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MATH IN FOCUS WITH RAJA NADEEM AMIN 03215862394
O & A Level campus
Trigonometric Functions
Q1
(N 2004/P1)
The function 𝑓(π‘₯) = 5sin2 π‘₯ + 3 cos2 π‘₯ is defined for the domain 0 ≤ π‘₯ ≤ πœ‹.
(i)
Express 𝑓(π‘₯) in the form π‘Ž + 𝑏 sin2 π‘₯, stating the values of ‘a’ and ‘b’.
(ii)
Hence find the values of x for which 𝑓(π‘₯) = 7𝑠𝑖𝑛π‘₯.
(iii)
State the range of 𝑓(π‘₯).
Q2
(J2005/P1/Q7)
A function f is defined by (π‘₯) = 3 − 2𝑠𝑖𝑛π‘₯ , for 00 ≤ x ≤3600 .
(i)
Find the range of 𝑓.
(ii)
Sketch the graph of 𝑦 = 𝑓(π‘₯)
A function g is defined by 𝑔(π‘₯) = 3 – 2𝑠𝑖𝑛π‘₯, for 00 ≤ x ≤A0 , where A is a constant.
(iii)
State the largest value of A for which g has an inverse.
(iv)
When A has this value, obtain an expression, in terms of x, for g-1(x).
Q3.
(J2006/P1/Q2)
[1]
[2]
𝑠𝑖𝑛 2π‘₯ + 3 π‘π‘œπ‘  2π‘₯ = 0,
Solve the equation
for 00 ≤ π‘₯ ≤ 1800 .
Q4.
[2]
[2]
[4]
(J2007/P1/Q8)
The function f is defined by 𝑓(π‘₯) = π‘Ž + π‘π‘π‘œπ‘ 2π‘₯, for 00 ≤ x ≤ π. It is given that
πœ‹
𝑓(0) = −1 and 𝑓 ( 2 ) = 7
(i)
(ii)
(iii)
Q5.
Find the values of a and b.
Find the x coordinates of the points where the curve 𝑦 = 𝑓(π‘₯) intersects the x-axis.
Sketch the graph of 𝑦 = 𝑓(π‘₯).
(J2007/P1/Q3)
1−tan2 π‘₯
Prove the identity 1+tan2 π‘₯ ≡ 1 − 2 sin2 π‘₯.
[3]
[3]
[2]
[4]
MATH IN FOCUS WITH RAJA NADEEM AMIN 03215862394
Q6.(j2008/P1/Q2)
(i) Show that the equation 2 tan2 πœƒ π‘π‘œπ‘  πœƒ = 3 can be written in the form
Q7.
2 cos 2 πœƒ + 3 π‘π‘œπ‘  πœƒ − 2 = 0.
[2]
(ii) Hence solve the equation 2 tan2 πœƒ π‘π‘œπ‘  πœƒ = 3, for 00 ≤ πœƒ ≤ 3600 .
[3]
(J2009/P1/Q4)
The diagram shows the graph of
𝑦 = π‘Ž 𝑠𝑖𝑛(𝑏π‘₯) + 𝑐 for 0 ≤ π‘₯ ≤ 2πœ‹.
(i) Find the values of a, b and c.
(ii) Find the smallest value of x in the interval 0 ≤ π‘₯ ≤ 2πœ‹ο€ for which 𝑦 = 0.
Q8.
[3]
[3]
(J2009/P1/Q1)
𝑠𝑖𝑛π‘₯
𝑠𝑖𝑛π‘₯
Prove the identity 1−𝑠𝑖𝑛π‘₯ − 1+𝑠𝑖𝑛π‘₯ ≡ 2 tan2 π‘₯
Q9.
[3]
(J2010/P1/Q1)
The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k,
(i) π‘‘π‘Žπ‘›(πœ‹ − π‘₯),
[1]
πœ‹
Q10.
(ii) π‘‘π‘Žπ‘› ( 2 – π‘₯),
[1]
(iii) sin x.
[2]
(J2010/P1/Q5)
The function 𝑓 is such that 𝑓(π‘₯) = 2 sin2 π‘₯ − 3 π‘π‘œπ‘ 2π‘₯ for 0 ≤ π‘₯ ≤ πœ‹.
(i)
(ii)
(iii)
Express 𝑓(π‘₯) in the form π‘Ž + 𝑏 cos 2 π‘₯, stating the values of a and b.
State the greatest and least values of 𝑓(π‘₯).
Solve the equation (π‘₯) + 1 = 0 .
[2]
[2]
[3]
MATH IN FOCUS WITH RAJA NADEEM AMIN 03215862394
Q11.
(J2011/P1/Q5)
π‘π‘œπ‘ πœƒ
(i)
Prove the identity π‘‘π‘Žπ‘›πœƒ(
(ii)
Hence solve the equation π‘‘π‘Žπ‘›πœƒ(
Q12.
1
≡ 1 + π‘ π‘–π‘›πœƒ
1−π‘ π‘–π‘›πœƒ )
π‘π‘œπ‘ πœƒ
1−π‘ π‘–π‘›πœƒ )
[3]
= 4 for 00 ≤ πœƒ ≤ 3600
(J2012/P11/Q1)
Solve the equation 𝑠𝑖𝑛 2π‘₯ = 2 π‘π‘œπ‘  2π‘₯, for 00 ≤ π‘₯ ≤ 1800 .
Q13.
[3]
[4]
(J2012/P11/Q7)
(a) The first two terms of an arithmetic progression are 1 and cos 2 π‘₯ respectively. Show that the sum of
the first ten terms can be expressed in the form π‘Ž – 𝑏𝑠𝑖𝑛2 π‘₯, where a and b are constants to be found.
[3]
1
(b) The first two terms of a geometric progression are 1 and 3 tan2 πœƒ respectively, where
πœ‹
0 < πœƒ < 2.
Q14.
(i)
Find the set of values of πœƒ for which the progression is convergent.
(ii)
Find the exact value of the sum to infinity when πœƒ = 6 .
πœ‹
[2]
[2]
(J2012/P12/Q5)
(i) Prove the identity
[2]
1
1
π‘‘π‘Žπ‘›π‘₯ +
≡
π‘‘π‘Žπ‘›π‘₯ 𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯
(ii)
Solve the equation
for 0 ≤ π‘₯ ≤ 1800 .
Q15.
(i)
(ii)
(iii)
1
= 1 + 3π‘‘π‘Žπ‘›π‘₯
𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯
[4]
The function f is such that 𝑓(π‘₯) = π‘Ž – 𝑏 π‘π‘œπ‘ π‘₯ for 00 ≤ x ≤3600, where a and b are positive constants.
The maximum value of 𝑓(π‘₯) is 10 and the minimum value is -2.
Find the value of a and b.
Solve the equation 𝑓(π‘₯) = 0.
Sketch the graph of 𝑦 = 𝑓(π‘₯).
[3]
[3]
[2]
MATH IN FOCUS WITH RAJA NADEEM AMIN 03215862394
Q16.
(J2014/P11/Q1)
The diagram shows part of the graph of y = a + bsin x. State the values of the constants a and b. [2]
Q17.
(J2015/P11/8)
π‘₯
The function 𝑓: π‘₯ → 5 + 3 cos (2) is defined for 0 ≤ π‘₯ ≤ 2πœ‹
(i)
(ii)
(iii)
(iv)
Q18.
Solve the equation𝑓(π‘₯) = 7, giving your answer correct to 2 decimal places.
Sketch the graph of 𝑦 = 𝑓(π‘₯).
Explain why f has an inverse.
Obtain an expression for 𝑓 −1 (π‘₯).
[3]
[2]
[1]
[3]
(J2016/P11/Q11)
πœ‹
πœ‹
The function f is defined by 𝑓 ∢ π‘₯ → 4 𝑠𝑖𝑛 π‘₯ − 1 for – 2 ≤ π‘₯ ≤ 2 .
(i)
(ii)
(iii)
(iv)
Q19.
State the range of f.
[2]
Find the coordinates of the points at which the curve 𝑦 = 𝑓(π‘₯) intersects the coordinate axes.[3]
Sketch the graph of 𝑦 = 𝑓(π‘₯).
[2]
−1
−1
Obtain an expression for 𝑓 (π‘₯), stating both the domain and range of 𝑓 .
[4]
(J2017/P11/Q3)
(a)
Prove the identity
1+π‘π‘œπ‘ πœƒ
π‘ π‘–π‘›πœƒ
(b) Hence solve the equation
π‘ π‘–π‘›πœƒ
2
+ 1+π‘π‘œπ‘ πœƒ ≡ π‘ π‘–π‘›πœƒ
1+π‘π‘œπ‘ πœƒ
π‘ π‘–π‘›πœƒ
π‘ π‘–π‘›πœƒ
[3]
3
+ 1+π‘π‘œπ‘ πœƒ = π‘ π‘–π‘›πœƒ for 0 ≤ πœƒ ≤ 360
[3]
MATH IN FOCUS WITH RAJA NADEEM AMIN 03215862394
Q20.
(J2017/P11/Q3)
The equation of a curve is 𝑦 = 2 π‘π‘œπ‘  π‘₯.
(i) Sketch the graph of 𝑦 = 2 π‘π‘œπ‘  π‘₯ for −πœ‹ ≤ π‘₯ ≤ πœ‹, stating the coordinates of the point of
intersection with the y-axis.
[2]
πœ‹
Points P and Q lie on the curve and have x-coordinates of 3 and πœ‹ respectively.
(ii) Find the length of 𝑃𝑄 correct to 1 decimal place.
[2]
The line through 𝑃 and 𝑄 meets the π‘₯ − π‘Žπ‘₯𝑖𝑠 at 𝐻(β„Ž, 0) and the 𝑦 − π‘Žπ‘₯𝑖𝑠 at 𝐾(0, π‘˜).
5πœ‹
(iii) Show that β„Ž = 9 and find the value of π‘˜.
Q21.
[3]
(J2017/P12/Q3)
(i)
Prove the identity
2
1
1−π‘ π‘–π‘›πœƒ
(π‘π‘œπ‘ πœƒ − π‘‘π‘Žπ‘› πœƒ) ≡ 1+π‘ π‘–π‘›πœƒ
(ii)
[3]
Hence solve the equation
2
1
1
(π‘π‘œπ‘ πœƒ − π‘‘π‘Žπ‘› πœƒ) = 2 for 00 ≤ πœƒ ≤ 3600
Q22.
Q23.
[3]
(J2017/P13/Q5)
2π‘ π‘–π‘›πœƒ+π‘π‘œπ‘ πœƒ
(i)
Show that the equation π‘ π‘–π‘›πœƒ+π‘π‘œπ‘ πœƒ = 2π‘‘π‘Žπ‘›πœƒ may be expressed as cos 2 πœƒ = 2 sin2 πœƒ
(ii)
(ii) Hence solve the equation π‘ π‘–π‘›πœƒ+π‘π‘œπ‘ πœƒ = 2π‘‘π‘Žπ‘›πœƒ for 00 < πœƒ < 1800 .
2π‘ π‘–π‘›πœƒ+π‘π‘œπ‘ πœƒ
Prove the identity
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
1 − cos 2 π‘₯ ≡ sin2 π‘₯
(1 − sin2 πœƒ)π‘‘π‘Žπ‘›πœƒ ≡ π‘π‘œπ‘ πœƒπ‘ π‘–π‘›πœƒ
1
sin2 πœƒ
cos2 πœƒ
− sin2 πœƒ ≡ 1
1
tan2 πœƒ ≡ cos2 πœƒ − 1
sin2 πœƒ −3 cos2 πœƒ+1
sin2 πœƒ−cos2 πœƒ
1
1
cos2 πœƒ
≡2
1
+ sin2 πœƒ ≡ cos2 πœƒ sin2 πœƒ
π‘π‘œπ‘ πœƒ
1
π‘‘π‘Žπ‘›πœƒ + π‘ π‘–π‘›πœƒ ≡ π‘ π‘–π‘›πœƒπ‘π‘œπ‘ πœƒ
1
1+π‘ π‘–π‘›πœƒ
1
2
+ 1−π‘ π‘–π‘›πœƒ ≡ cos2 πœƒ
[3]
[3]
MATH IN FOCUS WITH RAJA NADEEM AMIN 03215862394
1−tan2 π‘₯
(ix)
Prove the identity 1+tan2 π‘₯ ≡ 1 − 2 sin2 π‘₯
(x)
Prove the identity π‘π‘œπ‘ π‘₯ + 1+𝑠𝑖𝑛π‘₯ ≡ π‘π‘œπ‘ π‘₯
(xi)
Prove the identity 1−𝑠𝑖𝑛π‘₯ − 1+𝑠𝑖𝑛π‘₯ ≡ 2 tan2 π‘₯
(xii)
Prove the identity
1+𝑠𝑖𝑛π‘₯
π‘π‘œπ‘ π‘₯
𝑠𝑖𝑛π‘₯
𝑠𝑖𝑛π‘₯
2
(
1+π‘π‘œπ‘ πœƒ
(xiii)
Prove the identity
(xiv)
Prove the identity π‘‘π‘Žπ‘›πœƒ(
+
π‘ π‘–π‘›πœƒ
π‘ π‘–π‘›πœƒ
1+π‘π‘œπ‘ πœƒ
π‘π‘œπ‘ πœƒ
1−π‘ π‘–π‘›πœƒ )
≡
2
1
1 − π‘ π‘–π‘›πœƒ
− π‘‘π‘Žπ‘› πœƒ) ≡
π‘π‘œπ‘ πœƒ
1 + π‘ π‘–π‘›πœƒ
2
π‘ π‘–π‘›πœƒ
1
≡ 1 + π‘ π‘–π‘›πœƒ
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