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Trigonometry
a. Trigonometric functions: A function involving any trigonometric ratio (sin π‘₯ , cos π‘₯ , tan π‘₯ . ) is
said to be trigonometric function.
sin π‘₯
b. Basic Relation: 𝐬𝐒𝐧𝟐 𝒙 +𝐜𝐨𝐬 𝟐 𝒙 = 𝟏 and tan π‘₯ = cos π‘₯
c. Graphs of Trigonometric functions: sin π‘₯ , cos π‘₯ and tan π‘₯
d. Properties of graphs of trigonometric functions: Axis of graph, amplitude, period of graph,
minimum and maximum values of function.
Q.1
Sketch the graphs of following functions.
i.
2 + 3 sin 3π‘₯
iv.
βˆ’1 + 3 sin 2π‘₯
ii.
4 βˆ’ 3 cos2 π‘₯
v.
2 + cos π‘₯
iii.
2 tan 2π‘₯ βˆ’ 1
vi.
3 + 2 tan 2π‘₯
a. Trigonometric Equation: An equation involving a trigonometric ratio. A set of values of a
variable which satisfy the given equation is called solution.
b. Basic concepts of angle: Clockwise and anticlockwise angles. General and basic angles. Relation
of basic angle and general angle in different quadrants. Method to find basic angle from
trigonometric ratio. Sign of trigonometric ratios in quadrants.
c. Solution of simple linear trigonometric equation:
i.
Shift the trigonometric ratio on LHS of equation and find the sign of quadratic ratio.
ii.
Find the basic angle by using inverse trigonometric function.
iii.
Make the graph and find the general angles between given range.
iv.
If equation is given into two trigonometric ratios then change into one ratio.
v.
If range of the angle of equation is multiple then add 360 into general angles.
vi.
If the angle is sum, difference or product of constant number then whole angle is supposed to
be a single variable and write the range into new variable.
Q.2
Find all possible values of angle lying in given range.
i.
1 + 2 sin π‘₯ = 0, 0 ≀ π‘₯ ≀ 360
ii.
cos π‘₯ = βˆ’0.8 , 180 ≀ π‘₯ ≀ 360
iii.
tan x = 4, 180 ≀ π‘₯ ≀ 360
iv.
4 sin π‘₯ = 5 cos π‘₯, 0 ≀ π‘₯ ≀ 360
Q.3
Find all the solution in given interval π‘œ f each of the following equations.
1
i.
cos 2πœƒ = ,
0 ≀ πœƒ ≀ 360
3
ii.
sin 3πœƒ = βˆ’0.4 , 0 ≀ πœƒ ≀ 360
iii.
tan 3πœƒ = 2 ,
0 ≀ πœƒ ≀ 360
1
iv.
sin(2𝑑 βˆ’ 30°) = 2 , βˆ’180 ≀ 𝑑 ≀ 180
v.
tan(3𝑑 βˆ’ 180°) = βˆ’1, βˆ’180 ≀ 𝑑 ≀ 180
1
vi.
cos(2𝑑 βˆ’ 50°) = βˆ’ 2, βˆ’180 ≀ 𝑑 ≀ 180
a. Solution of Quadratic Trigonometric Equation:
i.
Change whole equation into one trigonometric ratio. Make factors of possible.
ii.
Change given trigonometric equation into algebraic quadratic equation and solve.
iii.
Write two linear trigonometric equations by using supposition and write answer into given
range.
Q.4
Find all possible angles between 0 and 360 which satisfy the following equations.
i.
4 sin π‘₯ cos π‘₯ = cos π‘₯
iv.
2 βˆ’ 2 cos 2 π‘₯ = sin π‘₯
2
ii.
2 cos π‘₯ βˆ’ cos π‘₯ = 1
v.
tan2 π‘₯ βˆ’ tan π‘₯ = 1
iii.
2 sin π‘₯ cos π‘₯ = tan π‘₯
vi.
4 sin2 π‘₯ βˆ’ 1 = 0
Compiled By : Sir Rashid Qureshi
www.levels.org.pk
1
vii.
b.
c.
i.
ii.
Q.5
10 sin2 π‘₯ βˆ’ 5 cos 2 π‘₯ + 2 = 4sin π‘₯
viii.
4 sin2 π‘₯ cos π‘₯ = tan2 π‘₯
Trigonometric Identity: If an equation is true for all values of variables then it is an identity.
Proof of identities:
Take LHS ( you can take RHS)
iii.
Use concept of rationalizing
Use basic identities to make RHS
denominator if possible.
Prove the following identities.
i.
ii.
iii.
iv.
v.
vi.
1
sin πœƒ
1
βˆ’ tan πœƒ ≑
sin2 πœƒ
1βˆ’cos πœƒ
1
cos πœƒ
1βˆ’ cos πœƒ
.
sin πœƒ
≑ 1 + cos πœƒ
+ tan πœƒ ≑
tan πœƒ sin πœƒ
1βˆ’cos πœƒ
1βˆ’sin πœƒ
cos πœƒ
cos πœƒ
1βˆ’sin πœƒ
1
= 1 + cos πœƒ
≑
1βˆ’2 sin2 πœƒ
cos πœƒ+sin πœƒ
cos πœƒ
1+sin πœƒ
= cos πœƒ βˆ’ sin πœƒ
d. To find the least/ greatest value of trigonometric function.
Use the following inequalities to find the range of trigonometric functions.
ο‚· βˆ’1 ≀ sin π‘₯ ≀ 1
ο‚· 0 ≀ sin2 π‘₯ ≀ 1
ο‚· βˆ’1 ≀ cos2 π‘₯ ≀ 1
ο‚· 0 ≀ cos 2 π‘₯ ≀ 1
e. Radian measure: Another unit of measuring angle. It is usually expressed in terms of πœ‹.
f. To find the values of unknown constants in trigonometric functions:
i.
Make two equations of constants.
ii.
Solve the equations simultaneously.
Q.6
Find the range of the following trigonometric Functions.
i.
ii.
iii.
iv.
Q.7
Q.8
Q.9
Q.10
i.
ii.
𝑓(π‘₯) = 3 βˆ’ 4 sin π‘₯
𝑔(π‘₯) = 4 cos π‘₯ βˆ’ 2
β„Ž(π‘₯) = 1 βˆ’ 2 sin2 (4π‘₯)
𝑓(π‘₯) = 3 + 4 cos 2 3π‘₯
𝑓: π‘₯ ⟼ π‘Ž + 𝑏 sin π‘₯ and 𝑓(90°) = 2, 𝑓(180°) = 3 then find the values of π‘Ž and 𝑏.
𝑔: π‘₯ ⟼ π‘š + 𝑛 cos 2π‘₯ and 𝑔(90°) = 2, 𝑓(180°) = 3 then find the values of π‘šand 𝑛.
Past Paper Questions
Solve the equation 3 tan(2π‘₯ + 15°) = 4 for 0° ≀ π‘₯ ≀ 360°.
The function 𝑓 is defined by 𝑓: π‘₯ ⟼ 5 βˆ’ 3 sin 2π‘₯ for 0 ≀ π‘₯ ≀ πœ‹.
Find the range of 𝑓.
Sketch the graph of 𝑦 = 𝑓(π‘₯).
Compiled By : Sir Rashid Qureshi
(2010)
(J2010)
www.levels.org.pk
2
Q.11
Q.12
Show that the equation 3(2 sin π‘₯ βˆ’ cos π‘₯) = 2(sin π‘₯ βˆ’ 3 cos π‘₯) can be written in the form of
3
tan π‘₯ = βˆ’ . Hence solve, 3(2 sin π‘₯ βˆ’ cos π‘₯) = 2(sin π‘₯ βˆ’ 3 cos π‘₯) for 0° ≀ π‘₯ ≀ 360°.
4
The function 𝑓: π‘₯ ⟼ π‘Ž + 𝑏 cos π‘₯ is defined for0 ≀ π‘₯ ≀ 2πœ‹.Given that 𝑓(0) = 10 and
2
i.
𝑓 (3 πœ‹) = 1, find the values of π‘Ž and 𝑏.
Find range of 𝑓.
ii.
The exact value of 𝑓 ( πœ‹).
Q.13
Q.14
Q.15
Q.16
5
6
(N2009)
sin π‘₯
sin π‘₯
βˆ’ 1+sin π‘₯ ≑ 2 tan2 π‘₯
(J2009)
+ 1+sin π‘₯ ≑ cos π‘₯.
(N2008)
Show that the equation 2 tan πœƒ cos πœƒ = 3 can be written in the form of
2 cos2 πœƒ + 3 cos πœƒ βˆ’ 2 = 0. Hence solve, 2 tan2 πœƒ cos πœƒ = 3 for 0° ≀ π‘₯ ≀ 360°.
(J2008)
Prove the identity
Prove the identity
1βˆ’sin π‘₯
1+sin π‘₯
cos π‘₯
2
Prove the identity.
1βˆ’tan2 π‘₯
1+tan2 π‘₯
cos π‘₯
2
2
≑ 1 βˆ’ 2 sin π‘₯
Answers:
1. Graphs
2.
i.
ii.
iii.
iv.
3.
i.
ii.
iii.
iv.
v.
vi.
4.
i.
ii.
iii.
iv.
v.
vi.
5.
6.
i.
ii.
iii.
iv.
7.
8.
9.
10.
11.
12.
15.
Compiled By : Sir Rashid Qureshi
www.levels.org.pk
3
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