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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
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
(a) sin x = 0.5
x = 30 or x = 180 - 30 = 30 or 150
sin is positive
angle is in Q1 or Q2
Value of sin x is 0.5
30 off x-axis
30 30
(b) tan x = -1
x = 180 - 45 or x = 360 - 45 = 135 or 315
tan is negative
angle is in Q2 or Q4
Value of tan x is -1
45 off x-axis
45°
45
General Solution of a Trig Function
cos θ = 0.643θ = cos-1 (0.643)θ ≈ 50°But cos 310° = 0.643 also
So there appears to be more than one
solution
So, how many solutions are there?
cos curve
-1
-0.5
0
0.5
1
-900 -540 -180 180 540 900
y=0.643
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
(a) sin x = 0.5
x = 30 or x = 180 - 30 = 30 or 150
general solution isx = 30 + n x 360 or x = 150 + n x
360
sin is positive
angle is in Q1 or Q2
Value of sin x is 0.5
30 off x-axis
30 30
Model Find all values of x (to the nearest
minute) where 0 ≤ x ≤ 360 for which
(a) sin2x = 0.25
(b) tan 3x = -1
(a) sin2x = 0.25
sin x = ± 0.5
x = 30 or x = 150 or x = 210 or x = 330
sin is pos or neg
angle is in Q1,Q2,Q3 or Q4
Value of sin x is 0.5
30 off x-axis
30 30
30 30
(b) tan 3x = -1
3x = 135° + 360n or 3x = 315 + 360n
x = 45 + 120n or x = 105 + 120n
45, 165, 285, 105, 225, 345
tan is negative
angle is in Q2 or Q4
Value of tan x is -1
45 off 3x-axis
45°
45
Derivatives of functions involving sin x and cos x
• Derivatives of functions involving sin x and cos x
Model : Find the derivative of(a) sin 2x(b) sin32x(c) sin2x cos3x
(d) sin(π-3x) • do some examples on Graphmatica
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
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 = 3u2 du = 2 cos 2x du dx dy = dy . du dx du dx = 3u2 . 2 cos 2x = 6 sin22x cos 2x
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
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
Model :
Find the gradient of the curve y = sin 2x at the point where x = π/3
1
cos2,
2cos2
2sin
32
3
dx
dyxWhen
xdx
dy
xy
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
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
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.
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
3t = 2nπ
t = 2nπ/3
t = 2π/3, 4π/3, 6π/3, …
i.e. first returns to starting point after 2π/3 secs
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
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
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
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 3π/6
= 0