Mathematics 9709/12 — October/November 2013

Cambridge AS Level · Pure Mathematics 1 · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Trigonometry · Differentiation · Coordinate Geometry · Integration · Functions · Circular Measure · +2 more

Q1TrigonometryFree sample

Given that $\cos x = p$ , where $x$ is an acute angle in degrees, find, in terms of $p$ ,

(i)

$\sin x$ ,

DifficultyEasy

Worked solution

Approach

Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ and substitute $\cos x = p$ . Since $x$ is acute, $\sin x$ is positive.

Working

\sin^2 x + \cos^2 x = 1

Substitute $\cos x = p$ :

\sin^2 x + p^2 = 1

\sin^2 x = 1 - p^2

Since $x$ is acute, $\sin x > 0$ , so

\sin x = \sqrt{1 - p^2}

Answer

\sin x = \sqrt{1 - p^2}

Final answer

sin x = sqrt(1 - p^2)

Detailed explanation

Walkthrough

We are told that $\cos x = p$ and that $x$ is acute. The identity $\sin^2 x + \cos^2 x = 1$ links sine and cosine for the same angle. Substituting $p$ for $\cos x$ gives $\sin^2 x = 1 - p^2$ . Because $x$ is acute, $\sin x$ must be positive, so we take the positive square root.

Key Takeaways

This question tests the fundamental Pythagorean identity $\sin^2 x + \cos^2 x = 1$ and the importance of the quadrant/acute condition when choosing the sign of a square root.

Common Mistakes

A common mistake is to write $\sin x = 1 - p^2$ instead of taking the square root. Another is to include $\pm$ ; since $x$ is acute, only the positive value is valid.

Things to Be Careful About

The sign of $\sin x$ is determined by the fact that $x$ is acute. Also, the answer must be in terms of $p$ only, with no $x$ remaining.

Techniques used

apply the Pythagorean identitytake the positive square root because the angle is acute

(ii)

$\tan x$ ,

DifficultyEasy

Worked solution

Approach

Use the identity $\tan x = \frac{\sin x}{\cos x}$ and substitute $\sin x = \sqrt{1 - p^2}$ and $\cos x = p$ .

Working

\tan x = \frac{\sin x}{\cos x}

Substitute the known values:

\tan x = \frac{\sqrt{1 - p^2}}{p}

Answer

\tan x = \frac{\sqrt{1 - p^2}}{p}

Final answer

tan x = sqrt(1 - p^2)/p

Detailed explanation

Walkthrough

From part (i), $\sin x = \sqrt{1 - p^2}$ . The identity $\tan x = \frac{\sin x}{\cos x}$ expresses tangent directly in terms of sine and cosine. Substituting $\sin x = \sqrt{1 - p^2}$ and $\cos x = p$ gives $\tan x = \frac{\sqrt{1 - p^2}}{p}$ .

Key Takeaways

This part tests the quotient identity $\tan x = \frac{\sin x}{\cos x}$ and the ability to substitute one expression into another.

Common Mistakes

A common mistake is to invert the fraction and write $\frac{p}{\sqrt{1 - p^2}}$ , which is actually $\tan(90^\circ - x)$ , not $\tan x$ . Another is to forget to use the result from part (i).

Things to Be Careful About

Keep the numerator and denominator in the correct order: sine over cosine. Since $x$ is acute, $p > 0$ , so there is no sign ambiguity in the denominator.

Techniques used

use the quotient identity tan x = sin x / cos xsubstitute the expression for sin x

(iii)

$\tan(90^\circ - x)$ .

DifficultyEasy

Worked solution

Approach

Use the complementary angle identity $\tan(90^\circ - x) = \cot x = \frac{1}{\tan x}$ , then substitute the result from part (ii).

Working

\tan(90^\circ - x) = \cot x = \frac{\cos x}{\sin x}

Substitute $\cos x = p$ and $\sin x = \sqrt{1 - p^2}$ :

\tan(90^\circ - x) = \frac{p}{\sqrt{1 - p^2}}

Equivalently, using part (ii):

\tan(90^\circ - x) = \frac{1}{\tan x} = \frac{p}{\sqrt{1 - p^2}}

Answer

\tan(90^\circ - x) = \frac{p}{\sqrt{1 - p^2}}

Final answer

tan(90° - x) = p/sqrt(1 - p^2)

Detailed explanation

Walkthrough

For an acute angle, $\tan(90^\circ - x) = \cot x = \frac{\cos x}{\sin x}$ . This is the reciprocal of $\tan x$ . Using $\cos x = p$ and $\sin x = \sqrt{1 - p^2}$ gives $\frac{p}{\sqrt{1 - p^2}}$ .

Key Takeaways

This part tests the complementary angle relationship $\tan(90^\circ - x) = \cot x$ , and the fact that it is the reciprocal of $\tan x$ .

Common Mistakes

A common mistake is to write $\tan(90^\circ - x) = \tan x$ , which is false. Another is to use $\frac{\sqrt{1 - p^2}}{p}$ instead of its reciprocal.

Things to Be Careful About

The expression $\tan(90^\circ - x)$ is undefined if $x = 90^\circ$ , but here $x$ is acute, so the expression is well-defined. Ensure the numerator and denominator are not swapped.

Techniques used

use the complementary angle identity tan(90° - x) = cot xexpress cot x as cos x / sin xsubstitute the given values

The rest of this paper

9 more questions

Q2Circular Measure6M
Q3Differentiation · Integration7M
Q4Coordinate Geometry · Trigonometry7M
Q5Coordinate Geometry7M
Q6Coordinate Geometry · Differentiation7M
Q7Series8M
Q8Trigonometry · Functions10M
Q9Differentiation · Integration10M
Q10Quadratics · Functions10M

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