Pure: Trigonometry Depth
A-level Pure Mathematics' trigonometry at depth covers the compound angle formulae, the double angle formulae used in both directions (expanding and simplifying), writing a sin(theta) + b cos(theta) in the harmonic form R sin(theta +- alpha) or R cos(theta -+ alpha) to find maxima, minima or to solve equations, proving identities, and solving equations that need an identity substitution first, always over a stated interval. The marks reward correctly applying the compound angle sign rule (which flips between the sine and cosine versions), stating R and alpha exactly (R as a surd or integer, alpha to the stated accuracy), and giving every solution within the interval the question specifies, in the ORIGINAL variable.
Before you start
Make sure you're comfortable with these topics first:
Method
- Learn and apply the compound angle formulae precisely: sin(A +- B) = sinAcosB +- cosAsinB; cos(A +- B) = cosAcosB -+ sinAsinB; tan(A +- B) = (tanA +- tanB)/(1 -+ tanAtanB). The subtracted sign flips between the sin and cos versions - this is the single most common slip.
- Derive the double angle formulae as the special case A = B: sin2A = 2sinAcosA; cos2A = cos^2A - sin^2A (also = 2cos^2A - 1 = 1 - 2sin^2A, useful for solving an equation in a single function); tan2A = 2tanA/(1-tan^2A).
- To write a sin(theta) + b cos(theta) in the form R sin(theta +- alpha) (or R cos(theta -+ alpha)), expand the target form with the compound angle formula, match coefficients to find R = sqrt(a^2+b^2) and the value of alpha from the ratio of a and b, and state which quadrant alpha is in from the signs of a and b.
- Use the R-form to read off the maximum value (R, when the trig function equals 1) and minimum value (-R) and where each occurs, or to solve an equation of the form a sin(theta) + b cos(theta) = c that has no other direct method.
- To prove an identity, work on the more complicated side only, converting everything to sin and cos if stuck, and use a Pythagorean identity or a compound/double angle substitution to reduce it to match the other side - never move terms across the equals sign as if solving an equation.
- When an equation mixes functions or angles (such as sin(2x) and sin(x), or sec^2(x) and tan(x)), substitute an identity so the whole equation is in a single function of a single angle, then solve it as a normal polynomial or trig equation.
- If the equation was solved in terms of a substituted variable (such as u = 2x - 30 degrees), convert the solution interval for u before finding u, then convert every valid u value back to the original variable at the very end, and check each still lies in the original given interval.
Worked example
Express 5cos(theta) - 12sin(theta) in the form Rcos(theta + alpha), where R > 0 and 0 < alpha < 90 degrees, giving alpha to 1 decimal place. Hence solve 5cos(theta) - 12sin(theta) = 6.5 for 0 <= theta <= 360 degrees, giving your answers to 1 decimal place.
- Expand Rcos(theta+alpha) = Rcos(theta)cos(alpha) - Rsin(theta)sin(alpha), and compare with 5cos(theta) - 12sin(theta) to get Rcos(alpha) = 5 and Rsin(alpha) = 12.
- Find R using R^2 = 5^2 + 12^2 = 169, so R = 13.
- Find alpha using tan(alpha) = 12/5 = 2.4, so alpha = 67.4 degrees (1 d.p.), since both Rcos(alpha) and Rsin(alpha) are positive, placing alpha in the first quadrant.
- Rewrite the equation as 13cos(theta+67.4) = 6.5, so cos(theta+67.4) = 0.5.
- Let phi = theta+67.4; since theta ranges from 0 to 360, phi ranges from 67.4 to 427.4. Solving cos(phi) = 0.5 in this range gives phi = 300 or phi = 420 (using the principal value 60, and 360-60=300, plus adding 360 for the next cycle).
- Convert back to theta: theta = phi - 67.4, giving theta = 232.6 degrees or theta = 352.6 degrees (1 d.p.).
Practice questions
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Q1Use the compound angle formula for cos(A-B) to show that cos(90 degrees - theta) = sin(theta).Show answer
Answer: cos(90-theta) = cos90 costheta + sin90 sintheta = (0)costheta + (1)sintheta = sintheta.
Q2Given that sin(A) = 3/5 with A acute, and cos(B) = -12/13 with B obtuse, find the exact value of sin(A+B).Show answer
Answer: -16/65. (cosA = 4/5 since A is acute; sinB = 5/13 since B is obtuse (sin positive). sin(A+B) = sinAcosB + cosAsinB = (3/5)(-12/13) + (4/5)(5/13) = -36/65 + 20/65.)
Q3Express 3sin(theta) + 4cos(theta) in the form Rsin(theta+alpha), R > 0, 0 < alpha < 90 degrees. Hence state the maximum value of 3sin(theta) + 4cos(theta), and the smallest positive value of theta at which it occurs, to 1 decimal place.Show answer
Answer: R = 5, alpha = 53.1 degrees (from tan(alpha) = 4/3). Maximum value = 5, occurring when theta + alpha = 90, i.e. theta = 36.9 degrees (1 d.p.).
Q4In triangle ABC, AB = 8 cm, BC = 6 cm, and angle BAC = 40 degrees. Find the two possible values of angle BCA, giving your answers to 1 decimal place.Show answer
Answer: 59.0 degrees or 121.0 degrees. (Sine rule: sin(C) = 8sin40/6 = 0.857 (3 s.f.); both the acute value and its supplement give a valid triangle angle sum.)
Q5Explain why, when using the sine rule to find an angle given two sides and a non-included angle, there may be two possible values, but this ambiguity does not occur when using the sine rule to find a side, or the cosine rule to find an angle.Show answer
Answer: sin(x) and sin(180-x) are equal, so solving sin(C) = k for an angle gives an acute and an obtuse candidate, both needing checking against the triangle's angle sum. Finding a side just substitutes into a formula, with no inverse trig function involved, so there is no ambiguity. The cosine rule gives cos(x) = k directly, and since cos is one-to-one (strictly decreasing) for 0 < x < 180, there is only one possible angle.
Q6Prove that (1 - cos(2theta))/sin(2theta) = tan(theta).Show answer
Answer: Using cos2theta = 1-2sin^2theta, the numerator 1-cos2theta = 2sin^2theta. The denominator sin2theta = 2sinthetacostheta. So the fraction equals 2sin^2theta/(2sinthetacostheta) = sintheta/costheta = tantheta.
Q7Using the small angle approximations sin(theta) ~= theta - theta^3/6 and cos(theta) ~= 1 - theta^2/2 (theta in radians), show that (1-cos(theta))/(theta sin(theta)) ~= 1/2 for small theta.Show answer
Answer: 1-costheta ~= theta^2/2. theta sintheta ~= theta(theta - theta^3/6) ~= theta^2, to leading order. So (1-costheta)/(theta sintheta) ~= (theta^2/2)/theta^2 = 1/2.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
Solve the equation sin(x + 60 degrees) = 2cos(x) for 0 <= x <= 360 degrees, giving your answers to 1 decimal place.
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(a) Show that cos(x + 45 degrees) + cos(x - 45 degrees) = sqrt(2)cos(x). (4) (b) Hence solve cos(x + 45 degrees) + cos(x - 45 degrees) = 1 for 0 <= x <= 360 degrees, giving your answers to 1 decimal place. (3)
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(a) Show that 4sin^2(x) + 3cos(x) can be written as -4cos^2(x) + 3cos(x) + 4. (3) (b) Hence show that the equation 4sin^2(x) + 3cos(x) = 6 has no solutions. (3)
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