WAVE FUNCTIONS What is a Wave Function Connection

WAVE FUNCTIONS What is a Wave Function Connection with Trig Identities Earlier Maximum and Minimum Values Solving Equations involving the Wave Function Exam Type Questions

The Wave Function Heart beat Many wave shapes, whether occurring as sound, light, water or electrical waves, can be described mathematically as a combination of sine and cosine waves. Spectrum Analysis Electrical

General shape for y = sinx + cosx 1. Like y = sin(x) shifted left The Wave Function 2. Like y = cosx shifted right 3. Vertical height different y = sin(x)+cos(x) y = sin(x) y = cos(x)

The Wave Function Whenever a function is formed by adding cosine and sine functions the result can be expressed as a related cosine or sine function. In general: With these constants the expressions on the right hand sides = those on the left hand side FOR ALL VALUES OF x

The Wave Function Worked Example: Re-arrange The left and right hand sides must be equal for all values of x. So, the coefficients of cos x and sin x must be equal: A pair of simultaneous equations to be solved

The Wave Function Find tan ratio Square and add note: sin(+) and cos(+)

The Wave Function Note: sin(+) and cos(+) 90 o 180 o S A T C 270 o 0 o

The Wave Example Expand Function equate coefficients Square and add Find tan ratio note: sin(+) and cos(+) 90 o 180 o S A T C 270 o 0 o

The Wave Function Finally:

The Wave Expand equate Function coefficients Example Square and add Find tan ratio noting sign ofo 90 sin(+) and cos(+) 180 o S A T C 270 o 0 o

The Wave Function Finally:

Maximum and Minimum Values Worked Example: b) Hence find: i) Its maximum value and the value of x at which this maximum occurs. ii) Its minimum value and the value of x at which this minimum occurs.

Maximum and Minimum Expand Values equate coefficients Square and add Find tan ratio o 90 note: sin(+) and cos(-) 180 o S A T C 270 o 0 o

Maximum and Minimum Values Maximum, we have:

Maximum and Minimum Values Minimum, we have:

Maximum and Minimum Values Example A synthesiser adds two sound waves together to make a new sound. The first wave is described by V = 75 sin to and the second by V = 100 cos to, where V is the amplitude in decibels and t is the time in milliseconds. Find the minimum value of the resultant wave and the value of t at which it occurs. For later, remember K = 25 k

Maximum and Minimum Expand Values equate coefficients Square and add Find tan ratio note: sin() and cos(+) 90 o 180 o S A T C 270 o 0 o

Maximum and Minimum Values remember K = 25 k =25 x 5 = 125 The minimum value of sin is -1 and it occurs where the angle is 270 o Therefore, the minimum value of Vresult is -125 Adding or subtracting 360 o leaves the sin unchanged

Maximum and Minimum Values Minimum, we have:

Solving Trig Equations Worked Example: Step 1: Compare Coefficients: Square &Add True for ALL x means coefficients equal.

Find tan ratio Solving Trig Equations note: sin(+) and cos(+) 90 o 180 o S A T C 270 o 0 o

Solving Trig Equations Step 2: Re-write the trig. equation using your result from step 1, then solve. 90 o 180 o S A T C 270 o 0 o

Solving Trig Equations Step 2:

Solving Trig Example Expand Equations equate coefficients Find tan ratio note: sin() and cos(-) 90 o Square and add 180 o S A T C 270 o 0 o

Solving Trig Equations 2 x – 213. 7 = 16. 1 o , (180 -16. 1 o), (360+180 -16. 1 o) 2 x – 213. 7 = 16. 1 o , 163. 9 o, 2 x = 229. 8 o , 310. 2 o, x = 114. 9 o , 188. 8 o, 376. 1 o, 589. 9 o, 294. 9 o, 523. 9 o, …. 670. 2 o, …. 368. 8 o, ….

Solving Trig Equations (From a past paper) Example A builder has obtained a large supply of 4 metre rafters. He wishes to use them to build some holiday chalets. The planning department insists that the gable end of each chalet should be in the form of an isosceles triangle surmounting two squares, as shown in the diagram. a) If θo is the angle shown in the diagram and A is the area m 2 of the gable end, show that 4 c) Find algebraically the value of θo for which the area of the gable end is 30 m 2. 4

Solving Trig Equations (From a past paper) Part (a) Let the side of the square frames be s. Use the cosine rule in the isosceles triangle: This is the area of one of the squares. The formula for the area of a triangle is Total area = Triangle + 2 x square: 4 s

Solving Trig Equations (From a past paper) Part (b) Find tan ratio note: sin(+) and cos(+) 90 o Square and add 180 o S A T C 270 o Finally: 0 o

Solving Trig Equations (From a past paper) Part (c) Find algebraically the value of θo for which the area is the 30 m 2 From diagram θo < 90 o ignore 2 nd quad 90 o 180 o S A T C 270 o 0 o

www. maths 4 scotland. co. uk Higher Maths Strategies The Wave Function Click to start

Maths 4 Scotland Higher The following questions are on The Wave Function Non-calculator questions will be indicated You will need a pencil, paper, ruler and rubber. Click to continue

Maths 4 Scotland Part of the graph of Higher y = 2 sin x + 5 cos x is shown in the diagram. a) Express y = 2 sin x + 5 cos x in the form k sin (x + a) b) where k > 0 and 0 a 360 Find the coordinates of the minimum turning point P. Expand ksin(x + a): Equate coefficients: Square and add a is in 1 st quadrant Dividing: (sin and cos are +) Put together: Minimum when: Hint P has coords. Previous Quit Next

Maths 4 Scotland Higher a) Write sin x - cos x in the form k sin (x - a) stating the values of k and a where b) k > 0 and 0 a 2 Sketch the graph of sin x - cos x for 0 a 2 showing clearly the graph’s maximum and minimum values and where it cuts the x-axis and the y-axis. Expand k sin(x - a): Equate coefficients: Square and add a is in 1 st quadrant Dividing: (sin and cos are +) Put together: Sketch Graph Hint Previous Table of exact values Quit Next

Maths 4 Scotland Express Higher in the form Expand kcos(x + a): Equate coefficients: Square and add a is in 1 st quadrant Dividing: (sin and cos are +) Put together: Hint Previous Quit Next

Maths 4 Scotland Higher Find the maximum value of and the value of x for which it occurs in the interval 0 x 2. Express as Rcos(x - a): Equate coefficients: Square and add a is in 4 th quadrant Dividing: (sin is - and cos is +) Put together: Max value: when Previous Table of exact values Hint Quit Next

Maths 4 Scotland Express Higher in the form Expand ksin(x - a): Equate coefficients: Square and add a is in 1 st quadrant Dividing: (sin and cos are both +) Put together: Hint Previous Quit Next

Maths 4 Scotland Higher The diagram shows an incomplete graph of Find the coordinates of the maximum stationary point. Max for sine occurs Sine takes values between 1 and -1 Max value of sine function: Max value of function: 3 Coordinates of max s. p. Hint Previous Quit Next

Maths 4 Scotland a) Higher Express f (x) in the form b) Hence solve algebraically Expand kcos(x - a): Equate coefficients: Square and add a is in 1 st quadrant Dividing: (sin and cos are both +) Put together: Solve equation. Hint Cosine +, so 1 st & 4 th quadrants Previous Quit Next

Maths 4 Scotland Higher Solve the simultaneous equations where k > 0 and 0 x 360 Use tan A = sin A / cos A Divide Find acute angle Determine quadrant(s) Sine and cosine are both + in original equations Solution must be in 1 st quadrant State solution Hint Previous Quit Next

Maths 4 Scotland Higher in the interval 0 x 360. Solve the equation Use R cos(x - a): Equate coefficients: Square and add a is in 2 nd quadrant Dividing: (sin + and cos - ) Put together: Solve equation. Cosine +, so 1 st & 4 th quadrants Previous Quit Hint Next

Maths 4 Scotland Higher You have completed all Previous 9 questions in this presentation Quit Back to start