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 cos(θ ∓ a)
where a > 0, b > 0, α is acute, R = √(a2 + b2) 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 = √(22 + 82) 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 1st or 2nd quadrant, there is no need to find the basic angle and obtain the angle in the 2nd quadrant since θ is an acute angle based on the context and diagram
Last Minute Revision for O Level Math?
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