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Type I : Mathematical Investigation
DIFFERENTIATING TRIGONOMETRIC FUNCTIONS
This assignment will be assessed against all 6 criteria
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Part 1
Below is a sector of radius r and angle  . If
and tan  .
You must clearly show your reasoning.
 is very small, find expressions for sin  , cos 
r
r
Solution :
If the angle  is very small, BC= BC =  r
(1) from the triangle
BC  .r

= .
r
AB
AC r
 1
cos  =
AB r
BC  .r

tan  =
= .
AC
r
sin  =
(2) From further calculation:
sin  =
BC  .r

= .
r
AB
cos  = 1 sin
2
1
2
1
2
 = 1    (1   )  1   2  ...
2
2
{this can be get from binomial theorem}
because  is very small, so we can just calculate the first two term .
cos  1-
1 2
 .
2
sin 
sin 
sin  1  sin 2   1   2

1




 (1   2  ...)
2
2
2
2
tan  = cos 
2
1  sin 
1
1
1  sin 
{this is also get from binomial theorem}
because  is very small, so we can just calculate the first two term .
tan  


1 2
2   2 2   3




(1.)

2
2
1 2
1 2
2  2 2
---------------------------------------------------------------------------------------------------------------------Part 2
(i)
With the aid of your GDC, investigate the results of Part 1.
(ii)
Investigate also lim
 0
sin 

and lim
 0
tan 

.
Solution:
{in this question,press “STAT”,choose “Edit”,then show below}
(i)
{put numbers in L1(  ),let L1  0 , put the formula sin(LI) in L2}
as the graph showing, when ,L1  0, L 2  L1 ,so   0, sin   
{put the formula cos(L1) in L3}
as the graph showing, when ,L1  0, L3  1 ,so   0, cos  1 .
{put the formula tan(L1) in L4}
as the graph showing, when ,L1  0, L3  L1 so  0, tan   
(ii)
{put the formula
sin L1
in L2}
L1
as the graph showing, when L1  0, L 2  1 ,so
lim
 0
sin 

=1
-----------------------------------------------------------------------------------------------------------------{put formula
tan L1
in L2}
L1
as the graph showing, when L1  0, L 2  1 , so lim
 0
tan 

=1
---------------------------------------------------------------------------------------------------------------------Part 3
Consider the right-angle triangle OTB containing the sector OAB of radius a.
By considering appropriate areas, show that 1 
Hence determine the lim
sin 
 0

sin 

 cos
, clearly showing your reasoning.
Solution:
In AOB, AQ  a sin  ,
In BOT , BT  a tan  ,
By considering appropriate ares, we know:
 area AOB  area sec torAOB  areaBOT
1
1
1
 a 2 sin   a 2  a 2 tan  {from the formula of triangle(area =1/2 ah),and the
2
2
2
formula of sectors(area = 1/2 rl)}
sin     tan 
s i n
co
s
1
1 cos 
 
sin   sin 
s i n   
product by sin  ,which is positive, as  
1 
sin 


2
:
 cos 
  0, cos  1, so 1 
 lim
 0
sin 

sin 

1
=1
---------------------------------------------------------------------------------------------------------------------Part 4
Consider the function f(x) = sin x ,x  [0, 2  ]. By investigating the gradient of this function ,
deduce f ’(x).
You may wish to draw appropriate graphs on your GDC.
Solution:
{in this question, should use “Y=” to enter the function, then “WINDOW” enter the range, and
“GRAPH” to draw the graphs, showing as below }
{as we known, the graph is like the graph of the function of y=cosx, so compare with the graph of
y=cosx}
as the graph showing, the graph of y=cosx is the same as the graph of f’(x), so f’(x)=cosx
------------------------------------------------------------------------------------------------------------------Part 5
Prove your conjecture in Part4 from first principles, showing clearly your reasoning.
Solution:
y  sin x
dy
sin( x  x)  sin( x)
 lim
dx x  0
x
2 cos( x 
x
) sin(
x
2
2
= lim
x  0
x
x
x
cos( x  ) sin( )
2
2
= lim
x
x  0
2
= lim cos( x 
x  0
∵
lim cos( x 
x
x  0
sin
lim
x  0
x
2
x
2
)
sin
) lim
x  0
x
2
x
2
) = cos x
2 1
x
2
dy
sin( x  x)  sin( x)
 lim
∴ f ' ( x) =
dx x  0
x
= lim cos( x 
x  0
x
2
sin
) lim
x  0
x
2
x
2
= cos x
---------------------------------------------------------------------------------------------------------------------Part 6
If h(x) = cos x, deduce h ‘(x)
Solution:
{in this question, should use “Y=” to enter the function, then “WINDOW” enter the range, and
“GRAPH” to draw the graphs, showing as below }
{As we known, the graph of the h’(x)is like the graph of the function of y=-sinx,so, compare with
the graph of y=-sinx}
as the graph showing, the graph of y=-sinx is the same as the graph of h’(x),so h’(x)= -sinx
Part 7
Prove your conjecture in part 6 from first principles, showing clearly your reasoning
Solution:
∴ h' ( x ) =
dy
cos( x  x)  cos( x)
 lim
dx x  0
x
According to the double angle identify:
cos  1  2 sin 2

2
∴ cos( x  x)  cos x = cos x cos x  sin x sin x  cos x
= cos x(cos x  1)  sin x sin x
=  sin x sin x  cos x(2 sin
=  sin x 2 sin
=2 sin
x
x
2
cos
( sin x cos
x
2
x
2
x
2
)
 cos x(2 sin 2
 cos x sin
2
2
dy
cos( x  x)  cos( x)
 lim
∴
dx x  0
x
x
x
 2 sin( x  ) sin( )
2
2
= lim
x  0
x
x
x
 sin( x  ) sin( )
2
2
= lim
x
x  0
2
 x 
 sin 2 

x


= lim  sin( x  )  lim 

x  0 
2  x  0  x 
 2 
x
2
x
2
)
)
=  sin x
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Part 8
Proce you conjecture in Part 6 by using the result of Part 5 or any other method.
Solution:
y  cos x
 cos x  sin(
 y  sin(
let u  (

2

2

2
 x)
 x)
 x), theny  sin u
du
dy
 1, and ,
 cos u
dx
du
dy dy du


: h' ( x ) 
dx du dx
so: differentiating gives:
 cos u  (1)
=  cos(

2
=  sin x
 x)
---------------------------------------------------------------------------------------------------------------------Part9
By using your results of Part 5 and Part 7, determine the derivative of g(x) = tan x .
Solution:
y  tan x 
sin x
cos x
let
u  sin x.so
du
 cos x
dx
dv
  sin x
dx
du
dv
v
u
dy
cos 2 x  sin 2 x
1
 dx 2 dx 

 sec 2 x
2
2
dx
v
cos x
cos x
v  cos x, so
lim (
: x0
{ sin x  cos x  1 }
2
2
y
dy
)
 sec 2 x
x
dx
 g ' ( x)  sec 2 x
----------------------------------------------------------------------------------------------------------------------
Part 10
Find g‘(x) in Part 9 from first principles, showing clearly your reasoning.
Solution:
y  tan x
y  y  tan( x  x) 
tan x  tan x
1  tan x tan x
d
g ( x)  lim tan( x  x)  tan( x)
x  0
dx
x
tan x  tan x
 tan x
∵ tan( x  x)  tan x =
1  tan x tan x
=
tan x  tan x  tan x  tan 2 x tan x
1  tan x tan x
=
tan x  tan 2 x tan x
1  tan x tan x
tan x(1  tan 2 x)
=
1  tan x tan x
sec 2 x  1  tan 2 x
∴ =
tan x sec 2 x
1  tan x tan x
 tan x sec 2 x
d
1
∴
g ( x)  lim 
 
x  0 1  tan x tan x
dx
x 



sec 2 x
 tan x 

  lim 
x  0 x  x  0 1  tan x tan x 


= lim 
 sec 2 x 

x 0 1  0 


=1  lim 
2
= sec x
=
1
cos 2 x
_____________________________________ THE END________________________________