Suppose that at any given time t (in second) the current i (in amperes) in an alternation current circuit is i=2cost+2sint
What is the peak current for this circuit. ( The largest magnitude)
Thanks in advance
Have you tried finding the maximum point of i by differentiating it? Alternatively, you can convert sin(t) + cos(t) to just sine using these formulas.
This is a calculus subforum. One of the best-known facts in calculus is Fermat's theorem.
Also, I assume you know how to find the maximum of $\displaystyle A\sin(t+\varphi)$ for some constants A and $\displaystyle \varphi$. The link in post #2 shows a formula for converting 2sin(t) + 2cos(t) into $\displaystyle Asin(t+\varphi)$ for some A and $\displaystyle \varphi$.
If this is not clear, please explain precisely what your difficulty is and what methods you know for finding maxima of functions.
So you have a problem like this and have never taken Calculus? That makes it a little harder but it can be done.
You know, I hope, that sin(a+ b)= cos(a)sin(b)+sin(a)cos(b). Taking b= x, you could use that immediately if you could have cos(a)= sin(a)= 2 but that is impossible because that would mean $\displaystyle cos^2(a)+ sin^2(a)= 4+ 4= 8$ which, of course, can't be true because $\displaystyle cos^2(a)+ sin^2(a)= 1$ for all a.
But we can fix that: rewrite 2sin(x)+ 2cos(x) as $\displaystyle 2\sqrt{2}(1/(\sqrt{2})cos(x)+ 1/(\sqrt{3})sin(x))$. Now, we can say that $\displaystyle cos(a)= sin(a)= 1/(\sqrt{2})$ because then $\displaystyle sin^2(a)+ cos^2(a)= 1/2+ 1/2= 1$ as required.
So we can write
$\displaystyle 2sin(x)+ 2cos(x)= 2\sqrt{2}(1/\sqrt{2}cos(x)+ 1/\sqrt{2}sin(x))$
$\displaystyle = 2\sqrt{2}(sin(\pi/4)cos(x)+ cos(\pi/4)sin(x))= 2\sqrt{2}sin(x+ \pi/4)$.
Now use the fact that cos(x) has a maximum value of 1.