1. Periodic Functions And Applications III
Significance of the constants A,B,C and D on the graphs of y = A sin(Bx+C) + D, y = A cos(Bx+C) + D
Application of periodic functions
Solution of simple trig equations within a specified domain
Derivatives of functions involving sin x and cos x
Applications of the derivatives of sin x and cos x in life-related situations

2. Periodic Functions And Applications III

3. No. 1-12 (parts a & b only), leave out no.10
REVISION
New Q
Page 351 Ex 10.1
No (parts a & b only), leave out no.10
FM Page Ex 5.8

4. Solving Trig(onometric) Equations
Model Find all values of x (to the nearest minute) where 0< x <360 for which
(a) sin x = 0.5
(b) tan x = -1

5. (a) sin x = 0.5 x = 30 or x = 180 - 30 = 30 or 150
sin is positive
 angle is in Q1 or Q2
(a) sin x = 0.5
x = 30 or x = 180 - 30
= 30 or 
30
30
Value of sin x is 0.5
 30 off x-axis

6. (b) tan x = -1 x = 180 - 45 or x = 360 - 45 = 135 or 315
tan is negative
 angle is in Q2 or Q4
(b) tan x = -1
x = 180 - 45 or x = 360 - 45
=  or 
45
45°
Value of tan x is -1
 45 off x-axis

7. New Q
Ex 10.3
Page ,6
FM Page Ex (orally)

8. General Solution of a Trig Function
cos θ = 0.643
θ = cos-1 (0.643)
θ ≈ 50°
But cos 310° = also
So there appears to be more than one solution
So, how many solutions are there?

9. θ = 50° + 360° x n θ = 310° + 360° x n y=0.643 cos curve cos θ = 0.634
θ = 50° or θ = 310°
or θ = 50° + 360° or θ = 310° + 360°
or θ = 50° + 2 x 360° or θ = 310° - 360°
or θ = 50° + 3 x 360° or θ = 310° - 2 x 360°
or θ = 50° - 360° or θ = 310° - 3 x 360°
or θ = 50° - 2 x 360° etc
θ = 50° + 360° x n θ = 310° + 360° x n

10. The General Solution for cos θ = 0.643
For all integer values of n

11. Model : Find all values of x (to the nearest minute) for which
(a) sin x = 0.5

12. (a) sin x = 0.5 x = 30 or x = 180 - 30 = 30 or 150
sin is positive
 angle is in Q1 or Q2
(a) sin x = 0.5
x = 30 or x = 180 - 30
= 30 or 
 general solution is
x = 30 + n x 360 or x = 150 + n x 360
30
30
Value of sin x is 0.5
 30 off x-axis

13. Model Find all values of x (to the nearest minute) where 0 ≤ x ≤ 360 for which
(a) sin2x = 0.25
(b) tan 3x = -1

14. sin is pos or neg
 angle is in Q1,Q2,Q3 or Q4
(a) sin2x = 0.25
sin x = ± 0.5
x = 30 or x = 150 or x = 210 or x = 330
30
30
30
30
Value of sin x is 0.5
 30 off x-axis

15. (b) tan 3x = -1 3x = 135° + 360n or 3x = 315 + 360n
tan is negative
 angle is in Q2 or Q4
(b) tan 3x = -1
3x = 135° + 360n or 3x = 315 + 360n
x = 45 + 120n or x = 105 + 120n
 45, 165, 285, 105, 225, 345
45
45°
Value of tan x is -1
 45 off 3x-axis

16. New Q
Page 371 Ex 10.5
1, 2, 4,10, 14
FM Page Ex with GC

17. Derivatives of functions involving sin x and cos x

18. Derivative of sin x and cos x
y = sin x
 dy = cos x dx
y = cos x
 dy = -sin x dx

19. Model : Find the derivative of (a) sin 2x (b) sin32x (c) sin2x cos3x (d) sin(π-3x)
do some examples on Graphmatica

20. ________________________________________
Model : Find the derivative of (a) sin 2x (b) sin32x (c) sin2x cos3x (d) sin(π-3x)
________________________________________
(a) y = sin 2x
= sin u where u = 2x
dy = cos u du = 2
du dx
dy = dy . du
dx du dx
= 2 cos u
= 2 cos 2x

21. ________________________________________
Model : Find the derivative of (a) sin 2x (b) sin32x (c) sin2x cos3x (d) sin(π-3x)
________________________________________
(b) y = sin32x ( = (sin 2x)3 )
= u3 where u = sin 2x
dy = 3u du = 2 cos 2x
du dx
dy = dy . du
dx du dx
= 3u2 . 2 cos 2x
= 6 sin22x cos 2x

22. ________________________________________
Model : Find the derivative of (a) sin 2x (b) sin32x (c) sin2x cos3x (d) sin(π-3x)
________________________________________
(c) y = sin2x cos3x
= uv where u = sin2x and v = cos3x
du = 2 sinx cosx dv = -3 sin3x
dx dx
dy = u dv + v du
dx dx dx
= -3 sin3x sin2x + cos3x  2 sin x cos x
= -3 sin3x sin2x + 2 cos3x sin x cos x

23. ________________________________________
Model : Find the derivative of (a) sin 2x (b) sin32x (c) sin2x cos3x (d) sin(π-3x)
________________________________________
(d) y = sin (π-3x)
= sin 3x
dy = 3 cos 3x dx

24. NEWQ P50 2.4 No. 1(a,b,d,f,h,j), 3 – 8 (all) FM Page 447 Ex 19.5 1,4,5

25. Model : Find the gradient of the curve y = sin 2x at the point where x = π/3

26. Model : Find the gradient of the curve y = sin 2x at the point where x = π/3

27. Trig functions and motion
Consider the motion of an object on the end of a spring dropped from a height of 1m above the equilibrium point which takes 2π seconds to return to the starting point. 1m 0m
-1m
s = cos t
v = -sin t
a = -cos t

28. Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (a) How far from the fixed point is the object at the start? (b) How long does it take for the object to return to its starting point? (c) Find the object’s velocity (i) at the start (ii) as it passes the fixed point (iii) after 2 sec (d) Find its acceleration as it passes the fixed point

29. Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (a) How far from the fixed point is the object at the start?
At the start, t = 0
When t = 0, s = 4 cos(3x0)
= 4 cos0
= 4 x 1
= 4
i.e. at the start, object is 4 metres from the fixed point.

30. Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (b) How long does it take for the object to return to its starting point?
Returns to starting point  s = 4
 4 cos3t = 4
 cos3t = 1
 t = 2nπ
 t = 2nπ/3
 t = 2π/3, 4π/3, 6π/3, …
i.e. first returns to starting point after 2π/3 secs

31. s = 4 cos3t  v = -12 sin3t (i) At the start When t = 0, v = -12 sin 0
Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (c) Find the object’s velocity (i) at the start (ii) as it passes the fixed point (iii) after 2 sec
s = 4 cos3t
 v = -12 sin3t
(i) At the start
When t = 0, v = -12 sin 0
= 0

32. Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (c) Find the object’s velocity (i) at the start (ii) as it passes the fixed point (iii) after 2 sec
(ii) at what time does it pass the fixed point
 s = 0
 4 cos 3t = 0
cos 3t = 0
3t = π/2, …
t = π/6, …
When t = π/6, v = -12 sin 3π/6
= -12

33. Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (c) Find the object’s velocity (i) at the start (ii) as it passes the fixed point (iii) after 2 sec
(iii) After 2 sec
v = -12 sin 3x2
= -12 sin 6
= 3.35

34. It passes the fixed point when t = π/6, a = -36 cos 3π/6 = 0
Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (d) Find its acceleration as it passes the fixed point
s = 4 cos3t
v = -12 sin3t
a = -36 cos3t
It passes the fixed point when t = π/6,
a = -36 cos 3π/6
= 0

35. NEWQ P59 Ex No. 1 – 4, 7