A1Arena Open the app

Trigonometry: identities and equations

Further Mathematics · WAEC and JAMB · SS2 and SS3

Trigonometry is the heaviest-scoring area of Further Mathematics and the one candidates prepare least. Nearly every question is solved by converting everything to a single ratio using an identity, then treating what is left as an ordinary equation.

What you need to know

  • The three Pythagorean identities are sin^2 A + cos^2 A = 1, 1 + tan^2 A = sec^2 A and 1 + cot^2 A = cosec^2 A. The second and third come from dividing the first by cos^2 A and by sin^2 A, so you only need to memorise one and derive the others.
  • When an equation mixes sin^2 x with cos x, replace sin^2 x with 1 - cos^2 x so that everything is in cos x. The result is a quadratic in cos x, which you factorise exactly as you would a quadratic in y.
  • Always reject any value of sin x or cos x lying outside the range from -1 to 1. A quadratic in cos x often has one valid root and one impossible one, and failing to reject the impossible root costs the mark.
  • The compound angle formulas are sin(A + or - B) = sin A cos B plus or minus cos A sin B, and cos(A + or - B) = cos A cos B minus or plus sin A sin B. Notice that cosine flips the sign while sine keeps it, which is the detail candidates reverse.
  • Use compound angles to find exact ratios of unusual angles: sin 75 = sin(45 + 30) = sin 45 cos 30 + cos 45 sin 30 = (sqrt(6) + sqrt(2))/4, which is about 0.9659.
  • The double angle formulas are sin 2A = 2 sin A cos A and cos 2A = cos^2 A - sin^2 A. The cosine form has two further versions, 2 cos^2 A - 1 and 1 - 2 sin^2 A, and choosing the right version is usually what makes a proof collapse neatly.
  • In a proof, start from the more complicated side and work towards the simpler one. Never move terms across the equals sign in a proof, because you are not solving an equation, you are showing that two expressions are already equal.
  • An expression of the form a cos x + b sin x can be written as R cos(x - alpha) or R sin(x + alpha), where R = sqrt(a^2 + b^2) and tan alpha = b/a. This turns a two-term equation into a single trigonometric equation you can solve directly.
  • The R method also gives maximum and minimum values instantly: a cos x + b sin x has maximum R and minimum -R, because the cosine factor can only range from -1 to 1.
  • For a basic equation, find the acute reference angle from your calculator, then place the answers in the correct quadrants using the ALL, SIN, TAN, COS rule. Cosine is positive in the first and fourth quadrants, sine in the first and second, tangent in the first and third.
  • The general solutions are: for sin x = k, x = n(180) + (-1)^n alpha; for cos x = k, x = n(360) plus or minus alpha; and for tan x = k, x = n(180) + alpha, where n is any integer and alpha is the reference angle.
  • When the equation involves a multiple angle such as tan 3x or sin 2x, solve for the multiple angle over the enlarged range first. For 0 to 360 degrees in x, you must search 0 to 1080 degrees in 3x, otherwise you will lose two thirds of the solutions.
  • In any triangle the sine rule is a/sin A = b/sin B = c/sin C, and the cosine rule is a^2 = b^2 + c^2 - 2bc cos A. Use the cosine rule when you have two sides and the included angle, or all three sides; use the sine rule otherwise.
  • The area of a triangle given two sides and the included angle is (1/2)ab sin C. This is the standard follow-up part after a sine or cosine rule calculation, so expect it.

Key terms

Identity
An equation that is true for every value of the variable, as opposed to an equation that is true only for particular values.
Reference angle
The acute angle between the terminal arm and the x-axis, used with the quadrant rule to generate all solutions.
General solution
A formula involving an integer n that generates every solution of a trigonometric equation, not just those in a stated range.
Amplitude
The maximum displacement of a sine or cosine curve from its central value, equal to R in the expression R sin(x + alpha).
Period
The horizontal length after which a trigonometric graph repeats, which is 360 degrees for sin x and cos x but 180 degrees for tan x.
Secant, cosecant and cotangent
The reciprocals of cosine, sine and tangent respectively, so sec A = 1/cos A, cosec A = 1/sin A and cot A = 1/tan A.

Formulae

  • sin^2 A + cos^2 A = 1
  • 1 + tan^2 A = sec^2 A
  • 1 + cot^2 A = cosec^2 A
  • tan A = sin A / cos A
  • sin(A + B) = sin A cos B + cos A sin B
  • sin(A - B) = sin A cos B - cos A sin B
  • cos(A + B) = cos A cos B - sin A sin B
  • cos(A - B) = cos A cos B + sin A sin B
  • tan(A + B) = (tan A + tan B)/(1 - tan A tan B)
  • sin 2A = 2 sin A cos A
  • cos 2A = cos^2 A - sin^2 A = 2cos^2 A - 1 = 1 - 2sin^2 A
  • tan 2A = 2 tan A/(1 - tan^2 A)
  • a cos x + b sin x = R cos(x - alpha), R = sqrt(a^2 + b^2), tan alpha = b/a
  • General solution for sin x = sin alpha: x = n(180) + (-1)^n alpha
  • General solution for cos x = cos alpha: x = n(360) + or - alpha
  • General solution for tan x = tan alpha: x = n(180) + alpha
  • Sine rule: a/sin A = b/sin B = c/sin C
  • Cosine rule: a^2 = b^2 + c^2 - 2bc cos A
  • Area of triangle = (1/2) ab sin C

Worked examples

Solve the equation 2 sin^2 x - 3 cos x = 0 for values of x from 0 to 360 degrees.

  1. Replace sin^2 x using the identity sin^2 x = 1 - cos^2 x: 2(1 - cos^2 x) - 3 cos x = 0.
  2. Expand and rearrange: 2 - 2cos^2 x - 3 cos x = 0, which becomes 2cos^2 x + 3 cos x - 2 = 0.
  3. Let c = cos x and factorise: 2c^2 + 3c - 2 = (2c - 1)(c + 2) = 0, so c = 1/2 or c = -2.
  4. Reject c = -2, because the cosine of an angle can never be less than -1.
  5. Solve cos x = 1/2. The reference angle is 60 degrees, and cosine is positive in the first and fourth quadrants.
  6. So x = 60 degrees or x = 360 - 60 = 300 degrees. Check x = 60: 2(sin 60)^2 - 3 cos 60 = 2(0.75) - 3(0.5) = 1.5 - 1.5 = 0.

Express 3 cos x - 4 sin x in the form R cos(x + alpha), then solve 3 cos x - 4 sin x = 2 for 0 to 360 degrees, correct to two decimal places.

  1. Compare with R cos(x + alpha) = R cos x cos alpha - R sin x sin alpha, so R cos alpha = 3 and R sin alpha = 4.
  2. Then R = sqrt(3^2 + 4^2) = sqrt(25) = 5 and tan alpha = 4/3, so alpha = 53.13 degrees.
  3. The expression is 5 cos(x + 53.13 degrees), so the equation becomes 5 cos(x + 53.13) = 2, that is cos(x + 53.13) = 0.4.
  4. The reference angle is cos inverse 0.4 = 66.42 degrees, and cosine is positive in the first and fourth quadrants, so x + 53.13 = 66.42 or 360 - 66.42 = 293.58.
  5. Therefore x = 66.42 - 53.13 = 13.29 degrees, or x = 293.58 - 53.13 = 240.45 degrees.
  6. Check x = 13.29: 3 cos(13.29) - 4 sin(13.29) = 2.9196 - 0.9194 = 2.00, as required.

The mistake to avoid

Giving only the first-quadrant answer. The calculator returns one angle, but almost every trigonometric equation on this syllabus has two or more solutions in the range 0 to 360 degrees, and the second one carries equal marks. The other killer is solving a multiple-angle equation over the original range instead of the enlarged range, which silently discards most of the answers.

In the exam

Write the range at the top of your working and keep looking back at it. For a multiple angle, multiply the range first, list every solution of the multiple angle across that enlarged range, then divide each one back down. In proofs, convert everything to sines and cosines as a default strategy; it is not always the slickest route but it almost always works.