Deriving R-Form Step by Step in Additional Mathematics

R-form becomes dependable when you can rebuild it from the angle-addition formula. This guide slows the derivation down and makes every decision visible: which form to choose, which coefficients correspond, how R appears and how the phase angle is selected.

Step 1: choose a target form

Begin with a sin x+b cos x. Suppose we want

a sin x+b cos x=R sin(x+α), with R>0.

The plus sign is only a chosen representation. If the coefficients suggest a negative phase, α itself can be negative, or the final form can be written as Rsin(x−β).

Step 2: expand the target

Use the addition formula:

sin(x+α)=sin x cosα+cos x sinα.

Multiplying by R gives

Rsin(x+α)=Rcosα sin x+Rsinα cos x.

Step 3: match like coefficients

Compare

a sin x+b cos x

with

Rcosα sin x+Rsinα cos x.

Therefore a=Rcosα and b=Rsinα.

Step 4: eliminate α to find R

Square both equations:

a²=R²cos²α and b²=R²sin²α.

Add:

a²+b²=R²(cos²α+sin²α)=R².

Hence R=√(a²+b²).

Step 5: recover α

Now use

cosα=a/R, sinα=b/R.

When a≠0, tanα=b/a is a convenient ratio, but choose the quadrant from the signs of sine and cosine rather than trusting the principal arctangent alone.

Worked derivation: 4sin x+3cos x

R=√(16+9)=5. Then cosα=4/5 and sinα=3/5, so α is acute.

4sin x+3cos x=5sin(x+α), where α=tan⁻¹(3/4).

Verification: 5sin(x+α)=5sin x(4/5)+5cos x(3/5)=4sin x+3cos x.

Worked derivation: −8sin x+6cos x

R=10. The coefficient conditions are cosα=−4/5 and sinα=3/5. Therefore α lies in quadrant II.

A calculator value from tan⁻¹(−3/4) alone may suggest a quadrant-IV angle. That angle has the wrong cosine sign. Use both coefficient equations to choose the correct phase.

Deriving cosine R-form

Suppose instead

a sin x+b cos x=Rcos(x−β).

Expand:

Rcos(x−β)=Rcosx cosβ+Rsinx sinβ.

Matching coefficients gives Rsinβ=a and Rcosβ=b. Squaring and adding again gives R=√(a²+b²).

Why equivalent phase forms exist

Trigonometric functions are periodic, so a phase angle is not unique without a chosen interval convention. Adding 2π to a sine or cosine phase produces the same function.

Different sine and cosine R-forms can also represent the same graph because sine and cosine themselves differ by a quarter-cycle shift.

Special coefficient cases

If b=0, the expression is already a multiple of sine. If a=0, it is already a multiple of cosine. The R-form machinery still works, but a long derivation may be unnecessary.

If a=b=0, then R=0 and the expression is the zero function. A phase angle is then irrelevant because there is no nonzero sinusoid to orient.

A verification protocol

1. Check R²=a²+b².
2. Expand the proposed shifted sine or cosine.
3. Check the sine coefficient.
4. Check the cosine coefficient.
5. Check the phase quadrant from both signs.

Why this derivation matters later

Once the expression is one sinusoid, its amplitude and phase are explicit. That representation can then be used to find extrema, solve equations or interpret a graph. The derivation is the bridge that makes those later uses trustworthy.

Practice

1. Derive sine R-form for 5sin x+12cos x.
2. Derive sine R-form for 8sin x−15cos x.
3. Write 12sin x+5cos x as Rcos(x−β).
4. For −3sin x−4cos x, state R and the signs of sinα and cosα in Rsin(x+α).
5. Explain why tanα alone can be insufficient to determine α.

Answers

1. R=13; cosα=5/13, sinα=12/13.
2. R=17; cosα=8/17, sinα=−15/17.
3. R=13; sinβ=12/13, cosβ=5/13.
4. R=5; cosα=−3/5 and sinα=−4/5, so α lies in quadrant III under a 0-to-2π convention.
5. Tangent has period π, so opposite quadrants have the same tangent sign and ratio.

Continue with Why R-Form Works, the existing R-Form guide, or return to the Additional Mathematics Hub.