Standard Grade and Higher still
www.mathsrevision.com
Wave Function Example
Write the equation below in the wave function format
f ( x)
sin( x)  cos ( x)
Comparing to the identity
k sin x  
k sin( x)  cos     k cos ( x) sin
We can express f ( x) sin( x)  cos ( x)
In the form of the wave equation
12  12
k
k sin  
2
k cos   
k sinx  
1
tan   
1

1
o
45
Hence we have
f ( x)

sin( x)  cos ( x)

o
2 sin x  45
Plotting each function we can see they are indeed equal.
Wave function
2
1
Amplitude
sin( x de g)
sin( x de g)  cos( x de g)
2 sin( x de g 45 de g)
90
0
90
180
270
360
450
540
1
2
x
de gree
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720
Standard Grade and Higher still
www.mathsrevision.com
Wave Function Example
Solve the expression below
f ( x)
sin( x)  cos ( x)
f ( x)
2 sin x  45


o


o
2 sin x  45
0  x  360
0.5
0.5
Dividing through by root 2 and taking inverse sine we get
x  45o
x  45o
sin  
1
0.5

 2
20.7 and 180  20.7  159.3 and all multiplies of 360 deg.
360  20.7  380.7
159.3 360  519.3
Hence we have
x = 20.7  45  24.3
159.3 45  114.3
We are only interested in the range
Hence answers are
380.7 35  345.7
519.3 45  474.3
0  x  360
114.3and 345.7
Solution to sin(x)+cos(x) = 0.5
2
345.7
114.3
114.3
1
2 sin( x de g 45 de g)
0.5
90
45
0
45
90
135
180
225
1
2
x
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270
315
360