sin (x+pi/4) = root2 cos x for 0<x<2pi

[fakinmath]

hey...this is for advanced functions chapter problem....nd i reallly don't get what to do ...can someone please help..
solve:
sin (x+pi/4) = root2 cos x for 0<x<2pi

 

[Deleted member 4993]

Re: sin (x+pi/4) = root2 cox x fro 0<x<2pi

hey...this is for advanced functions chapter problem....nd i reallly don't get what to do ...can someone please help..
solve:
sin (x+pi/4) = root2 cox x fro 0<x<2pi <<< What is that one?

Use:

sin (A+B) = sinA * cosB + cosA * sinB

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[fakinmath]

i got stuck at the beginning...i did:
sin (x+y) = sinx cosy + cosx siny
sinx coxpi/4 + cosx sinpi/4 = root cos
since cospi/4 = sin pi/4
cospi/4 (sinx+cosx) = root 2 cosX
i don't know what to do next

 

[Deleted member 4993]

i got stuck at the beginning...i did:
sin (x+y) = sinx cosy + cosx siny
sinx coxpi/4 + cosx sinpi/4 = root cos
since cospi/4 = sin pi/4 = 1/?2

[sin(x) + cos(x)]/?2 = ?2 * cos(x)

sin(x) + cos(x) = 2 * cos(x)

now continue...


cospi/4 (sinx+cosx) = root 2 cosX
i don't know what to do next