6. Expressing a sin θ ± b cos θ in the form R sin(θ ± α)

by M. Bourne

In electronics, we often get expressions involving the sum of sine and cosine terms. It is more convenient to write such expressions using one single term.

Our Problem:

Express a sin θ ± b cos θ in the form

R sin(θ ± α),

where a, b, R and α are positive constants.

Solution:

First we take the "plus" case, (θ + α), to make things easy.

Let

a sin θ + b cos θR sin (θ + α)

(The symbol " ≡ " means: "is identically equal to")

Using the compound angle formula from before (Sine of the sum of angles),

sin(A + B) = sin A cos B + cos A sin B,

we can expand R sin (θ + α) as follows:

R sin (θ + α)
 R (sin θ cos α + cos θ sin α)

 R sin θ cos α + R cos θ sin α

So

a sin θ + b cos θR cos α sin θ + R sin α cos θ

Equating the coefficients of `sin\ θ` and `cos\ θ` in this identity, we have:

For sin θ: a = R cos α ..........(1) in green above

For cos θ: b = R sin α .........(2) in red above

Eq. (2) ÷ Eq.(1):

`b/a=(R\ sin\ alpha)/(R\ cos\ alpha)=tan\ alpha`

So

`alpha=arctan\ b/a`

(α is a positive acute angle and a and b are positive.)

Now we square each of Eq. (1) and Eq. (2) and add them to find an expression for R.

[Eq. (1)]2 + [Eq. (2)]2:

a2 + b2

= R2 cos2 α + R2 sin2 α

= R2(cos2 α + sin2 α)

= R2 (since cos2 A + sin2 A = 1)

So

`R=sqrt(a^2+b^2`

(We take only the positive root)

In summary, if

`alpha=arctan\ b/a`

and

`R=sqrt(a^2+b^2)`

then we have expressed a sin θ + b cos θ in the form required:

a sin θ + b cos θ = R sin(θ + α)

You will notice that this is very similar to converting rectangular to polar form in Polar form of Complex Numbers. We can get α and R using calculator, similar to the way we did it in the complex numbers section.

The Minus Case

Similarly, for the minus case, we equate a sin θb cos θ with the expansion of R sin (θα) as follows (note the minus signs carefully):

a sin θb cos θR cos α sin θR sin α cos θ

Once again we will obtain (try it yourself!):

`alpha=arctan\ b/a`

and

`R=sqrt(a^2+b^2)`

Our equation for the minus case is:

a sin θb cos θ = R sin(θα)

Equations of the type a sin θ ± b cos θ = c

To solve an equation in the form

a sin θ ± b cos θ = c,

express the LHS in the form R sin(θ ± α) and then solve

R sin(θ ± α) = c.

Exercises - Sine Form

1. (a) Express   4 sin θ + 3 cos θ  in the form R sin(θ + α).

(b) Using your answer from part (a), solve the equation

4 sin θ + 3 cos θ = 2     for `0^@ ≤ θ < 360^@`.

2. Solve the equation

`sin\ theta-sqrt2\ cos\ theta=0.8`,     for `0^@ ≤ θ < 360^@`.

3. The current i (in amperes) at time t in a particular circuit is given by

i = 12 sin t + 5 cos t.

Find the maximum current and the first time that it occurs.

4. Solve 7 sin 3θ − 6 cos 3θ = 3.8 for `0^@ ≤ θ < 360^@`.

5. The current i amperes in a certain circuit after t seconds is given by

`i=2\ sin(t-pi/3)-cos(t+pi/2)`

Find the maximum current and the earliest time it occurs.
(Note: t > 0)

The Cosine Form

We can also express our sum of a sine term and a cosine term using cosine rather than sine. It may be more convenient to do so in some situations.

The expressions obtained are similar to those we obtained for the sine case, but note the differences as we proceed.

For a, b and R positive and α acute, our equivalent expression is given by:

a sin θ + b cos θR cos (θα)

This time there is a difference in the way we obtain α, compared to before.

Expanding R cos (θα) using our result for the expansion of cos(A − B) gives us:

R cos (θα) = R cos θ cos α + R sin θ sin α

Rearranging and equating co-efficients gives us

a sin θ + b cos θR cos α cos θ + R sin α sin θ

So:

`a = R\ sin\ α\ \ \ ... (1)`

`b = R\ cos\ α\ \ \ ... (2)`

Dividing (1) by (2) gives us:

`a/b=(R sin alpha)/(R cos alpha)=tan alpha`

Therefore:

`alpha=arctan\ a/b`

(Note the fraction is `a/b` for the `"cosine"` case, whereas it is `b/a` for the `"sine"` case.)

We find R the same as before:

`R=sqrt(a^2+b^2)`

So the sum of a sine term and cosine term have been combined into a single cosine term:

a sin θ + b cos θR cos(θα)

Once again, a, b, R and α are positive constants and α is acute.

The Cosine Minus Case

If we have a sin θb cos θ and we need to express it in terms of a single cosine function, the formula we need to use is:

a sin θb cos θ ≡ −R cos (θ + α)

Once again, a, b and R are positive.

α is acute and given by:

`alpha=arctan\ a/b`

R is given by:

`R=sqrt(a^2+b^2`

Cosine Exercises

1. Express 7 sin θ + 12 cos θ in the form R cos (θα), where `0 <= alpha < pi/2`.

2. Express 2.348 sin θ − 1.251 cos θ in the form −R cos (θ + α), where `0 <= alpha < pi/2`.

Summary

Here is a summary of the expressions and conditions that we have found in this section.

Original Expression Combined Expression α
a sin θ + b cos θ R sin (θ + α) `alpha=arctan (b/a)`
a sin θb cos θ R sin (θα) `alpha=arctan (b/a)`
a sin θ + b cos θ R cos (θα) `alpha=arctan (a/b)`
a sin θb cos θ -R cos (θ + α) `alpha=arctan (a/b)`

In each case, a, b and R are positive and α is an acute angle.

R is given by:

`R=sqrt(a^2+b^2)`

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