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9709/11

Mathematics 9709/11 — May/June 2015

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

10
questions
75
marks
105
minutes

Topics Coordinate Geometry · Differentiation · Quadratics · Trigonometry · Series · Circular Measure · +2 more

Q1TrigonometryFree sample

Given that θ\theta is an obtuse angle measured in radians and that sin⁡θ=k\sin \theta = k, find, in terms of kk, an expression for

(i)

cos⁡θ\cos \theta,

1M
DifficultyEasy
Worked solution

Approach

Since θ\theta is obtuse, it lies in the second quadrant, where cos⁡θ<0\cos \theta < 0. Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 with sin⁡θ=k\sin \theta = k to find cos⁡θ\cos \theta.

Working

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

Substitute sin⁡θ=k\sin \theta = k:

k2+cos⁡2θ=1k^2 + \cos^2 \theta = 1 cos⁡2θ=1−k2\cos^2 \theta = 1 - k^2 cos⁡θ=±1−k2\cos \theta = \pm \sqrt{1 - k^2}

Since θ\theta is obtuse, cos⁡θ<0\cos \theta < 0, so:

cos⁡θ=−1−k2\cos \theta = -\sqrt{1 - k^2}

Answer

cos⁡θ=−1−k2\cos \theta = -\sqrt{1 - k^2}
Final answer

cos theta = -sqrt(1 - k^2)

Detailed explanation

Walkthrough

We know sin⁡θ=k\sin \theta = k and θ\theta is obtuse. An obtuse angle lies between 90∘90^\circ and 180∘180^\circ, so it is in the second quadrant. In the second quadrant, sine is positive and cosine is negative. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 relates sine and cosine. Substitute kk for sin⁡θ\sin \theta, rearrange to get cos⁡2θ=1−k2\cos^2 \theta = 1 - k^2, then take square roots. The square root gives two signs, but because θ\theta is obtuse we choose the negative sign. Hence cos⁡θ=−1−k2\cos \theta = -\sqrt{1 - k^2}.

Key Takeaways

The identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 lets you find one trigonometric ratio from another. The quadrant determines the sign of the result.

Common Mistakes

Forgetting the negative sign and writing cos⁡θ=1−k2\cos \theta = \sqrt{1 - k^2}. Also forgetting to take the square root of 1−k21 - k^2.

Things to Be Careful About

"Obtuse" means the angle is in the second quadrant, so cosine is negative. The answer must be in terms of kk only, with no θ\theta remaining. The mark scheme states this is cao.

Techniques used
apply the Pythagorean identitychoose the sign according to the quadrant
(ii)

tan⁡θ\tan \theta,

2M
DifficultyMedium-Easy
Worked solution

Approach

Use the identity tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, substituting sin⁡θ=k\sin \theta = k and the value of cos⁡θ\cos \theta found in part (i).

Working

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

Substitute sin⁡θ=k\sin \theta = k and cos⁡θ=−1−k2\cos \theta = -\sqrt{1 - k^2}:

tan⁡θ=k−1−k2\tan \theta = \frac{k}{-\sqrt{1 - k^2}} tan⁡θ=−k1−k2\tan \theta = -\frac{k}{\sqrt{1 - k^2}}

Answer

tan⁡θ=−k1−k2\tan \theta = -\frac{k}{\sqrt{1 - k^2}}
Final answer

tan theta = -k/sqrt(1 - k^2)

Detailed explanation

Walkthrough

The tangent identity is tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}. We already know sin⁡θ=k\sin \theta = k, and from part (i) cos⁡θ=−1−k2\cos \theta = -\sqrt{1 - k^2}. Substitute both into the identity. Dividing by a negative square root gives a negative result, so tan⁡θ=−k1−k2\tan \theta = -\frac{k}{\sqrt{1 - k^2}}. This is the exact expression in terms of kk.

Key Takeaways

Tangent is the ratio of sine to cosine. Once sine and cosine are known, tangent follows immediately. The signs must be consistent with the quadrant.

Common Mistakes

Using cos⁡θ=+1−k2\cos \theta = +\sqrt{1 - k^2}, which would give the wrong sign. Also incorrectly simplifying k−1−k2\frac{k}{-\sqrt{1 - k^2}} to k1−k2\frac{k}{\sqrt{1 - k^2}}.

Things to Be Careful About

The mark scheme awards M1 for using tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and A1FT for the correct simplified expression following the candidate's cosine. The final answer should have the negative sign in front of the fraction.

Techniques used
use the tangent ratio identitysubstitute the expression for cosinesimplify the quotient
(iii)

sin⁡(θ+π)\sin(\theta + \pi).

1M
DifficultyEasy
Worked solution

Approach

Use the identity sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin \theta, which follows from the sine graph or the unit circle, and substitute sin⁡θ=k\sin \theta = k.

Working

sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin \theta

Since sin⁡θ=k\sin \theta = k:

sin⁡(θ+π)=−k\sin(\theta + \pi) = -k

Answer

sin⁡(θ+π)=−k\sin(\theta + \pi) = -k
Final answer

sin(theta + pi) = -k

Detailed explanation

Walkthrough

The identity sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin \theta comes from the unit circle: adding π\pi radians rotates the point by 180∘180^\circ, changing the yy-coordinate from sin⁡θ\sin \theta to −sin⁡θ-\sin \theta. Since sin⁡θ=k\sin \theta = k, the result is −k-k. We do not need to know the value of θ\theta itself.

Key Takeaways

The sine function satisfies sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin \theta. This is a useful symmetry for simplifying angles shifted by π\pi.

Common Mistakes

Writing sin⁡(θ+π)=sin⁡θ\sin(\theta + \pi) = \sin \theta or incorrectly expanding as sin⁡θ+sin⁡π\sin \theta + \sin \pi. The identity sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A + B) \neq \sin A + \sin B.

Things to Be Careful About

The angle is measured in radians, but the identity holds for any angle. The answer is simply −k-k, and the mark scheme says cao.

Techniques used
apply the sine phase shift identitysubstitute the given value of sine

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