Dividing these two equations  = 36.9 o

12 R 2 sin 2  = 9 andR 2 cos 2  = 16 Adding these two equations R 2 sin 2 a + R 2 cos 2  = = 25 R 2 (sin 2  + cos 2  ) = 25 R = 5 as sin 2  + cos 2  = 1 Rsin  = 3 Rcos  = 4 Squaring these two equations

13 Hence 3cosx + 4sinx = Rsin(x +  ) = 5sin(x ) This is a sine graph which has been translated horizontally – and stretched vertically by a factor of 5 This is evident from the graph on the right

14 Using the Rsin(x ±  ) or Rcos(x ±  ) form to solve equations of the form acosx + bsinx = c Solve 3cosx + 4sinx = 4 3cosx + 4sinx = 5cos(x ) Shown previously So 5cos(x – 53.1) = 4 Let y = x – 53.1 So cosy =  x – 53.1 = 36.8, -36.8, find 1st two answers and add 360 x = 89.9, 16.3, add 53.1 to both sides y = cos -1  = 36.8, -36.8, 323.1


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