4037/21

Additional Mathematics 4037/21May/June 2013

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

12
questions
80
marks
120
minutes

Topics Calculus · Trigonometry · Straight-line graphs · Equations, inequalities and graphs · Series · Simultaneous equations · +5 more

Q14MTrigonometryFree sample

Prove that

(1+sinθcosθ)2+(1sinθcosθ)2=2+4tan2θ\left(\frac{1 + \sin \theta}{\cos \theta}\right)^2 + \left(\frac{1 - \sin \theta}{\cos \theta}\right)^2 = 2 + 4\tan^2 \theta
DifficultyMedium-Easy
Worked solution

Approach

Start from the left-hand side, expand both squares, combine over the common denominator cos2θ\cos^2 \theta, then use sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 and sec2θ=1+tan2θ\sec^2 \theta = 1 + \tan^2 \theta to reach the right-hand side.

Working

(1+sinθcosθ)2+(1sinθcosθ)2=(1+sinθ)2cos2θ+(1sinθ)2cos2θ\left(\frac{1 + \sin \theta}{\cos \theta}\right)^2 + \left(\frac{1 - \sin \theta}{\cos \theta}\right)^2 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} + \frac{(1 - \sin \theta)^2}{\cos^2 \theta} =(1+2sinθ+sin2θ)+(12sinθ+sin2θ)cos2θ=2+2sin2θcos2θ= \frac{(1 + 2\sin \theta + \sin^2 \theta) + (1 - 2\sin \theta + \sin^2 \theta)}{\cos^2 \theta} = \frac{2 + 2\sin^2 \theta}{\cos^2 \theta}

Using sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, so that 2+2sin2θ=2(sin2θ+cos2θ)+2sin2θ=4sin2θ+2cos2θ2 + 2\sin^2 \theta = 2(\sin^2 \theta + \cos^2 \theta) + 2\sin^2 \theta = 4\sin^2 \theta + 2\cos^2 \theta:

=4sin2θ+2cos2θcos2θ=4sin2θcos2θ+2cos2θcos2θ=4tan2θ+2= \frac{4\sin^2 \theta + 2\cos^2 \theta}{\cos^2 \theta} = \frac{4\sin^2 \theta}{\cos^2 \theta} + \frac{2\cos^2 \theta}{\cos^2 \theta} = 4\tan^2 \theta + 2

which is the required result.

Answer

(1+sinθcosθ)2+(1sinθcosθ)2=2+4tan2θ\left(\frac{1 + \sin \theta}{\cos \theta}\right)^2 + \left(\frac{1 - \sin \theta}{\cos \theta}\right)^2 = 2 + 4\tan^2 \theta

as required.

Final answer

Proved: the left-hand side simplifies to 2 + 4 tan^2 theta

Detailed explanation

Walkthrough

The left-hand side has two fractions with the same denominator cosθ\cos \theta, squared. Expanding each numerator with (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2, the middle terms +2sinθ+2\sin \theta and 2sinθ-2\sin \theta cancel when added, leaving 2+2sin2θ2 + 2\sin^2 \theta over cos2θ\cos^2 \theta. The Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 lets us rewrite 2+2sin2θ2 + 2\sin^2 \theta as 4sin2θ+2cos2θ4\sin^2 \theta + 2\cos^2 \theta, which splits into 4sin2θcos2θ+2cos2θcos2θ\frac{4\sin^2 \theta}{\cos^2 \theta} + \frac{2\cos^2 \theta}{\cos^2 \theta}. The first term is 4tan2θ4\tan^2 \theta (since tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}) and the second is simply 22, giving exactly the printed right-hand side.

Key Takeaways

  • Expanding (1+sinθ)2(1 + \sin \theta)^2 and (1sinθ)2(1 - \sin \theta)^2 together makes the cross terms cancel — a very common pattern in identity proofs.
  • The Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 is the main tool for converting between sin2\sin^2 and cos2\cos^2.
  • Dividing term by term by cos2θ\cos^2 \theta converts everything into tan2θ\tan^2 \theta and constants.

Common Mistakes

  • Expanding (1±sinθ)2(1 \pm \sin \theta)^2 as 1±sin2θ1 \pm \sin^2 \theta, forgetting the middle term ±2sinθ\pm 2\sin \theta — this destroys the proof.
  • Cancelling sinθ\sin \theta or dividing through by cosθ\cos \theta at some stage, which can lose solutions or break the identity's generality.
  • Working backwards from the RHS: the mark scheme allows it only if done rigorously; starting forward from the LHS is safest.
  • Not showing every intermediate line — the scheme awards B1 for each distinct correct step (2+2sin2θcos2θ\frac{2 + 2\sin^2 \theta}{\cos^2 \theta}, the Pythagoras use, the tan\tan conversion, and completion), so skipped algebra loses marks.

Things to Be Careful About

  • This is an "AG" (answer given) style part: you must derive forward to exactly the printed target and show sufficient correct detail for all four marks.
  • Keep the angle θ\theta consistent throughout — the scheme penalises inconsistent notation (recoverable, but avoid it).
  • Every mark corresponds to a visible line of working; do not jump straight from the expanded form to 2+4tan2θ2 + 4\tan^2 \theta without showing the Pythagorean substitution.
Techniques used
expand the squares of the two fractionscombine over a common denominatorapply the Pythagorean identity sin^2 + cos^2 = 1use sec^2 = 1 + tan^2 to reach the target

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