trig
posted by Kalen .
write y=sinxcosx 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) ...?

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 = 01
ksina= 1
sina = 1/k
let x = 90°
k(1)(cosa) + k(0) = 10
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)