Download GCSE-Trigonometry3-4GraphsAndEquations

GCSE Trigonometry Parts 3 and 4 –
Trigonometric Graphs and Equations
Dr J Frost ([email protected])
Objectives: (from the IGCSE FM specification)
Last modified: 28th December 2016
Sin Graph
What does it look like?
-360
-270
-180
-90
?
90
180
270
360
Sin Graph
What do the following graphs look like?
-360
-270
-180
-90
90
180
270
Suppose we know that sin(30) = 0.5. By thinking about symmetry in the graph,
how could we work out:
sin(150) = 0.5?
sin(-30) = -0.5?
sin(210) = -0.5
?
360
Cos Graph
What do the following graphs look like?
-360
-270
-180
-90
?
90
180
270
360
Cos Graph
What does it look like?
-360
-270
-180
-90
90
180
270
Suppose we know that cos(60) = 0.5. By thinking about symmetry in the graph,
how could we work out:
cos(120) = -0.5
?
cos(-60) = 0.5?
cos(240) = -0.5
?
360
Tan Graph
What does it look like?
-360
-270
-180
-90
?
90
180
270
360
Tan Graph
What does it look like?
-360
-270
-180
-90
90
180
270
Suppose we know that tan(30) = 1/√3. By thinking about symmetry in the
graph, how could we work out:
tan(-30) = -1/√3
?
tan(150) = -1/√3
?
360
Solving Trig Equations
Solve sin 𝑥 = 0.6 in the range 0 ≤ 𝑥 < 360
𝒙 = 𝐬𝐢𝐧−𝟏 𝟎. 𝟔 =?𝟑𝟔. 𝟖𝟕°, 𝟏𝟒𝟑.?𝟏𝟑°
Angle Law #1:
𝐬𝐢𝐧 𝒙 = 𝐬𝐢𝐧(𝟏𝟖𝟎 − 𝒙)
0.6
-360
-270
-180
-90
?
𝟑𝟔. 𝟖𝟕°
90
180
270
360
Solving Trig Equations
Solve 3cos 𝑥 = 2 in the range 0 ≤ 𝑥 < 360
2
cos 𝑥 =?
3
𝟐
−𝟏
𝒙 = 𝒄𝒐𝒔
= ?𝟒𝟖. 𝟏𝟗°, 𝟑𝟏𝟏.?𝟖𝟏°
𝟑
Angle Law #2:
𝒄𝒐𝒔 𝒙 = 𝒄𝒐𝒔(𝟑𝟔𝟎 − 𝒙)
𝟐
𝟑
-360
-270
-180
-90
𝟒𝟖. 𝟏𝟗°
90
?
180
270
360
Solving Trig Equations
Solve sin 𝑥 = −0.3 in the range 0 ≤ 𝑥 < 360
𝒙 = 𝐬𝐢𝐧−𝟏 −𝟎. 𝟑 = −𝟏𝟕.
? 𝟒𝟔°, 𝟏𝟗𝟕. 𝟒𝟔°,?𝟑𝟒𝟐. 𝟓𝟒
Angle Law #3:
Sin and cos repeat every 360°
−𝟏𝟕. 𝟒𝟔°
-360
-270
-180
-90
90
-0.3
?
180
?
270
360
Laws of Trigonometric Functions
!
sin 𝑥 = sin 180?− 𝑥
cos 𝑥 = cos 360?− 𝑥
?
sin and cos repeat every 360°
?
tan repeats every 180°
Solutions generally come in pairs (e.g. 30°
and 150° for sin) for each ‘cycle’ of sin/cos.
We can therefore get the next pair of
solutions by going to the next cycle.
Test Your Understanding
Solve cos 𝑥 = 0.9 in the range 0 ≤ 𝑥 < 360
𝑥 = cos −1 0.9 = 𝟐𝟓. 𝟖𝟒°
?
360 − 25.84° = 𝟑𝟑𝟒. 𝟏𝟔°
Solve tan 𝑥 = 1 in the range 0 ≤ 𝑥 < 360
𝑥 = tan−1 1 = 𝟒𝟓°
?
45° + 180° = 𝟐𝟐𝟓°
Set 4 Paper 2 Q14
sin 𝜃 = −0.68 → 𝜃 = −42.84°
? = 𝟐𝟐𝟐. 𝟖𝟒°
180 − −42.84
−42.84 + 360 = 𝟑𝟏𝟕. 𝟏𝟔°
Exercise 1
1
Solve the following in the range 0 ≤ 𝑥 ≤ 360
a
sin 𝑥 = 0.5 → 𝒙 = 𝟑𝟎°,?𝟏𝟓𝟎°
3
b
cos 𝑥 =
→ 𝒙 = 𝟑𝟎°, 𝟑𝟑𝟎°
?
2
c tan 𝑥 = 3
?
→ 𝒙 = 𝟔𝟎°, 𝟐𝟒𝟎°
→ 𝒙 = 𝟓. 𝟕𝟒°, 𝟏𝟕𝟒.
d sin 𝑥 = 0.1
? 𝟐𝟔°
e
4 cos 𝑥 = 3 → 𝟒𝟏. 𝟒𝟏°,?𝟑𝟏𝟖. 𝟓𝟗°
6 tan 𝑥 = 5 → 𝟑𝟗. 𝟖𝟏°,?𝟐𝟏𝟗. 𝟖𝟏°
f
sin 𝑥 = 0.4 → 𝟐𝟑. 𝟓𝟖°,?𝟏𝟓𝟔. 𝟒𝟐°
g
2
Solve the following in the range 0 ≤ 𝑥 ≤ 360
a sin 𝜃 = −0.5 → 𝟐𝟏𝟎°, 𝟑𝟑𝟎°
?
b cos 𝜃 = −0.5 → 𝟏𝟐𝟎°, 𝟐𝟒𝟎°
?
c tan 𝜃 = −1
→ 𝟏𝟑𝟓°, 𝟑𝟏𝟓°
?
d sin 𝜃 = −0.4 → 𝟐𝟎𝟑. 𝟓𝟖°,?𝟑𝟑𝟔. 𝟒𝟐°
e cos 𝜃 = −0.7 → 𝟏𝟑𝟒. 𝟒°, 𝟐𝟐𝟓. 𝟔°
?
f tan 𝜃 = −0.2 → 𝟏𝟔𝟖. 𝟔𝟗°,?𝟑𝟒𝟖. 𝟔𝟗°
Trigonometric Identities
1
Using basic trigonometry to find
these two missing sides…
sin?𝜃

cos?
1
2
Then 𝒕𝒂𝒏 𝜽 =
𝒔𝒊𝒏 𝜽
?
𝒄𝒐𝒔 𝜽
Pythagoras gives
you...
These two identities are all
you will need for IGCSE FM.
𝒔𝒊𝒏𝟐 𝜽 + 𝒄𝒐𝒔𝟐 𝜽 = 𝟏
?
sin2 𝜃 is a shorthand for sin 𝜃 2 . It does NOT mean
the sin is being squared – this does not make sense
as sin is not a quantity that we can square!
Application #1: Solving Harder Trig Equations
Solve sin 𝑥 = 2 cos 𝑥 in the range 0 ≤ 𝑥 < 360°
The problem here is that we have two different trig functions. Is there
anything we could divide by to get just one trig function?
sin 𝑥 2 cos 𝑥
=?
cos 𝑥
cos 𝑥
tan 𝑥? = 2
𝑥 = 63.43°,
? 243.43°
Bro Tip: In general, when you have a
mixture of sin and cos, divide
everything by cos.
1.
2.
3.
4.
sin 𝑥 = sin 180 − 𝑥
cos 𝑥 = cos 360 − 𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
sin 𝑥
A. tan 𝑥 = cos 𝑥
B. sin2 𝑥 + cos2 𝑥 = 1
Test Your Understanding
Solve 2 s𝑖𝑛 𝑥 = cos 𝑥 in the range 0 ≤ 𝑥 < 360°
2 tan 𝑥 = 1
1
tan 𝑥 = ?
2
𝑥 = 26.57°, 206.57°
Solve cos 𝑥 = sin 𝑥 in the range 0 ≤ 𝑥 < 360°
1 = tan 𝑥
𝑥 = 45°, 225°
?
1.
2.
3.
4.
sin 𝑥 = sin 180 − 𝑥
cos 𝑥 = cos 360 − 𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
sin 𝑥
A. tan 𝑥 = cos 𝑥
B. sin2 𝑥 + cos2 𝑥 = 1
Application #1: Solving Harder Trig Equations
June 2013 Paper 2 Q22
Solve tan2 𝜃 + 3 tan 𝜃
= 0 in the range 0 ≤ 𝑥 < 360°
This looks a bit like a quadratic. What would be our usual strategy to solve!
tan 𝜃 tan 𝜃 + ?3 = 0
?
tan 𝜃 = 0 𝑜𝑟
𝜃 = 0, 180,
?
tan 𝜃 = −3
108.43°, 288.4°
1.
2.
3.
4.
sin 𝑥 = sin 180 − 𝑥
cos 𝑥 = cos 360 − 𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
sin 𝑥
A. tan 𝑥 = cos 𝑥
B. sin2 𝑥 + cos2 𝑥 = 1
More Examples
Solve 2sin2 𝜃 − sin 𝜃 = 0 in the range 0 ≤ 𝑥 < 360°
sin 𝜃 2 sin 𝜃 − 1 = 0
1
sin 𝜃 = 0 𝑜𝑟? sin 𝜃 =
2
𝜃 = 0, 180°, 30°, 150°
2
1
4
Solve cos 𝜃 = in the range 0 ≤ 𝑥 < 360°
1
1
cos 𝜃 =
𝑜𝑟 cos 𝜃 = −
2
2
?
𝜃 = 60°, 300°, 120°, 240°
Test Your Understanding
Solve cos 2 𝜃 + cos 𝜃 = 0 in the range 0 ≤ 𝑥 < 360°
cos 𝜃 cos 𝜃 + 1 = 0
cos 𝜃 = 0 𝑜𝑟? cos 𝜃 = −1
𝜃 = 90°, 270°, 180°
Expand and simplify (2𝑠 + 1)(𝑠 − 1). Hence or otherwise, solve
2 sin2 𝜃 − sin 𝜃 − 1 = 0 for 0° ≤ 𝜃 < 360°
2 sin 𝜃 + 1 sin 𝜃 − 1 = 0
1
sin 𝜃 = − 𝑜𝑟
? sin 𝜃 = 1
2
𝜃 = 210°, 330°, 90°
Exercise 2
1 Solve the following in the range
0° ≤ 𝑥 < 360°
a sin 𝜃 = 3 cos 𝜃
? 𝟓𝟕°
→ 𝜽 = 𝟕𝟏. 𝟓𝟕°, 𝟐𝟓𝟏.
b 2 sin 𝜃 = 3 cos 𝜃 → 𝜽 = 𝟓𝟔. 𝟑𝟏°, 𝟐𝟑𝟔.
? 𝟑𝟏°
2
a
b
c
Solve the following by first factorising. 0° ≤ 𝑥 < 360°
cos 2 𝜃 − cos 𝜃 = 0
→ 𝜽 = 𝟗𝟎, 𝟐𝟕𝟎, 𝟎?
tan2 𝜃 − 3 tan 𝜃 = 0
→ 𝜽 = 𝟎, 𝟏𝟖𝟎, 𝟕𝟏. ?
𝟓𝟕, 𝟐𝟓𝟏. 𝟓𝟕°
sin 𝑥 cos 𝑥 + sin 𝑥 = 0 → 𝜽 = 𝟎°, 𝟏𝟖𝟎° ?
3 Solve the following: 0° ≤ 𝑥 < 360°
3
→ 𝜽 = 𝟔𝟎°, 𝟏𝟐𝟎°,?𝟐𝟒𝟎°, 𝟑𝟎𝟎°
a sin2 𝜃 = 4
3
b cos 2 𝜃 =
→ 𝜽 = 𝟑𝟎°, 𝟏𝟓𝟎°, ?
𝟐𝟏𝟎°, 𝟑𝟑𝟎°
4
→ 𝜽 = 𝟔𝟎°, 𝟏𝟐𝟎°, ?
𝟐𝟒𝟎°, 𝟑𝟎𝟎°
c tan2 𝜃 = 3
4
a
b
N
By factorising these ‘quadratics’, solve in the range 0 ≤ 𝑥 < 360
3 cos 2 𝜃 + 2 cos 𝜃 − 1 = 0 → 𝜽 = 𝟕𝟎. 𝟓𝟑°, 𝟏𝟖𝟎°,
? 𝟐𝟖𝟗. 𝟒𝟕°
6 sin2 𝜃 − sin 𝜃 − 1 = 0
→ 𝜽 = 𝟑𝟎°, 𝟏𝟓𝟎°, 𝟏𝟗𝟗.
? 𝟒𝟕°, 𝟑𝟒𝟎. 𝟓𝟑°
sin 𝜃 cos 𝜃 + sin 𝜃 + cos 𝜃 = −1 → 𝐬𝐢𝐧 𝜽 + 𝟏 𝐜𝐨𝐬 𝜽 + 𝟏 ?
= 𝟎 → 𝜽 = 𝟏𝟖𝟎°, 𝟐𝟕𝟎°
Review of what we’ve done so far


partly

Application of identities #2: Proofs
Prove that 1 − tan 𝜃 sin 𝜃 cos 𝜃 ≡ cos 2 𝜃
Recall that ≡ means
‘equivalent to’, and just
means the LHS is always
equal to the RHS for all
values of 𝜃.
sin 𝜃
𝐿𝐻𝑆 = 1 −
sin 𝜃 cos 𝜃
cos 𝜃 ?
sin2 𝜃 𝑐𝑜𝑠𝜃
?
=1−
cos 𝜃
2
= 1 − sin 𝜃 ?
= cos 2 𝜃 = 𝑅𝐻𝑆
?
1.
2.
3.
4.
sin 𝑥 = sin 180 − 𝑥
cos 𝑥 = cos 360 − 𝑥
𝑠𝑖𝑛, 𝑐𝑜𝑠 repeat every 360
𝑡𝑎𝑛 repeats every 180
𝐬𝐢𝐧 𝒙
We want to use
these…
A. 𝐭𝐚𝐧 𝒙 = 𝐜𝐨𝐬 𝒙
B. 𝐬𝐢𝐧𝟐 𝒙 + 𝐜𝐨𝐬 𝟐 𝒙 = 𝟏
Another Example
June 2012 Paper 1 Q16
Prove that tan 𝜃 +
𝐿𝐻𝑆 =
1
tan 𝜃
≡
1
sin 𝜃 cos 𝜃
sin 𝜃 cos 𝜃
+
cos 𝜃 ?
sin 𝜃
sin2 𝜃
cos 2 𝜃
=
+
sin 𝜃 cos 𝜃 ?sin 𝜃 cos 𝜃
sin2 𝜃 + cos 2 𝜃
=
?𝜃
sin 𝜃 cos
1
=
= 𝑅𝐻𝑆
sin 𝜃 cos 𝜃?
Bro Tip: Whenever you
have a fraction in a proof
question, always add the
fractions.
Test Your Understanding
Prove that
tan 𝑥 cos 𝑥
1−cos2 𝑥
≡1
sin 𝑥
cos 𝑥 sin 𝑥
cos
𝑥
=
=?
=1
2
sin
𝑥
sin 𝑥
AQA Worksheet
Prove that tan2 𝜃 ≡
1
cos2 𝜃
−1
tan2 𝜃
sin2 𝜃
=
cos 2 𝜃
1 − cos 2 𝜃
=?
cos 2 𝜃
1
cos 2 𝜃
=
−
cos 2 𝜃 cos 2 𝜃
Exercise 3
1 Simplify 3 sin 𝑥 sin 𝑥 + 2 − 3 2 sin 𝑥 − cos 2 𝑥
= 𝟑 𝐬𝐢𝐧𝟐 𝒙 + 𝟔 𝐬𝐢𝐧 𝒙 − 𝟔 𝐬𝐢𝐧 𝒙 + 𝟑 𝐜𝐨𝐬 𝟐 𝒙
?𝟑
= 𝟑 𝐬𝐢𝐧𝟐 𝒙 + 𝐜𝐨𝐬 𝟐 𝒙 =
2 Write out the following in terms of sin 𝑥:
2
2
a) cos 𝑥 tan 𝑥
b) tan 𝑥 cos 3 𝑥
c)
cos 𝑥 2 cos 𝑥
𝐬𝐢𝐧𝟐 𝒙
= 𝒄𝒐𝒔 𝒙 𝐜𝐨𝐬𝟐 𝒙 = 𝐬𝐢𝐧𝟐 𝒙
𝐬𝐢𝐧 𝒙
= 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝟑 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝟐 𝒙 =
− 3 tan 𝑥 = 𝟐 𝐜𝐨𝐬 𝟐 𝒙 − 𝟑 𝐬𝐢𝐧 𝒙
𝟐
?
?
3 Prove the following:
a. tan 𝑥 1 − sin2 𝑥 ≡ sin 𝑥
b.
c.
d.
e.
f.
1−cos2 𝑥
1−sin2 𝑥
≡ tan2 𝑥
1 + sin 𝑥 1 − sin 𝑥 ≡ cos 2 𝑥
2 sin 𝑥 cos 𝑥
2𝑥
≡
2
−
2
sin
tan 𝑥
1
cos 𝜃
− cos 𝜃 ≡ sin 𝜃 tan 𝜃
2 sin 𝜃 − cos 𝜃 2 + sin 𝜃 + 2 cos 𝜃
2
≡5
𝐬𝐢𝐧 𝒙 𝟏 − 𝐬𝐢𝐧𝟐 𝒙
=𝟐?
𝟏 − 𝐬𝐢𝐧𝟐 𝒙 − 𝟑 𝐬𝐢𝐧 𝒙
Using triangles to change between sin/cos/tan
3
Given that sin 𝜃 = and that 𝜃 is
5
acute, find the exact value of:
𝟒
a) cos 𝜃 = ?
b) tan 𝜃 =
𝟓
𝟑
𝟒
?
?5
?3
Represent as
a triangle
𝜃
4
Test Your Understanding
1
5
13
Given that cos 𝜃 = and that
𝜃 is acute, find the value of:
𝟏𝟐
a) sin 𝜃 = ?
b) tan 𝜃
𝟏𝟑
𝟏𝟐
= ?
𝟓
2
5
Given that tan 𝜃 = and that
2
𝜃 is acute, find the value of:
a) sin 𝜃 =
b) cos 𝜃 =
𝟓
?
𝟑
𝟐
?
𝟑