9709/33

Mathematics 9709/33May/June 2017

Cambridge A-Level · Pure Mathematics 3 · worked solutions for every part, with the mark scheme

11
questions
75
marks
105
minutes

Topics Integration · Trigonometry · Algebra · Differentiation · Logarithmic and Exponential Functions · Numerical Solution of Equations · +3 more

Q13MTrigonometryFree sample

Prove the identity

cotxtanxcotx+tanxcos2x\frac{\cot x - \tan x}{\cot x + \tan x} \equiv \cos 2x
DifficultyMedium-Easy
Worked solution

Approach

Rewrite cotx\cot x and tanx\tan x in terms of cosx\cos x and sinx\sin x, combine the fractions in the numerator and denominator, then use cos2x+sin2x=1\cos^2 x + \sin^2 x = 1 and the double-angle formula cos2x=cos2xsin2x\cos 2x = \cos^2 x - \sin^2 x.

Working

Let

L=cotxtanxcotx+tanx.L = \frac{\cot x - \tan x}{\cot x + \tan x}.

Using cotx=cosxsinx\cot x = \frac{\cos x}{\sin x} and tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}:

L=cosxsinxsinxcosxcosxsinx+sinxcosxL = \frac{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}}

Combine each pair of fractions:

L=cos2xsin2xsinxcosxcos2x+sin2xsinxcosxL = \frac{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}}{\frac{\cos^2 x + \sin^2 x}{\sin x \cos x}}

Cancel the common factor sinxcosx\sin x \cos x:

L=cos2xsin2xcos2x+sin2xL = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}

Using cos2x+sin2x=1\cos^2 x + \sin^2 x = 1:

L=cos2xsin2xL = \cos^2 x - \sin^2 x

Finally, by the double-angle formula:

L=cos2xL = \cos 2x

Answer

cotxtanxcotx+tanxcos2x\frac{\cot x - \tan x}{\cot x + \tan x} \equiv \cos 2x
Final answer

LHS ≡ cos 2x

Detailed explanation

Walkthrough

We need to prove an identity, so we start with the left-hand side and transform it until it equals the right-hand side.

First, rewrite cotx\cot x and tanx\tan x using the basic definitions:

cotx=cosxsinx,tanx=sinxcosx.\cot x = \frac{\cos x}{\sin x}, \qquad \tan x = \frac{\sin x}{\cos x}.

This is the first mark: the expression is now written entirely in terms of cosx\cos x and sinx\sin x.

Substitute these into the numerator and denominator. Both become fractions. To combine them, give each fraction the common denominator sinxcosx\sin x \cos x:

cotxtanx=cos2xsin2xsinxcosx\cot x - \tan x = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}

and

cotx+tanx=cos2x+sin2xsinxcosx.\cot x + \tan x = \frac{\cos^2 x + \sin^2 x}{\sin x \cos x}.

Dividing the first fraction by the second cancels the common denominator sinxcosx\sin x \cos x, leaving

cos2xsin2xcos2x+sin2x.\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}.

Now use the Pythagorean identity cos2x+sin2x=1\cos^2 x + \sin^2 x = 1. This is the second mark. The denominator becomes 1, so the expression simplifies to

cos2xsin2x.\cos^2 x - \sin^2 x.

Finally, recognise that cos2xsin2x\cos^2 x - \sin^2 x is the double-angle formula for cos2x\cos 2x. This gives the required result and earns the final mark.

Key Takeaways

  • The definitions tanx=sinxcosx\tan x = \frac{\sin x}{\cos x} and cotx=cosxsinx\cot x = \frac{\cos x}{\sin x} allow many trig expressions to be rewritten in terms of sinx\sin x and cosx\cos x.
  • The Pythagorean identity cos2x+sin2x=1\cos^2 x + \sin^2 x = 1 is often used to simplify denominators or numerators.
  • Recognising cos2xsin2x=cos2x\cos^2 x - \sin^2 x = \cos 2x is essential for proving double-angle identities.
  • When simplifying a compound fraction, combine the numerator and denominator separately, then cancel common factors.

Common Mistakes

  • Dividing fractions incorrectly: a/bc/b=ac\frac{a/b}{c/b} = \frac{a}{c}, not abcb\frac{a}{b} \cdot \frac{c}{b}.
  • Sign errors: the numerator must be cos2xsin2x\cos^2 x - \sin^2 x, not sin2xcos2x\sin^2 x - \cos^2 x.
  • Writing cos2x+sin2x=0\cos^2 x + \sin^2 x = 0 or replacing it by something other than 1.
  • Stopping at cos2xsin2x\cos^2 x - \sin^2 x without applying the double-angle formula.
  • The mark scheme requires method: an unsupported final answer would not earn full marks.

Things to Be Careful About

  • The identity is only defined where both tanx\tan x and cotx\cot x are defined, i.e. where cosx0\cos x \neq 0 and sinx0\sin x \neq 0.
  • Keep the common denominator sinxcosx\sin x \cos x until it cancels; do not cancel it too early.
  • Use exactly cos2x+sin2x=1\cos^2 x + \sin^2 x = 1; this is the Pythagorean identity needed for the M1 mark.
  • The final step must state cos2x\cos 2x to earn the A1 mark.
Techniques used
rewrite cot and tan in terms of sin and coscombine compound fractionsapply Pythagorean identityapply double angle formula

The rest of this paper

10 more questions
  • Q2Algebra4M
  • Q3Logarithmic and Exponential Functions4M
  • Q4Integration4M
  • Q5Differentiation · Trigonometry6M
  • Q6Numerical Solution of Equations7M
  • Q7Differentiation · Integration8M
  • Q8Differential Equations9M
  • Q9Algebra · Integration10M
  • Q10Vectors10M
  • Q11Complex Numbers10M
Loading the full paper…