15
cosec x
11 (a) (i) Show that = cos x . [3]
cot x + tan x
cosec 3y
(ii) Hence solve = 0.5 for 0 G y G r radians, giving your answers in
cot 3y + tan 3y
terms of r . [3]
Question 11(b) is printed on the next page.
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16
(b) Solve 2 sin z + 8 cos 2 z = 5 for 0° 1 z 1 360° . [4]
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To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International
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© UCLES 2021 0606/12/F/M/21
14
10 (a) Solve tan `a + 45°j =-
1
for 0° G a G 360° . [3]
2
1 1
(b) (i) Show that - = a sec 2 i , where a is a constant to be found. [3]
sin i - 1 sin i + 1
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15
1 1 r r
(ii) Hence solve - =- 8 for - G z G radians. [5]
sin 3z - 1 sin 3z + 1 3 3
Question 11 is on the next page.
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8
8
f(x)
0 2r 4r 2r 8r x
-1 3 3 3
-2
-3
-4
-5
-6
8r
The diagram shows the graph of f (x) = a cos bx + c for 0 G x G radians.
3
(a) Explain why f is a function. [1]
(b) Write down the range of f. [1]
(c) Find the value of each of the constants a, b and c. [4]
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8
8 (a) Solve 3 cot 2 x - 14 cosec x - 2 = 0 for 0° 1 x 1 360°. [5]
sin 4 y - cos 4 y
(b) Show that = tan y - 2 cos y sin y. [4]
cot y
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11
10 Solve the equation
(a) 5 sec 2 A + 14 tan A - 8 = 0 for 0° G A G 180°, [4]
5 sin b4B - l + 2 = 0 for - G B G
r r r
(b) radians. [4]
8 4 4
© UCLES 2021 0606/23/M/J/21 [Turn over
4
2 (i) On the axes below, sketch the graph of y = 5 cos 4x - 3 for - 90° G x G 90°.
y
6
-90° -45° 0 45° 90° x
-2
-4
-6
-8
- 10
[4]
(ii) Write down the amplitude of y. [1]
(iii) Write down the period of y. [1]
© UCLES 2021 0606/13/O/N/21
(a) Solve 2 sin bx + l = 3
r
1. for 0 1 x 1 r radians. [3]
4
(b) Solve 3 sec y = 4 cosec y for 0° 1 y 1 360° . [3]
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(c) Solve 7 cot z - tan z = 2 cosec z for 0° 1 z 1 360° . [6]
© UCLES 2021 0606/23/O/N/21 [Turn over
10
(a) Solve 2 sin bx + l = 3 for 0 1 x 1 r radians.
r
9 [3]
4
(b) Solve 3 sec y = 4 cosec y for 0° 1 y 1 360° . [3]
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11
(c) Solve 7 cot z - tan z = 2 cosec z for 0° 1 z 1 360° . [6]
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8
1 1
7 (i) Show that - = 2 cosec x cot x . [4]
1 - cos x 1 + cos x
1 1
(ii) Hence solve the equation - = sec x for 0 G x G 2r radians. [4]
1 - cos x 1 + cos x
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1 1
10 (a) (i) Show that - = 2 cot 2 i . [3]
sec i - 1 sec i + 1
1 1
(ii) Hence solve - = 6 for - 90° 1 x 1 90° . [5]
sec 2x - 1 sec 2x + 1
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15
(b) Solve cosec b y + l = 2 for 0 G y G 2r radians, giving your answers in terms of r.
r
[4]
3
Question 11 is printed on the next page.
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