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January 30, 2015

Posted by **lindsay** on Saturday, April 21, 2012 at 5:22pm.

- trigonometry -
**Reiny**, Saturday, April 21, 2012 at 5:56pmlet 3cosx - 2sinx = Rcos(x+a)

Rcos(x+a) = R(cosxcosa - sinxsina)

= Rcosxcosa - Rsinxsina

so we have the identity

Rcosxcosa - Rsinxsina = 3cosx-2sinx

this must be valid for any x

so let's pick x's that simplify this

let x = 0

then

Rcos0cosa - Rsin0sins = 3cos0 - 2sin0

Rcosa = 3

cosa = 3/R

let x = 90°

Rcos90cosa - Rsin90sina = 3cos90 - 2sin90

-Rsina = -2

sina = 2/R

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

4/R^2 + 9/R^2 = 1

R^2 = 13

R = √13

also : sina/cosa = (2/R) / (3/R) = 23

tana = 2/3

a = arctan (2/3) = 33.69°

thus 3cosx - 2sinx = √13cos(x + 33.69°)

check by taking any angle x

let x = 26°

LS =1.8196...

RS = √13 cos(5969) = 1.8196

........ how about that !!

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