Proof by mathematical induction — Question 16

QCAA 2020, Paper 1 · 6 marks

Q16 · 2020 · Technology-freeComplex familiar6 marks

QUESTION 16 (6 marks)

Given

\cos(\theta)\ne0

for all

n\in\mathbb{Z}^+

, use mathematical induction to prove

\cos\theta-\cos3\theta+\cos5\theta-\cdots+(-1)^{n+1}\cos((2n-1)\theta)=\frac{1-(-1)^n\cos(2n\theta)}{2\cos\theta}.

WORKED SOLUTION

6 marks

ANSWER

The identity holds for all

n\in\mathbb{Z}^+

Worked solution

Let

P(n)

denote the proposition. For

n=1

, the right-hand side is

[1+\cos(2\theta)]/[2\cos\theta]=\cos\theta

, so

P(1)

is true. Assume

P(k)

is true. Adding the next term

(-1)^{k+2}\cos((2k+1)\theta)

to the assumed expression and using

2\cos A\cos B=\cos(A-B)+\cos(A+B)

gives

\frac{1-(-1)^{k+1}\cos(2(k+1)\theta)}{2\cos\theta}.

Thus

P(k+1)

is true. By mathematical induction the formula holds for

n=1,2,\ldots

correctly proves the initial statement

[1 mark]

correctly states the assumption and the proof requirement for the inductive step

[1 mark]

uses assumption in the proof of the inductive step

[1 mark]

determines a simplified expression based on the use of a common denominator and a suitable trigonometric product identity

[1 mark]

determines a simplified expression based on the recognition that

(-1)^k=(-1)^{k+2}

[1 mark]

completes proof and communicates a suitable conclusion

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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