In the solution of part 4, why does the trig-function sin change back to cos after the derivative has been madeto find the velocity of the block? v = -1.00 sin (wt +) to the next portionof step 4 which is: v = -1.00 cos (wt) Shouldnt the trig-sign sin stay the same? pleaseexplain thankyou, does the phase constant have anything to do withit? thanks In the solution of part 4, why does the trig-function sin change back to cos after the derivative has been madeto find the velocity of the block? v = -1.00 sin (wt +) to the next portionof step 4 which is: v = -1.00 cos (wt) Shouldnt the trig-sign sin stay the same? pleaseexplain thankyou, does the phase constant have anything to do withit? thanks
Explanation / Answer
Yes, it's because of the phase constant. It's a trig identity.When = 90 degrees (or/2) then . sin (wt+) is the same as cos (wt)