Posted by **Kalen** on Thursday, May 17, 2012 at 6:49pm.

write y=sinx-cosx in the form y=ksin (x+a), where the measure of a is in radians. answers: sqrt2sin (x+3pi/4), sqrt2sin (x+5pi/4), sqrt2sin (x+pi/4), or sqrt2sin (x+7pi/4) ...?

- trig -
**Reiny**, Thursday, May 17, 2012 at 7:03pm
ksin(x+a) = k(sinxcosa) + cosxsina)

= ksinxcosa + kcosxsina

then

ksinxcosa + kcosxsina = sinx - cosx , which must be an identity, so true for all x

let x =0

k(0)cosa + k(1)sina = 0-1

ksina= 1

sina = 1/k

let x = 90°

k(1)(cosa) + k(0) = 1-0

cosa = 1/k

sina/cosa = (1/k) / (1/k)

tana = 1

a = 45° or π/4

then is sina = 1/k

√2/2 = 1/k

k = 2/√2 = √2

then

y = sinx - cosx

= √2sin(x+π/4)

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