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4037/22

Additional Mathematics 4037/22 — May/June 2018

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

12
questions
80
marks
120
minutes

Topics Calculus · Circular measure · Trigonometry · [Legacy] Set language and notation · Factors of polynomials · Simultaneous equations · +6 more

Q1TrigonometryFree sample
(i)

Show that cos⁡θcot⁡θ+sin⁡θ=cosec⁡θ\cos \theta \cot \theta + \sin \theta = \operatorname{cosec} \theta.

3M
DifficultyMedium-Easy
Worked solution

Approach

Rewrite cot⁡θ\cot\theta as cos⁡θsin⁡θ\frac{\cos\theta}{\sin\theta}, combine everything over the common denominator sin⁡θ\sin\theta, then use cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 to reach cosec⁡θ\operatorname{cosec}\theta.

Working

cos⁡θcot⁡θ+sin⁡θ=cos⁡2θsin⁡θ+sin⁡θ\cos \theta \cot \theta + \sin \theta = \frac{\cos^2\theta}{\sin\theta} + \sin\theta

Over a common denominator:

=cos⁡2θ+sin⁡2θsin⁡θ= \frac{\cos^2\theta + \sin^2\theta}{\sin\theta}

Using cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1:

=1sin⁡θ=cosec⁡θ= \frac{1}{\sin\theta} = \operatorname{cosec}\theta

as required.

Answer

cos⁡θcot⁡θ+sin⁡θ=cosec⁡θ\cos \theta \cot \theta + \sin \theta = \operatorname{cosec}\theta
Final answer

Shown: cos θ cot θ + sin θ = cosec θ

Detailed explanation

Walkthrough

The left-hand side mixes three different trigonometric functions, so the first job is to express everything in terms of just sine and cosine. The identity cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta} does this, turning the first term into cos⁡2θsin⁡θ\frac{\cos^2\theta}{\sin\theta}. Now both terms can sit over the common denominator sin⁡θ\sin\theta: writing sin⁡θ=sin⁡2θsin⁡θ\sin\theta = \frac{\sin^2\theta}{\sin\theta} gives the single fraction cos⁡2θ+sin⁡2θsin⁡θ\frac{\cos^2\theta + \sin^2\theta}{\sin\theta}. The Pythagorean identity collapses the numerator to 11, leaving 1sin⁡θ\frac{1}{\sin\theta}, which is exactly the definition of cosec⁡θ\operatorname{cosec}\theta. This matches the printed target, so the identity is proved.

Key Takeaways

  • To prove an identity, convert all functions to sines and cosines first.
  • A common denominator lets separate terms merge into one fraction.
  • cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 and cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta = \frac{1}{\sin\theta} are the workhorse identities here.

Common Mistakes

  • Starting from the right-hand side or manipulating both sides at once — identities must be derived forward from one side to the other.
  • Forgetting to put sin⁡θ\sin\theta over the common denominator before adding, giving an invalid fraction.
  • Cancelling sin⁡θ\sin\theta incorrectly instead of combining fractions.
  • The mark scheme caps this part at 2 marks if the working is not fully correct or does not complete to cosec⁡θ\operatorname{cosec}\theta — every algebraic line must be shown.

Things to Be Careful About

This is an "AG" (answer given) proof: you must show sufficient correct detail, deriving forward to exactly cosec⁡θ\operatorname{cosec}\theta. Each of the three steps (using cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}, forming the common denominator, applying cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1) earns a mark, so none may be skipped.

Techniques used
rewrite cotangent in terms of sine and cosinecombine terms over a common denominatorapply the Pythagorean identity cos^2 + sin^2 = 1
(ii)

Hence solve cos⁡θcot⁡θ+sin⁡θ=4\cos \theta \cot \theta + \sin \theta = 4 for 0∘≤θ≤90∘0^\circ \leq \theta \leq 90^\circ.

2M
DifficultyEasy
Worked solution

Approach

Using the result of part (i), the equation becomes cosec⁡θ=4\operatorname{cosec}\theta = 4, so sin⁡θ=14\sin\theta = \frac{1}{4}. Take the inverse sine within 0∘≤θ≤90∘0^\circ \leq \theta \leq 90^\circ.

Working

From part (i), cos⁡θcot⁡θ+sin⁡θ=cosec⁡θ\cos \theta \cot \theta + \sin \theta = \operatorname{cosec}\theta, so:

cosec⁡θ=4⇒sin⁡θ=14\operatorname{cosec}\theta = 4 \quad\Rightarrow\quad \sin\theta = \frac{1}{4}

Since 0∘≤θ≤90∘0^\circ \leq \theta \leq 90^\circ, there is only the principal value:

θ=sin⁡−1(14)=14.4775…∘\theta = \sin^{-1}\left(\frac{1}{4}\right) = 14.4775\ldots^\circ

Answer

θ=14.5∘ (1 d.p.)\theta = 14.5^\circ \text{ (1 d.p.)}
Final answer

14.5°

Detailed explanation

Walkthrough

The word "Hence" signals that the identity proved in part (i) must be used. Substituting it turns the equation into cosec⁡θ=4\operatorname{cosec}\theta = 4. Taking reciprocals gives sin⁡θ=14\sin\theta = \frac{1}{4}. Because the interval is only 0∘≤θ≤90∘0^\circ \leq \theta \leq 90^\circ, the sine curve takes the value 14\frac{1}{4} exactly once there, in the first quadrant, so the single principal value from the inverse sine function is the whole answer. Evaluating gives 14.4775…∘14.4775\ldots^\circ, which rounds to 14.5∘14.5^\circ to one decimal place (the standard accuracy for angles in degrees).

Key Takeaways

  • "Hence" questions require the earlier result to actually be used.
  • cosec⁡θ=k\operatorname{cosec}\theta = k means sin⁡θ=1k\sin\theta = \frac{1}{k}.
  • On a restricted interval such as 0∘0^\circ to 90∘90^\circ, only the principal inverse-sine value is needed.

Common Mistakes

  • Re-solving from scratch without using the part (i) identity, contrary to the instruction "Hence".
  • Giving extra solutions outside the interval, or forgetting that no second-quadrant solution exists here because of the restriction.
  • Rounding too early; the mark scheme accepts answers which round to 14.5∘14.5^\circ but requires 4 or more figures before rounding.
  • The answer is "not from wrong working" (nfww): an incorrect route scores nothing even if the final number looks right.

Things to Be Careful About

Angles in degrees are given to 1 decimal place. Keep full calculator precision (14.47751…∘14.47751\ldots^\circ) until the end, then round once. Only values arising from correct working earn the accuracy mark.

Techniques used
substitute the identity proved in part (i)solve sin θ = k for θ in a restricted interval

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