# Expressing asinx + bcosx in the forms

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Expressing asinx + bcosx in the forms
Rsin(x ± a) or Rcos(x ± a) 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.

If it is considered to be a cosine curve then it has
been translated horizontally and stretched vertically by a factor of 530 5 The object is to find the horizontal translation and vertical stretch.

If it is considered to be a sine curve then it has
been translated horizontally and stretched vertically by a factor of -370 5 The object is to find the horizontal translation and vertical stretch.

If the curve is taken to be a translated cosine curve then its equation will be of the form
3cosx + 4sinx = Rcos(x - a) Where a 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.

3 cosx + sinx = Rcos(x - a) = R(cosxcosa + sinxsina) Using cos(A - B) cosx Matching up the left and right hand side then Rcosa = 3 Rsina = 4 4 = (Rcosa) + (Rsina) sinx

Rsina = 4 Rcosa = 3 Dividing these two equations a = 53.1o

Rsina = 4 Rcosa = 3 Squaring these two equations R2 sin2a = and R2 cos2a = 9 Adding these two equations R2 sin2a + R2cos2a = = 25 R2(sin2a + cos2a) = 25 R = 5 as sin2a + cos2a = 1

Hence 3cosx + 4sinx = Rcos(x - a)
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

If the curve is taken to be a translated sine curve then its equation will be of the form
3cosx + 4sinx = Rsin(x + a) Where a 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.

3 cosx + sinx = Rsin(x + a) = R(sinxcosa + cosxsina) Using sin(A + B) sinx Matching up the left and right hand side then Rcosa = 4 Rsina = 3 4 = (Rcosa) + (Rsina) cosx

Rsina = 3 Rcosa = 4 Dividing these two equations a = 36.9o

Rsina = 3 Rcosa = 4 Squaring these two equations R2 sin2a = 9 and R2 cos2a = 16 Adding these two equations R2 sin2a + R2cos2a = = 25 R2(sin2a + cos2a) = 25 R = 5 as sin2a + cos2a = 1

Hence 3cosx + 4sinx = Rsin(x + a)
This is a sine graph which has been translated horizontally –36.90 and stretched vertically by a factor of 5 This is evident from the graph on the right

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

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