# Simplify in cos2a+cos4a+i(sin4a-sin2a)/cos2a+cos4a-i(sin4a-sin2a)

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### 1 Answer

You should convert the sum `cos2a+cos4a` and the difference `sin4a-sin2a` into a product such that:

`cos2a+cos4a = 2[cos(2a+4a)/2]*[cos(2a-4a)/2]`

`cos2a+cos4a = 2cos 3a*cos (-a)`

You need to remember that cosine function is even, hence `cos (-a) = cos a.` `sin4a-sin2a = 2 [cos(2a+4a)/2]*[sin(2a-4a)/2]`

`sin4a-sin2a = 2cos 3a*sin(-a)`

You need to remember that sine function is odd, hence `sin (-a) = -sin a` .

`sin4a-sin2a = -2cos 3a*sin a`

Hence, you need to write the fraction using the products instead of sums such that:

`(2cos 3a*cos a- 2i*cos 3a*sin a)/(2cos 3a*cos a + 2i*cos 3a*sin a)`

Factoring out `2 cos 3a` yields:

`(2 cos 3a*(cos a - i*sin a))/(2 cos 3a*(cos a+ i*sin a))`

Reducing like terms yields:

`(cos a - i*sin a)/(cos a + i*sin a)`

You need to multiply the fraction by `cos a - i*sin a` such that:

`(cos a - i*sin a)^2/(cos^2 a + sin^2 a)`

You need to use the basic formula of trigonometry such that:

`cos^2 a + sin^2 a = 1`

`(cos a - i*sin a)^2/(cos^2 a + sin^2 a) = (cos a - i*sin a)^2`

Expanding the binomial yields:

`(cos a - i*sin a)^2 = cos^2 a - 2i*sin a*cos a - sin^2 a`

`(cos a - i*sin a)^2 = (cos^2 a - sin^2 a) - i*sin 2a`

`(cos a - i*sin a)^2 = cos 2a - i*sin 2a`

**Hence, reducing fraction to its lowest terms yields `cos 2a - i*sin 2a.` **