# 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

Rsin(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

asinθ+bcosθ≡Rsin (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:

Rsin (θ+α)≡R(sinθcosα+ cosθsinα)≡

Rsinθcosα+Rcosθsinα

So

sinaθ+cosbθ≡sinRcosαθ+cosRsinαθ

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

For sin

θ:a=Rcosα..........(1) in green aboveFor cos

θ:b=Rsinα.........(2) in red above

Eq. (2) ÷ Eq.(1):

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

So

`alpha=arctan\ b/a`

(

αis a positive acute angle andaandbarepositive.)

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}:

a^{2}+b^{2}=

R^{2}cos^{2}α+R^{2}sin^{2}α=

R^{2}(cos^{2}α+ sin^{2}α)=

R^{2}(since cos^{2}A + sin^{2}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:

asinθ+bcosθ=Rsin(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 *θ* − *b* cos *θ* with the expansion of *R* sin (*θ* − *α*) as follows (note the minus signs carefully):

sinaθ−cosbθ≡sinRcosαθ−cosRsinαθ

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:

asinθ−bcosθ=Rsin(θ−α)

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

To solve an equation in the form

asinθ±bcosθ=c,

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

Rsin(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 cosθ= 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 sint+ 5 cost.

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

4. Solve 7 sin 3*θ* − 6 cos 3*θ* = 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:

asinθ+bcosθ≡Rcos (θ−α)

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

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

Rcos (θ−α) =Rcosθcosα+Rsinθsinα

Rearranging and equating co-efficients gives us

sinaθ+cosbθ≡cosRcosαθ+sinRsinαθ

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:

asinθ+bcosθ≡Rcos(θ−α)

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

## The Cosine Minus Case

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

asinθ−bcosθ≡ −Rcos (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 (*θ* − *α*), 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 θ − b cos θ |
R sin (θ − α) |
`alpha=arctan (b/a)` |

a sin θ + b cos θ |
R cos (θ − α) |
`alpha=arctan (a/b)` |

a sin θ − b 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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