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Trigonometry Type R

Tuition given in the topic of A-Maths Tuition Notes & Tips from the desk of Miss Loi at 2:05 pm (Singapore time)

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Type R Logo

The pride of racers and the probable source of
noise pollution in Miss Loi’s estate at 2am

Like Honda’s Type R variant of cars that appeal to Ah Beng racer boys who like to bully weave in front of Miss Loi’s demure little car on the expressways, Trigonometry has its own “Type R” model in the form of the R-Formula that was unveiled in this year’s New AMaths Syllabus.

As such students are warned to listen out for the roar of the R-Formula whenever you encounter a trigonometric question:

  1. with an expression that contains a constant, like this:

    Solve for 0° < x < 360°,
    2 cot x = 3 + 2 cosec x
  2. or where you’re asked to find an expression’s max/min value, like in this rather cheong hei (long-winded) question:

    Trignometry R-Formula Question Diagram

    The diagram shows two right-angled triangles with their right angles at B and N. The sides OB and AB are of length 6 cm and 2 cm respectively. The lines OB is inclined at angle θ to ON. The line AM is perpendicular to ON.

    1. Express OM in the form OM = R cos (θ + α) and calculate the values of R and α.
    2. Find the value of θ for which OM = 5 cm.
    3. In the diagram, state the line with a length of R cm and the angle with a value α.
    4. Express AM in terms of R and (θ + α) and show that the area of triangle OAM is 10 sin 2(θ + α).
    5. Given that θ can vary, find the maximum value of the area of triangle OAM and the corresponding value of θ when this occurs.

To handle the new R-Formula questions, students need to recall the following expressions, as they will not be appearing on your formula sheet during your actual exam:

The R-Formula

R-Formula Triangle

  • a cos θ ± b sin θ = R cos (θα) – look out for that inverted plus/minus!
  • a sin θ ± b cos θ = R sin (θ ± α)

where R = sqrt{a^2 + b^2} and tan alpha = b/a
and a > 0, b > 0, α is acute

Maximum/Minimum Values of R-Formula Expressions

  • The maximum value is R which occurs when cos/sin (θ ± α) = 1
  • the minimum value –R which occurs when cos/sin (θ ± α) = -1

Deriving The R-Formula

Do note that the R-Formula highlighted above is NOT a given, especially when the relevant part of the question carries a high weightage of marks e.g.

Express 4 cos θ – 3 sin θ in the form of R cos (θ+α) [6 Marks]

There’s a bit of debate on this but don’t think you can get away with all 6 marks here by simply substituting the values into the R-Formula! So to be a little kiasu, it’s advisable to derive the whole thing as per the following steps:

4 cos θ – 3 sin θ
= R cos (θ+α)
= R [cosθcosα-sinθsinα] (Addition Formula)
= (R cosα)cosθ-(R sinα)sinθ

Comparing coefficients between LHS & RHS,
R cosα = 4 —– (1)
R sinα = 3 —– (2)

(2)/(1): tan α = 3/4 ⇒ α = 36.9°

R2cos2α+R2sin2α = 42+32
R2[cos2α+sin2α] = 25 ⇒ R = 5

∴ 4 cos θ – 3 sin θ = 5 cos(θ+36.9°)

Important Notes

Warning! To reiterate what was stated above, a and b MUST be positive. DON’T do this:

(-3) cos θ + (-2) sin θ = 3 → WRONG!

Instead, move the terms around to ensure that the first term is always positive:
3 cos θ + 2 sin θ = -3 → CORRECT! YAY!

Warning! Please, please, please, please, please for the addition cases (i.e. a cos θ + b sin θ or a sin θ + b cos θ) you can simply re-arrange the sine and cosine terms depending on how you wish to express the R function in your question.


3 cos θ + 2 sin θ → if you require R cos (θα)
2 sin θ + 3 cos θ → if you require R sin (θ + α)

Hailed as the pinnacle of O-Level Trigonometry, a typical R-Formula question may also require your skills on lesser trigonometry variants such as the Addition Formulae, or the Double Angle Formulae, or the Factor Formulae, together with your knowledge of basic angles (e.g. in question 1 above when you’re asked to find all angles within the range of blah blah blah).

But once you tame it, you’ll be rewarded with a straightforward ride and zoom away with plenty of marks.

So are you ready to test-drive your R-Formula with the two questions above? Miss Loi doesn’t expect you to crash this in your O-Level in two weeks’ time!

*Zooms away for Jφss Sticks Sessions (Type-R variant)*

N.B. Once again, Miss Loi does NOT advocate nor encourage speeding or any form of street racing in Singapore.


Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

  1. Leave A Comment!
Comments & Reactions


  1. Anoneemers's Avatar
    Anoneemers commented in tuition class


    I'm trying this for fun =) (for lbmfh practice actually)

    Miss Loi is now marking your answer in a LMBFH manner too 😛 ...

    1)Is there only 1 answer? My only answer is 247.4

    Yes you're right there's only one answer! To elaborate, we have

    2 cot x = 3 + 2 cosec x

    So seeing that lonely constant 3 in the equation should signal to you that you're going to start the Trigonometry R-Formula engine *VRROOOOOM*

    Expanding ...

    {2/{tan x}} - {2/{sin x}} = 3
    2({{cos x}/{sin x}} - {1/{sin x}}) = 3
    2 cos x - 2 = 3 sin x
    2 cos x - 3 sin x = 2
    (Note: we put the [2 cos x] as the first term so as to keep a (in the R-Formula) positive)

    To express the LHS above in R cos (x + α) form, we have

    R = √(22+32) = √13
    tan α = 3/2 ⇒ α = 56.3°
    (Note: Both a & b has to be > 0 when you're obtaining α! DON'T do a tan α = -3/2!)

    So we now have:

    √13 cos (x + 56.3°) = 2
    ⇒ cos (x + 56.3°) = 2/√13

    This is where many students get pwned and forget that All Science Teachers are Crazy (ASTC), as the question asks for ALL values of x in the range of 0° < x < 360°.

    So doing a cos-1 2/√13 in your calculator will only yield a basic angle of (x + 56.3°) = 56.3°. But since 2/√13 is positive, we see from the ASTC diagram below that:

    Basic Angle

    x + 56.3° = 56.3°, 303.7° (1st and 4th quadrant)
    x = 0°, 247.4°

    But we shall take only x = 247.4° since 0° is not in the range of 0° < x < 360°!


    Once again, to elaborate on your answer (next time show workings okay? Tsk tsk.) ...

    Typically in a question like this, we first need to express OM in terms of sin θ and cos θ BEFORE we can use the R-Formula. From the following diagram,

    Trigonometry R-Formula Answer Diagram

    ... you should be able to see that ∠MAB = θ, and since OM = ON - MN
    OM = 6 cos θ - 2 sin θ
    (using your TOA CAH SOH)

    Using the R-Formula,
    OM = R cos (θ + α) ----- (1)
    R = √(62+22) = √40
    α = tan-1 2/6 = 18.435°
    OM = √40 cos (θ + 18.435°)


    Think you've made a careless mistake here but Miss Loi won't know in her lifetime the cause of it since your workings are absent!

    In anycase, simply equate OM to 5cm and find the value of θ when this occurs:
    √40 cos (θ + 18.435°) = 5
    cos (θ + 18.4°) = 5/√40
    (θ + 18.435°) = 37.761°, 322.239°

    Once again don't forget to A-S-T-C thingy BUT BUT BUT in this case we're sure from the diagram that θ is acute so we can safely discard the 322.239° and conclude that:
    θ = 37.761° - 18.435° = 19.3° when OM = 5cm

    (iii)AO, angle AOB

    This part requires a bit of 阴阳眼 (The Third Eye) which Anoneemers has demonstrated that he obviously possesses one 😉

    Trigonometry R-Formula Answer Diagram 2

    You should be able to 'see' this from the diagram above 😉

    (iv)AM=Rsin(θ + α)=√40sin(θ+18.43)

    Area of OAM
    =10sin 2(θ+18.43)

    WOAH you have workings now! Yes, from the diagram in Part iii., you can easily see that

    AM = R sin (θ+α) ----- (2) simple right???

    But if you read carefully again, Part iv. only requires you to express in terms of R and (θ+α) (vs Part i. which required you to calculate the values of R and α). So strictly speaking you shouldn't replace R and α with their actual values in your answer at this point.

    Now area of ΔOAM = 0.5 x OM x AM. Using the expressions in (1) and (2),
    Area of ΔOAM
    = 0.5 x R cos (θ + α) x R sin (θ+α)
    = (0.5)R2(0.5) 2 cos (θ + α) sin (θ+α)
    (when you see a sine and cosine term with the same angle together - you should instinctively think of the Double Angle Formula for sine!)

    Sub in R = √40 obtained from Part i.
    = (√402/4) sin 2(θ + α)
    = 10 sin 2(θ + α) YAY!


    Marks are usually served to you on the platter in the Maximum/Minimum parts of R-Formula questions so DON'T EVER 对不起你自己 and skip this part!

    From the notes above: The maximum value is R which occurs when cos/sin (θ ± α) = 1.

    So in our case the maximum area is simple 10 cm2 (the coefficient of our expression in Part iv) and this occurs when sin 2(θ + α) = 1
    ⇒ sin 2(θ + 18.435°) = 1
    ⇒ 2(θ + 18.435°) = 90°
    θ = 26.6° (since θ must be acute)

    By the way how do u type all those square roots and teeter symbol i had to copy and paste =.=
    It's my 1st time making

  2. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    Hey Anoneemers, Miss Loi has commented on your test drive of the Trigo R-Formula questions.

    Regarding the symbols, Miss Loi was taught from the start to reference this table. But after awhile it became pretty intuitive e.g. for α you type &alpha;, for β you type &beta; etc.

    But having said that, copy and pasting still seem to be the quickest!

  3. mathslover's Avatar
    mathslover commented in tuition class


    What? This thing is in O level A maths nowadays? We're moving closer to this day...

    Used to take a long time remembering this when I did A levels, and until now still cannot figure out the logic behind all those steps used to derive the Rsin and Rcos.

  4. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    mathslover: As far as Miss Loi can recall (please correct if she's wrong though), the R-formula was only removed from the A-Math syllabus between 2002-2007, as she could still remember herself utilizing the R-Formula diligently when she was still a kawaii teeny Sec Four school girl.

    So her kaypoh guess is you're around 18-23 yrs-old? Oh wait ... but you know how to do Plane Geometry questions!

    The plot thickens ...

  5. mathslover's Avatar
    mathslover commented in tuition class


    Miss Loi: Actually, my O level year not within the range. Its the extraordinary year where matrices and vectors are out.

    Guess either u remember wrongly, or i was sleeping in A maths class..

    Plane Geometry is new? I kinda faintly remembered doing something called Circular Measure which include tangent/chord/segment/sector...

    No plot lah Miss Loi, you watch too much HK drama put your kawaii notes below the prawns so big, anyone can do the question le. 🙂

  6. Rose's Avatar
    Rose commented in tuition class


    Help! I can't get the answer for this question, 4sin3x + 3cos3x=√3.
    Tried doing the question for 2-3 times but still can't get the answer.
    I don't understand the minimum and maximum thing 🙁 🙁

  7. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    Hello Rose, yours is really a classic a sin θ ± b cos θ = c trigonometric equation question.

    First understand that the R-Formula part i.e. converting the LHS into a R-Formula expression, is straightforward:

    Sub a=4, b=3 for *R sin (θ±α) (do check with your Cher if you need to derive the R-Formula using Addition Formula, especially if the part carries a lot of marks),

    R = √(a2+b2) = √(42+32) = 5

    α = tan−1(b/a) = tan−1(3/4) ≈ 36.870°

    Thus your LHS: 4 sin 3x + 3 cos 3x can be expressed as:
    5 sin (3x + 36.870°)

    *Note: Since both terms on LHS are positive, and assuming the question didn't explicitly state which form to express in, you may also express it in the form of R cos(θ−α) but in this case your a=3 and b=4!

    To address your confusion of the "minimum and maximum thing", as Miss Loi has no further detail of your question, let's assume that we're looking for the max/min values of 5 sin (3x + 36.870°).

    Consider the typical graph of a sine function y = a sin θ which you learnt in Sec 3:

    Typical Sine Graph

    Looking at your R-Formula expression of 5 sin (3x + 36.870°), can you see that it's also a sine graph when you let a=5 and θ = 3x+36.870°? Hence from the diagram above, the max value is 5 and the min value is −5.

    Now to solve for x in your equation (the part where many students get KIA because they often forget about their basic angles):

    We have:

    5 sin (3x + 36.870°) = √3
    sin (3x + 36.870°) = sqrt{3}/5

    SATC Diagram

    Basic angle β = sin−1sqrt{3}/5 = 20.268°

    Assuming your question asks for all values of x in the range 0° ≤ x ≤ 360°,

    Since sin (3x + 36.870°) is positive,
    ⇒ 3x + 36.870°
    = 20.268°, 159.732°, 380.268°, 519.732°, 740.268°, 879.732°

    Note that since it's 3x in the expression, we need to go 3 rounds round the SATC diagram i.e. β + 0°, β + 360°, β + 720°

    3x = −16.602°, 122.862°, 343.398°, 482.862°, 703.398°, 842.862°

    x ≈ 354.5°, 50.0°, 114.5°, 161.0°, 234.5°, 281.0°

    Does the answer tally with yours?

  8. Rose's Avatar
    Rose commented in tuition class


    Thank you!
    Erm... Can you teach me how to find the max and min values of (5sinx - 12cosx)²?

  9. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    Hello again Rose and a interesting but nonetheless a pretty straightforward question you have!

    To find the max/min values of (5 sin x − 12 cos x)², you may first express the expression within the brackets using the R-Formula i.e:

    (5 sin x − 12 cos x)² = [R sin (xα)]²

    As discussed earlier throughout this blog post,
    we know R = √(5²+12²) = 13
    ⇒ (5 sin x − 12 cos x)² = [13 sin (xα)]²

    Since max/min value of sin (xα) = ±1, we can deduce that the max/min value of 5 sin x − 12 cos x = ±13 BUT we're dealing with a SQUARED expression (5 sin x − 12 cos x)² so

    Max value = 13² = 169
    Min value = 0 (since the square of a −ve term is always +ve)

    Happy New Year!

  10. AMath Student's Avatar
    AMath Student commented in tuition class


    I am weak in trigonometry, and realise that it stands a huge amount of marks in paper 2. I'm a Sec 4 student and MYE is coming :S

  11. soha's Avatar
    soha commented in tuition class


    i m weak in math s ive an exam on 1 june please guide me wat can id o to get a seat in bitsat

  12. Jukie Primestein's Avatar
    Jukie Primestein commented in tuition class


    Miss Loi is teta = 90°- alpha in the first triangle?

  13. Jukie Primestein's Avatar
    Jukie Primestein commented in tuition class


    sorry I was referring to the second diagram I didnot see the first one?

  14. Jukie Primestein's Avatar
    Jukie Primestein commented in tuition class


    Can you please send me a few more variation of the roar R-formula. Based on a few times I encountered it it is not always clear if you are to find a maximum or minimum! Oh another question I have not seen any proof to convince me than cosine of a negative number is positive, but please don't get me wrong I know it is correct however a proof maybe through a diagram would really do the trick!

  15. Jukie Primestein's Avatar
    Jukie Primestein commented in tuition class


    which trig identity do you refer to as the factor formula?

  16. weijie's Avatar
    weijie commented in tuition class


    why cannot r be negative 5 in the exaple given ?

  17. Ray's Avatar
    Ray commented in tuition class


    Hi Miss Loi,

    I love your site and your sense of humour, definitely very localised.

    Anyway, a quick question to your type-R question 2, the part on maximum area of the triangle.

    Should we apply the maximum value of R at the length stage? meaning to AM and OM individually instead of applying to the area straghit?

    The maximum value of AM and OM exist when cos (θ + α) =1, Hence, maximum value for AM and OM should be √40. Hence, 1/2 times AM times OM should give us 20cm sq.?

    I'm confused. Pls clarify if you can? Thanks! =]

    • 2011

      @Ray: Hello and welcome to Jφss Sticks, Ray!

      From parts 2(i) & 2(iii) of the workings here,
      OM = √40 cos (θ+18.43°)
      AM = √40 sin (θ+18.43°)

      This means that both the lengths of OM and AM are dependent on the same angle θ, which, looking at the equations above, means that both cannot be maximum at the same value of θ

      i.e. When OM is maximized when cos (θ + α) = 1, AM is actually minimized at this same value of θ since sin (θ + α) = 0.

      So the key here is we need an intermediate value of θ that will maximize the area instead of the individual lengths hence we have to obtain the ONE R-formula for the area using the double-angle formula etc.

  18. Agnes's Avatar
    Agnes commented in tuition class


    Hi. I was wondering if you can help me with these questions.

    8 cos x - 15 sin x = 17 cos (x + 1.08) -------- (R formula)

    1. Find the minimum and maximum value of 20 - 8 cos x + 15 sin x and the corresponding x values for 0 ≤ x ≤ 2π.
    2. Solve the equation |8 cos x - 15 sin x| = 6 for 0 ≤ x ≤ 2π.

    Thanks in advance!

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