Recent content by HighCommander4

I get it now. A, B, C and D are constants so they don't depend on the value of x. So I can finally present a well-formed solution to this problem. A big thank you for all your help!

What I mean to say is, how do I know that A = C and B = D for x = pi/4 or x = 1.23456 or any other value of x?

Working with Asin(x)+Bcos(x)=Csin(x)+Dcos(x), x = 0 yields B = D x = pi/2 yields A = C But how does that prove that A = C and B = D in general?

I follow your logic, but I haven't done integration yet either. I'm basically just starting my first-year calculus course and I'm expected to solve this problem using what I already learned in high school. Is there a way of solving it using the more basic techniques taught in high school?

Thank you for your continued help. Squaring and adding does indeed make it much easier. I now get r = sqrt(7), but for theta I still get arcsin(sqrt(3/7)). Is that as good as it gets? I've done linear independence of vectors in R2 and R3, but not of functions like sine and cosine. If this...

Thank you for your reply. I'm guessing you meant to set sin(x) + 2cos(x - pi/6) equal to r sin(x + theta). I tried that and followed your instructions and this is what I got: sin(x) + 2cos(x - pi/6) = r sin(x + theta) sin(x) + 2[cos(x)cos(pi/6) + sin(x)sin(pi/6)] = r[sin(x)cos(theta) +...

I'm not sure if this is considered pre- or post-calculus (I am doing it for a calculus course, but I doubt it involves actual calculus) but I'll go ahead and post here. Homework Statement I am required to re-write the function sin x + 2 cos (x - pi/6) in the form r sin (x + theta)...