# How to Use Trigo R-Formula

Trigonometry (Trigo) R-formula is a must-know formula in O-levels A-Math. In this revision note, you will learn the Trigo R-formula and how to apply it to solve common O-levels A-Math questions.

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## R−Formula

The R−Formula allows an expression involving the sum (or difference) of sine and cosine functions to be expressed as a **single** trigonometric function in sine or cosine. In general,

*a* sin θ ± *b* cos θ = R sin(θ ± *a*)

*a* cos θ ± *b* sin θ = R sin(θ ∓ *a*)

where* a* > 0, *b* > 0, α is acute, R = √(*a*^{2} +* b*^{2}) and α= tan^{−1}(*b*/*a*).

## R-Formula Trigo Example

The diagram below shows an L-shaped rod, XYZ, placed against an upright wall. Given that XY = 2 m, YZ = 8 m and angle OZY = θ°.

(i) Show that OZ = (2 sin θ + 8 cos θ) m.

(ii) Express OZ in the form R sin (θ – α), where α is acute.

(iii) Hence state the maximum value of OZ and the value of θ for which OZ is maximum.

(iv) Find the value of θ for which OZ = 6 m.

**(i) Show that OZ = (2 sin θ + 8 cos θ) m.**

Let W be on YZ such that XW be parallel to OZ. Let A be on XW and B be on OZ such that YB is perpendicular to OB. Hence angle XWY = angle OZY = θ (corresponding angles, XW // OZ)

In the right−angled triangle XYW,

Angle YXW = 180° − angle XWY − 90° = 90° − θ (sum of angles in triangle = 180°)

In right−angled triangle XYA,

Angle XYA = 180° − angle YXW − 90° = θ (sum of angles in triangle = 180°)

sin∠XYA = sin θ =AX/2

AX = 2 sin θ = OB

In right−angled triangle YBZ,

cos∠YZB = cos θ = BZ/8

BZ = 8 cos θ

OZ = OB + BZ = 2 sin θ + 8 cos θ (shown)

**(ii) Express OZ in the form R sin (θ – α), where α is acute.**

Using the Trigo R-Formula,

α = tan^{-1}(8/2) = 76.0° (1 dp)

2 sin θ + 8 cos θ = √(2^{2} + 8^{2}) sin(θ + 76.0°) = √68 sin (θ + 76.0°) (ans)

**(iii) Hence state the maximum value of OZ and the value of θ for which OZ is maximum.**

−1 ≤ sin(θ + 76.0°) ≤ 1

−√68 ≤ √68 sin(θ + 76.0°) ≤ √68

The maximum value of OZ is √68 = 8.25 *m* (3 *sf* ) (ans)

OZ is maximum when sin(θ + 76.0°) = 1.

θ + 76.0° = 90°

θ = 14.0° (*ans*)

**(iv) Find the value of θ for which OZ = 6 m.**

√68 sin(θ + 76.0°) = 6

sin(θ + 76.0°) = 6/√68

θ + 76.0° = 46.6861°

θ = 29.3° (1 *dp*) (ans)

Note: While it can be inferred that θ may lie in the 1^{st} or 2^{nd} quadrant, there is no need to find the basic angle and obtain the angle in the 2^{nd} quadrant since θ is an acute angle based on the context and diagram

## Last Minute Revision for O Level Math?

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