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 θ ≡ R 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 θ ≡ R 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):
So
(α is a positive acute angle and a and b are positive.)
Now we square each of Eq. (1) and Eq. (2) and add themm 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
(We take only the positive root)
In summary, if
and
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.
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):
a sin θ − b cos θ ≡ R cos α sin θ − R sin α cos θ
Once again we will obtain (try it yourself!):
and
Our equation for the minus case is:
a sin θ − b cos θ = R sin(θ − α)
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
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
.
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 θ ≡ R cos (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:
R cos (θ − α) = R cos θ cos α + R sin θ sin α
Rearranging things a bit and equating co-efficients gives us
a sin θ + b cos θ ≡ R cos α cos θ − R sin α sin θ
So this time we have:
a = -R sin α ..... (1)
b = -R cos α ..... (2)
Dividing (1) by (2) gives us:
Therefore (and note that this is different to before):
We find R the same as before:
So the sum of a sine term and cosine term have been combined into a single cosine term:
a sin θ + b cos θ ≡ R cos(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:
a sin θ − b cos θ ≡ −R cos (θ + α)
Once again, a, b and R are positive.
α is acute and given by:
R is given by:
Cosine Exercises
1. Express 7 sin θ + 12 cos θ in the form R cos (θ − α), where 0 ≤ α < π/2.
2. Express 2.340 sin θ − 1.251 cos θ in the form -R cos (θ + α), where 0 ≤ α < π/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) | |
| a sin θ − b cos θ | R sin (θ − α) | |
| a sin θ + b cos θ | R cos (θ − α) | |
| a sin θ − b cos θ | -R cos (θ + α) |
In each case, a, b and R are positive and α is an acute angle.
R is given by:
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