In the solution of part 4, why does the trig-function sin change back to cos aft

Question

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)

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