# Write the following trigonometric identities problem as instructed:Maxwell's theory of electromagnetic waves established that electromagnetism moved through space in waves. under certain...

Write the following trigonometric identities problem as instructed:

Maxwell's theory of electromagnetic waves established that electromagnetism moved through space in waves. under certain conditions, an equation that Maxwelled used to explain the motion of such waves would take the form **B/A = (sin2a - sin 2b) / (sin2a +sin2b) **where:

**a **= angle of incdience of the original wave**b** = angle of incidence of the wave when reflected**A** = amplitude of the original wave**B** = amplitude of the wave when reflected

Express **B/A** in terms of sin**a**, cos**a, **sin**b**, and cos**b**; in other words, eliminate the double angles.

The multiple-choice answers to this question:

a) (sin**b **cos**b **+ sin**b** cos**b**) / (sin**b** cos**b** - sin**b** cos**b**)

b) (sin**a** cos**a** - sin**b** cos**b**) / (sin**a** cos**a** + sin**b** cos**b**)

c) (sin**a** cos**a** + sin**b** cos**b**) / (sin**a** cos**a** - sin**b** cos**b**)

d) (sin**b** cos**b** - sin**a** cos**b**) / sin**a** cos**a** +sin**b** cos**b**)

### 2 Answers | Add Yours

The original equations given by Maxwell is: B/A = (sin 2a - sin 2b) / (sin 2a + sin 2b)

To convert this to a form that does not use double angles we use the following relation: sin 2x = 2*sin x * cos x

Applying this to the expression for B/A

B/A = (sin2a - sin 2b) / (sin2a +sin2b)

=> (2*sin a*cos a - 2*sin b*cos b)/(2*sin a*cos a + 2*sin b*cos b)

cancel the common factor 2

=> (sin a*cos a - sin b*cos b)/(sin a*cos a + sin b*cos b)

We get a result that matches option b

**The correct option is option B**

We'll manage the right side of the identity and we'll transform the difference from numerator and the sum from denominator, into products.

sin2a - sin 2b = 2cos[(2a+2b)/2]*sin[(2a-2b)/2]

sin2a - sin 2b = 2cos[(a+b)]*sin[(a-b)] (1)

sin2a +sin2b = 2sin[(2a+2b)/2]*cos[(2a-2b)/2]

sin2a +sin2b = 2sin[(a+b)]*cos[(a-b)] (2)

We'll divide (1) by (2) and we'll get:

B/A = 2cos[(a+b)]*sin[(a-b)]/2sin[(a+b)]*cos[(a-b)]

B/A = cos[(a+b)]*sin[(a-b)]/sin[(a+b)]*cos[(a-b)]

cos (a+b) = cosa*cos b - sin a*sin b

sin (a-b) = sin a*cos b - sin b*cos a

cos (a+b)*sin (a-b) = sin a*cos a*(cos b)^2 - sin b*cos b*(cos a)^2 - sin b*cos b*(sin a)^2 + sin a*cos a*(sin b)^2

We'll group the 1st and the last terms and the middle terms and we'll get:

sin a*cos a*[(cos b)^2 + (sin b)^2] - sin b*cos b*[(cos a)^2 + (sin a)^2]

But (cos b)^2 + (sin b)^2 = 1 and (cos a)^2 + (sin a)^2 = 1

cos (a+b)*sin (a-b) = sin a*cos a - sin b*cos b (3)

sin[(a+b)] = sin a*cos b + sin b*cos a

cos[(a-b)] = cos a*cos b + sin a*sin b

sin[(a+b)]*cos[(a-b)] = sin a*cos a*(cos b)^2 + sin b*cos b*(sin a)^2 + sin b*cos b*(cos a)^2 + sin a*cos a*(sin b)^2

sin[(a+b)]*cos[(a-b)] = sin a*cos a*[(cos b)^2 + (sin b)^2] + sin b*cos b*[(cos a)^2 + (sin a)^2]

sin[(a+b)]*cos[(a-b)] = sin a*cos a + sin b*cos b (4)

B/A = (3)/(4)

B/A = (sin a*cos a - sin b*cos b)/(sin a*cos a + sin b*cos b )

**As we can notice, the option b) is the correct option: **

**B/A = (sin a*cos a - sin b*cos b)/(sin a*cos a + sin b*cos b )**