I wish I knew the answer to that one.

Beacuse of way of teachings

@@FlippingPhysics really brother you saved me many times... Your way of teaching is really awsome

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By definition, the same angle in both units should have the same sine and the same cosine, when you configure your calculator for a mode that is consistent with the angle unit. 60 degrees and pi/3 radians are both the same angle. So sin(60 deg) must equal sin(pi/3 radians), due to the fact that 60 deg = pi/3 radians. The conversion factor is 180 degrees = pi radians. To convert 60 degrees to radians, multiply by 1 in a fancy way. 1 = (pi radians)/(180 degrees). 60 deg*(pi rad)/(180 deg), degrees cancel, and we are left with pi/3 radians.

It takes the background of differential equations to derive this from first principles. However, knowing the solution in advance, we can show why this is the case, and only need the background of Calculus and Trigonometry. We desire to satisfy the equation x"(t) + K*x(t) = 0, which is the form that our equation of motion takes, when deriving it from Newton's laws. Note that K is not necessarily the spring constant, but rather the constant of proportionality between acceleration and position. In the mass/spring example, K=k/m. Solution known in advance: x(t)=A*cos(ω*t+ϕ) Use angle sum identities to convert this into sine and cosine: cos(ω*t+ϕ) = cos(ω*t) * cos(ϕ) - sin(ω*t)*sin(ϕ) Define constants B and C, such that B = A*cos(ϕ), and C = -A*sin(ϕ). Since ϕ is a constant, it can be embedded in as part of the leading constants of each trig term. We now can write our formula for x(t) as: x(t)=B*cos(ω*t) + C*sin(ω*t) Take the first derivative of x(t): x'(t) = -B*ω*sin(ω*t) + C*ω*cos(ω*t) Now take the derivative again: x"(t) = -B*ω^2*cos(ω*t) - C*ω^2*sin(ω*t) Factor out -ω^2: x"(t) = -ω^2* (B*cos(ω*t) + C*sin(ω*t)) Recognize the factor on the right side is equal to our original function for x(t), therefore: x"(t) = -ω^2 * x(t) Plug this in to the differential equation we are trying to solve: x"(t) + K*x(t) = 0 -ω^2 * x(t) + K*x(t) = 0 And you can see that when ω^2=K, our differential equation is satisfied by any linear combination of sin(ω*t) and cos(ω*t). All linear combinations of sine and cosine of the same input, can be re-written as A*cos(ω*t+ϕ). The essential difference between sine and cosine is a phase shift, as is the difference between cosine and any linear combination of sine and cosine.

@@carultch how old are you bud?

@@lawliet2263 34. Any particular reason you are asking?

@@carultch i read your other comments and could tell you're either a professor in physics or just who knows physics at a completely different level than an average joe

Best of luck on your test tomorrow! (lots of my videos are on IB topics)