Question

How do you Rewrite 2sin(x)−2cos(x) as Asin(x+ϕ)? x=? ϕ=?

Answer 1

See below.

Explanation:

Considering the general case

`a sin x+ b cos x = ?` we can proceed with calling

`c= sqrt(a^2+b^2)` and

`a sin x+ b cos x = c(a/c sin x+b/c cosx)`

but there exists by construction `phi` such that

`a/c = cos phi` and `b/c = sin phi` so

`c(a/c sin x+b/c cosx)=c(sin x cos phi + cos x sin phi)` and

`c(sin x cos phi + cos x sin phi) = c sin(x + phi)`

In our case

`2(sinx-cosx)= 2(a sin x + b cos x)` then

`{(a = 1),(b = -1):} rArr c = sqrt2` and

`cos phi = 1/sqrt2`
`sin phi = -1/sqrt2`

and also

`phi = arctan phi = arctan(-1/sqrt2,1/sqrt2) = 3/4 pi`

and finally

`2(sinx-cosx)= 2 sqrt2 sin(x + 3/4 pi)`

Answer 2

`2sqrt2sin (x - pi/4)`
`A = 2sqrt2`
`phi = - pi/4`

Explanation:

Use trig identity:
`sin a - cos a = sqrt2sin (x - pi/4)`
In this case, replace a by x -->
`2sin x - 2cos x = 2(sin x - cos x) = 2(sqrt2sin (x - pi/4))`
The answers are:
`A = 2sqrt2`
`phi = - pi/4`