This is a method of solving equations in the form asinx+bcosx = c where a and b are constants and c is another expression.

It involves rewriting letting asinx + bcosx = rsin(x+y) (or you could use cos(x+y)) where y is acute and then finding values for r and y, then with only one trig function to deal with the equation can be solved more easily.

For example

consider 2sinx + 3cosx = 3

Let 2sinx + 3cosx = rsin(x+y)

Now expand the sin(x+y) to get

2sinx + 3cosx = rsinx cosy + rcosx siny

Since y is constant and therefore cosy and sin y are constant we can compare the coefficients to get

2 = rcosy —–(1)
3 = rsiny ——(2)

We can solve these to find values for r and y.
To find y consider (2)/(1) to get

3/2 = tany
since sin/cos = tan and the r’s cancel
so y = 56.3 °

To find r consider (1)2+(2)2 to get
22+32 = r2
since sin2+cos2 = 1
so r =√13

So we can write

2sinx + 3 cosx = √13 cos(x+56.3) = 3

so x = cos^-1(3/√13) -56.3

so x = cos-1(3/√13) -56.3

since cos-1(3/√13) = 33.7 for solutions between 0° and 90°

x = ±33.7 -56.3 + 180n where n is an integer

In General

asinx + bcosx = √(a2+b2) sin(x+tan-1(b/a))

If you have any questions, comments or corrections please leave them as a comment below

By David Woodford