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UNIT 3 Lesson 3
Derivatives of Sine and Cosine
1
DERIVATIVE OF THE SINE FUNCTION
π’š = 𝐬𝐒𝐧 𝒙
π’…π’š
=?
𝒅𝒙
2
Find the derivative of f(x) = sin x from first principles:
𝐬𝐒𝐧 𝒙 + 𝒉 βˆ’ 𝐬𝐒𝐧 𝒙
π’‰β†’πŸŽ
𝒙+𝒉 βˆ’π’™
𝒇′ 𝒙 = π₯𝐒𝐦
Sum Identity: sin (A + B) = sin A cos B + cos A sin B
𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒉 + 𝐜𝐨𝐬 𝒙 𝐬𝐒𝐧 𝒉 βˆ’ 𝐬𝐒𝐧 𝒙
π’‰β†’πŸŽ
𝒉
𝒇′ 𝒙 = π₯𝐒𝐦
𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒉 βˆ’ 𝐬𝐒𝐧 𝒙 + 𝐜𝐨𝐬 𝒙 𝐬𝐒𝐧 𝒉
π’‰β†’πŸŽ
𝒉
𝒇′ 𝒙 = π₯𝐒𝐦
𝐬𝐒𝐧 𝒙 (𝐜𝐨𝐬 𝒉 βˆ’ 𝟏)) + 𝐜𝐨𝐬 𝒙 𝐬𝐒𝐧 𝒉
π’‰β†’πŸŽ
𝒉
𝒇′ 𝒙 = π₯𝐒𝐦
𝒇′
𝐬𝐒𝐧 𝒙 (𝐜𝐨𝐬 𝒉 βˆ’ 𝟏))
𝐜𝐨𝐬 𝒙 𝐬𝐒𝐧 𝒉
𝒙 = π₯𝐒𝐦
+ π₯𝐒𝐦
π’‰β†’πŸŽ
𝒉
π’‰β†’πŸŽ
𝒉
𝐜𝐨𝐬 𝒉 βˆ’ 𝟏
𝐬𝐒𝐧 𝒉
+ π₯𝐒𝐦 𝐜𝐨𝐬 𝒙 π₯𝐒𝐦
π’‰β†’πŸŽ
π’‰β†’πŸŽ
π‘β†’πŸŽ 𝒉
𝒉
𝒇′ 𝒙 = π₯𝐒𝐦 𝐬𝐒𝐧 𝒙 × π₯𝐒𝐦
π’‰β†’πŸŽ
continued
3
𝐜𝐨𝐬 𝒉
𝐬𝐒𝐧 𝒉
𝒇 𝒙 = 𝐬𝐒𝐧 𝒙 π₯𝐒𝐦
+ 𝐜𝐨𝐬 𝒙 π₯𝐒𝐦
π’‰β†’πŸŽ 𝒉
π’‰β†’πŸŽ 𝒉
β€²
From the limits that we developed in the last lesson, we know
that
𝐜𝐨𝐬 𝒉 βˆ’ 𝟏
𝐬𝐒𝐧 𝒉
π₯𝐒𝐦
=𝟎
π₯𝐒𝐦
=𝟏
π’‰β†’πŸŽ
π’‰β†’πŸŽ 𝒉
𝒉
𝒇′ 𝒙 = 𝐬𝐒𝐧 𝒙 𝟎 + 𝐜𝐨𝐬 𝒙(𝟏)
𝒇′ 𝒙 = 𝐜𝐨𝐬 𝒙
Our first trigonometric derivative is:
𝒅
𝐬𝐒𝐧 𝒙 = 𝐜𝐨𝐬 𝒙
𝒅𝒙
4
EXAMPLE 1: Find the derivative of 𝒇 𝒙 = πŸ‘ 𝐬𝐒𝐧 𝒙
𝒇′
𝒙 = πŸ‘ 𝐜𝐨𝐬 𝒙
π’”π’Šπ’ 𝜽 =
π’š
=π’š
𝟏
𝒄𝒐𝒔 𝜽 =
𝒙
=𝒙
𝟏
Find the slope of the tangent at
𝒕𝒂𝒏 𝜽 =
,
𝝅
𝒙=
πŸ‘
𝝅
𝒙=
𝟐
𝝅
𝝅
𝟏
πŸ‘
𝒇
= πŸ‘ 𝐜𝐨𝐬
=πŸ‘
=
πŸ‘
πŸ‘
𝟐
𝟐
𝝅
𝝅
β€²
𝒇
= πŸ‘ 𝐜𝐨𝐬
= πŸ‘(𝟎) = 𝟎
𝟐
𝟐
𝒙=𝝅
𝒇′ 𝝅 = πŸ‘ 𝐜𝐨𝐬 𝝅 = πŸ‘ βˆ’πŸ = βˆ’πŸ‘
π’š
𝒙
β€²
5
EXAMPLE 2: Find the derivative of π’ˆ 𝒙 = 𝐬𝐒𝐧(πŸ‘π’™)
Use the Chain Rule
π’ˆβ€² 𝒙 = (𝐜𝐨𝐬(πŸ‘π’™))(πŸ‘) = πŸ‘ 𝐜𝐨𝐬(πŸ‘π’™)
Find the slope of the tangent at
𝒙=𝟎
π’ˆβ€² 𝟎 = πŸ‘ 𝐜𝐨𝐬 πŸ‘ × πŸŽ = πŸ‘ cos 𝟎 = πŸ‘ 𝟏 = πŸ‘
𝝅
𝒙=
πŸ”
π’ˆβ€²
𝝅
𝝅
𝝅
= πŸ‘ 𝐜𝐨𝐬 πŸ‘ ×
= πŸ‘ 𝐜𝐨𝐬 = πŸ‘ 𝟎 = 𝟎
πŸ”
πŸ”
𝟐
𝝅
𝒙=
πŸ‘
π’ˆβ€²
𝝅
𝝅
= πŸ‘ 𝐜𝐨𝐬 πŸ‘ ×
= πŸ‘ 𝐜𝐨𝐬 𝝅 = πŸ‘ βˆ’πŸ = βˆ’πŸ‘
πŸ‘
πŸ‘
6
EXAMPLE 3: Differentiate h(x) = – 2sin(5x)
𝒉′(𝒙) = βˆ’πŸ 𝐜𝐨𝐬(πŸ“π’™)(πŸ“)
𝒉′(𝒙) = βˆ’πŸπŸŽ 𝐜𝐨𝐬(πŸ“π’™)
.
Find the slope of the tangent at
𝝅
𝒙=
πŸ“
𝝅
𝝅
𝒉′
= βˆ’πŸπŸŽ 𝐜𝐨𝐬 πŸ“ ×
= βˆ’πŸπŸŽ 𝐜𝐨𝐬 𝝅 = βˆ’πŸπŸŽ βˆ’πŸ = 𝟏𝟎
πŸ“
πŸ“
7
EXAMPLE 4: Find the derivative of π’š = 𝐬𝐒𝐧 𝒙
𝟐
Find the slope of the tangent at
𝒙=𝝅
π’…π’š
= 𝟐 𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒙
𝒅𝒙
π’…π’š
= 𝟐 𝐬𝐒𝐧 𝝅 𝐜𝐨𝐬 𝝅
𝒅𝒙
π’…π’š
= 𝟐 βˆ’πŸ 𝟎 = 𝟎
𝒅𝒙
8
EXAMPLE 5: Find the derivative of π’š = πŸ‘ 𝐬𝐒𝐧𝟐 𝒙 = πŸ‘ 𝐬𝐒𝐧 𝒙
π’…π’š
= πŸ” 𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒙
𝒅𝒙
Find the slope of the tangent at
𝒙=𝝅
πŸ” 𝐬𝐒𝐧 𝝅 𝐜𝐨𝐬 𝝅 = πŸ” 𝟎 βˆ’πŸ = 𝟎
9
𝟐
EXAMPLE 6: Find the derivative of 𝑦 = sin2 3π‘₯ = sin(3π‘₯)
2
π’…π’š
= 𝟐 𝐬𝐒𝐧(πŸ‘π’™)(πŸ‘) 𝐜𝐨𝐬(πŸ‘π’™)
𝒅𝒙
π’…π’š
= πŸ” 𝐬𝐒𝐧(πŸ‘π’™) 𝐜𝐨𝐬(πŸ‘π’™)
𝒅𝒙
Find the slope of the tangent at
𝝅
π’…π’š
𝝅
𝝅
𝒙=
= πŸ” 𝐬𝐒𝐧(πŸ‘ × ) 𝐜𝐨𝐬(πŸ‘ × )
𝟏𝟐
𝒅𝒙
𝟏𝟐
𝟏𝟐
π’…π’š
𝝅
𝝅
= πŸ” 𝐬𝐒𝐧 𝐜𝐨𝐬
𝒅𝒙
πŸ’
πŸ’
𝟏
𝟏
πŸ”
πŸ”×
×
= =πŸ‘
𝟐
𝟐 𝟐
10
EXAMPLE 7: Differentiate 𝑓 π‘₯ = 4 sin2 π‘₯ 2 + 2
𝒇 𝒙 = πŸ’ π’”π’Šπ’πŸ π’™πŸ + 𝟐
𝟐
𝒇 𝒙 = πŸ’ 𝐬𝐒𝐧 𝒙 + 𝟐
REWRITE
𝟐
Using the Chain Rule
β†’
𝒇′ 𝒙 = 𝟐(πŸ’) 𝐬𝐒𝐧 π’™πŸ + 𝟐
𝐜𝐨𝐬 π’™πŸ + 𝟐 (πŸπ’™)
𝒇′ 𝒙 = πŸπŸ”π’™ 𝐬𝐒𝐧 π’™πŸ + 𝟐
𝐜𝐨𝐬 π’™πŸ + 𝟐
Using the sine double angle identity
2sin A cos A = sin 2A
𝒇′ 𝒙 = πŸ–π’™ 𝟐 𝐬𝐒𝐧 π’™πŸ + 𝟐
𝐜𝐨𝐬 π’™πŸ + 𝟐
𝒇′ 𝒙 = πŸ–π’™ 𝐬𝐒𝐧 𝟐 π’™πŸ + 𝟐
11
DERIVATIVE OF THE COSINE FUNCTION
What is
d
cos x
dx
?
12
To find the derivative of the function f(x) = cos x, we can
use a complementary angle identity in combination
with the derivative of f(x) = sin x.
COMPLEMENTARY ANGLES (sum is
A
90o
or
𝝅
)
𝟐
c
b
B
C
a
Angles A and B are complementary angles.
𝝅
𝝅
A= –B
and B = – A
𝟐
𝟐
𝒂
𝒂
𝝅
𝐜𝐨𝐬 𝑩 = = 𝐬𝐒𝐧 𝑨 = = 𝐬𝐒𝐧( – 𝑩)
𝒄
𝒄
𝟐
𝒂
𝒂
𝝅
𝐬𝐒𝐧 𝑨 = = 𝒄𝒐𝒔 𝑩 = = 𝒄𝒐𝒔( – 𝑨)
𝒄
𝒄
𝟐
13
Find the derivative of π’š = 𝐜𝐨𝐬 𝒙
𝝅
π’š = 𝐜𝐨𝐬 𝒙 = 𝐬𝐒𝐧
βˆ’π’™
𝟐
We now differentiate using the Chain Rule:
𝑑𝑦
πœ‹
= cos βˆ’ π‘₯ (βˆ’1)
𝑑π‘₯
2
𝑑𝑦
πœ‹
= βˆ’cos βˆ’ π‘₯
𝑑π‘₯
2
𝑑𝑦
= βˆ’ sin π‘₯
𝑑π‘₯
This is the second trigonometric derivative:
𝒅
𝐜𝐨𝐬 𝒙 = βˆ’ 𝐬𝐒𝐧 𝒙
𝒅𝒙
14
EXAMPLE 8: Differentiate f(x) = sin (x) cos (x) and use the
derivative to find the slope of the tangent to the function when
πœ‹
π‘₯=
3
Using the Product Rule
𝒇′ 𝒙 = 𝐬𝐒𝐧 𝒙 βˆ’ 𝐬𝐒𝐧 𝒙 + 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒙
2
3
𝒇′ 𝒙 = βˆ’ 𝐬𝐒𝐧𝟐 𝒙 + 𝐜𝐨𝐬𝟐 𝒙
𝒇′
𝒙 =
βˆ’ 𝐬𝐒𝐧𝟐
𝝅
𝒇′
=βˆ’
πŸ‘
πŸ‘
𝟐
𝝅
𝝅
𝟐
+ 𝐜𝐨𝐬
πŸ‘
πŸ‘
𝟐
𝟏
+
𝟐
𝟐
1
πŸ‘ 𝟏
𝟏
=βˆ’ + =βˆ’
πŸ’ πŸ’
𝟐
15
EXAMPLE 9: Differentiate 𝑓 π‘₯ =
sin 3π‘₯
cos π‘₯ 2
Using the Quotient Rule and Chain Rule
𝟐 πŸ‘ 𝐜𝐨𝐬 πŸ‘π’™ βˆ’ 𝐬𝐒𝐧 πŸ‘π’™ βˆ’ 𝐬𝐒𝐧 π’™πŸ (πŸπ’™)
𝐜𝐨𝐬
𝒙
𝒇′ 𝒙 =
𝐜𝐨𝐬 π’™πŸ 𝟐
𝟐 )(𝐜𝐨𝐬 πŸ‘π’™) + πŸπ’™(𝐬𝐒𝐧 πŸ‘π’™) 𝐬𝐒𝐧 π’™πŸ
πŸ‘(𝐜𝐨𝐬
𝒙
𝒇′ 𝒙 =
𝐜𝐨𝐬 π’™πŸ 𝟐
16
2
2
EXAMPLE 10: Differentiate f  x  ο€½ 3x cos  2 x 
Using the Product and Chain Rules.
g  x  ο€½ 3x
2
g '  x  ο€½ 6x
h  x  ο€½ cos  2 x  ο€½  cos 2 x 
2
2
h '  x  ο€½ 2cos  2 x   ο€­ sin  2 x    2 
h '  x  ο€½ ο€­4  cos  2 x  sin  2 x  
f '  x  ο€½ cos 2 (2 x)[6 x]  3x 2  ο€­4  cos  2 x  sin  2 x  
f '  x  ο€½ 6 x cos 2 (2 x) ο€­ 12 x 2 cos  2 x  sin  2 x 
17
2
2
EXAMPLE 10: Differentiate f  x  ο€½ 3x cos  2 x 
Using the Product and Chain Rules.
g  x  ο€½ 3x
2
g '  x  ο€½ 6x
Alternate approach
h  x  ο€½ cos  2 x  ο€½  cos 2 x 
2
2
h '  x  ο€½ 2cos  2 x   ο€­ sin  2 x    2 
h '  x  ο€½ ο€­4  cos  2 x  sin  2 x  
Using the sine double angle identity sin 2A = 2sin A cos A
h '  x  ο€½ ο€­2  2cos  2 x  sin  2 x   ο€½ ο€­2sin  4 x 
f '  x  ο€½ 3x 2  ο€­2sin 4 x   cos2  2 x  6 x 
f '  x  ο€½ ο€­6 x2 sin  4 x   6 x cos 2  2 x 
18
Assignment Questions
Differentiate each of the following trigonometric functions.
1 a) y = 4cos x
dy
ο€½ ο€­4sin x
dx
1 b) f(x) = 3sin(5x2)
f '  x  ο€½ 3cos  5x 2  10 x 
f '  x  ο€½ 30 x cos  5x 2 
sin 2 x
1 c) y ο€½
cos2 x 
2
cos
2
x
2sin
x
cos
x
ο€­
sin
x  ο€­2sin 2 x 


dy
ο€½
dx
cos 2  2 x 
19
Assignment Questions
Differentiate each of the following trigonometric functions.
d) f(x) = –
2cos3(7x)
ο€½ ο€­2  cos  7x  
3
REWRITE
f '  x  ο€½ ο€­6  cos  7 x    ο€­ sin  7 x  7  
2
f '  x  ο€½ 42cos2  7 x  sin  7 x 
e) y = x sin x
y ' ο€½ x  cos x   sin x 1
y ' ο€½ x cos x  sin x
20
Assignment Questions
2. Find the slope of the tangent to the function
y = cos2(2x) at the point
xο€½

2
.
dy
ο€½ 2  cos  2 x    2   ο€­ sin  2 x  
dx
dy
ο€½ ο€­4cos  2 x  sin  2 x 
dx
The slope of the tangent at
xο€½
dy
ο€½ ο€­2  2cos  2 x  sin  2 x  
dx
dy
ο€½ ο€­2  2sin  2 x  cos  2 x   ο€½ ο€­2sin  4 x 
dx
dy
   οƒΆοƒΆ
ο€½ ο€­2sin  4  οƒ· οƒ· ο€½ ο€­2sin  2  ο€½ ο€­2  0  ο€½ 0
dx
  2 οƒΈοƒΈ

2
is 0
21
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