UCL Department of Mathematics 1

UCL DEPARTMENT OF MATHEMATICS

C3 Revision

Trigonometry

UCL Department of Mathematics 2

Your Toolkit

UCL Department of Mathematics 3

UCL Department of Mathematics 4

June 2009 Qn. 2

(a) Use the identity cos2 θ + sin2 θ = 1 to prove that tan2 θ = sec2 θ – 1.

(2)

(b) Solve, for 0 θ < 360°, the equation 2 tan2 θ + 4 sec θ + sec2 θ = 2.

(6)

UCL Department of Mathematics 5

June 2006 Qn. 6

(a) Using sin2 + cos2

1, show that cosec2 – cot2 1.

(2)

(b) Hence, or otherwise, prove that cosec4 – cot4 cosec2 + cot2 .

(2)

(c) Solve, for 90 < < 180, cosec4 – cot4 = 2 – cot .

(6)

UCL Department of Mathematics 6

Jan 2009 Qn 6

(a) (i) By writing 3θ = (2θ + θ), show that sin 3θ = 3 sin θ – 4 sin3 θ.

(4)

(ii) Hence, or otherwise, for 0 < θ < 3

, solve 8 sin3 θ – 6 sin θ + 1 = 0.

Give your answers in terms of π.

(5)

(b) Using sin (θ – ) = sin θ cos – cos θ sin , or otherwise, show that sin 15 = 4

1(6 – 2).

(4)

UCL Department of Mathematics 7

Jan 2009 Qn. 8

(a) Express 3 cos θ + 4 sin θ in the form R cos (θ – α), where R and α are constants, R > 0 and 0 < α < 90°.

(4)

(b) Hence find the maximum value of 3 cos θ + 4 sin θ and the smallest positive value of θ for which this

maximum occurs.

(3)

The temperature, f(t), of a warehouse is modelled using the equation

f (t) = 10 + 3 cos (15t)° + 4 sin (15t)°,

where t is the time in hours from midday and 0 t < 24.

(c) Calculate the minimum temperature of the warehouse as given by this model.

(2)

(d) Find the value of t when this minimum temperature occurs. (3)

UCL Department of Mathematics 8

EXAM QUESTIONS

1. June 2011 question 6

(a) Prove that

2sin

2cos

2sin

1 = tan , 90n, n ℤ.

(4)

(b) Hence, or otherwise,

(i) show that tan 15 = 2 – 3,

(3)

(ii) solve, for 0 < x < 360°,

cosec 4x – cot 4x = 1.

(5)

2. January 2012 question 5

Solve, for 0 180°,

2 cot2 3 = 7 cosec 3 – 5.

Give your answers in degrees to 1 decimal place.

(10)

3. June 2012 question 5

(a) Express 4 cosec2 2θ − cosec2 θ in terms of sin θ and cos θ.

(2)

(b) Hence show that

4 cosec2 2θ − cosec2 θ = sec2 θ .

(4)

(c) Hence or otherwise solve, for 0 < θ < ,

4 cosec2 2θ − cosec2 θ = 4

giving your answers in terms of .

(3)

UCL Department of Mathematics 9

4. June 2011 question 8

(a) Express 2 cos 3x – 3 sin 3x in the form R cos (3x + ), where R and are constants, R > 0 and 0 < < 2

.

Give your answers to 3 significant figures.

(4)

f(x) = e2x cos 3x.

(b) Show that f ′(x) can be written in the form

f ′(x) = Re2x cos (3x + ),

where R and are the constants found in part (a).

(5)

(c) Hence, or otherwise, find the smallest positive value of x for which the curve with equation y = f(x) has a

turning point.

(3)

5. June 2010 question 7

UCL Department of Mathematics 10

6. June 2012 question 8

f(x) = 7 cos 2x − 24 sin 2x.

Given that f(x) = R cos (2x + α), where R > 0 and 0 < α < 90,

(a) find the value of R and the value of α.

(3)

(b) Hence solve the equation

7 cos 2x − 24 sin 2x = 12.5

for 0 x < 180, giving your answers to 1 decimal place.

(5)

(c) Express 14 cos2 x − 48 sin x cos x in the form a cos 2x + b sin 2x + c, where a, b, and c are constants to be

found.

(2)

(d) Hence, using your answers to parts (a) and (c), deduce the maximum value of

14 cos2 x − 48 sin x cos x.

(2)

7. January 2010 question 8

Solve

cosec2 2x – cot 2x = 1

for 0 x 180.

(7)

UCL Department of Mathematics 11

8. June 2009 question 6

(a) Use the identity cos (A + B) = cos A cos B – sin A sin B, to show that

cos 2A = 1 − 2 sin2 A

(2)

The curves C1 and C2 have equations

C1: y = 3 sin 2x

C2: y = 4 sin2 x − 2 cos 2x

(b) Show that the x-coordinates of the points where C1 and C2 intersect satisfy the equation

4 cos 2x + 3 sin 2x = 2

(3)

(c) Express 4cos 2x + 3 sin 2x in the form R cos (2x – α), where R > 0 and 0 < α < 90°, giving the value of α

to 2 decimal places.

(3)

(d) Hence find, for 0 x < 180°, all the solutions of

4 cos 2x + 3 sin 2x = 2,

giving your answers to 1 decimal place.

(4)