Question

why differentiation of sinx = cosx?

Answer 1

If you use the limit definition of differentiation,
you could see why this is true :)
it's a little tricky though.

Answer 2

\[\large\rm (\sin x)'=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}\]

Answer 3

We need to apply our Angle Sum Formula for Sine,\[\large\rm \sin(a+b)=\sin a \cos b+\sin b \cos a\]

Answer 4

\[\large\rm (\sin x)'=\lim_{h\to0}\frac{\sin x \cos h+\sin h \cos x-\sin x}{h}\]

Answer 5

Break it up into two fractions,\[\large\rm (\sin x)'=\lim_{h\to0}\frac{\sin x \cos h-\sin x}{h}+\frac{\sin h \cos x}{h}\]Factor a sin x out of the first fraction,
and write the cos x in front of the second fraction,\[\large\rm (\sin x)'=\lim_{h\to0}\quad \sin x\left(\frac{\cos h-1}{h}\right)1+\cos x\left(\frac{\sin h }{h}\right)\]

Answer 6

Rewrite the limit of this sum as the sum of limits,\[\rm (\sin x)'=\lim_{h\to0}~\sin x\left(\frac{\cos h-1}{h}\right)1+\lim_{h\to0}~\cos x\left(\frac{\sin h }{h}\right)\]You can pull the x stuff out of the limits,\[\rm (\sin x)'=\sin x\left(\lim_{h\to0}\frac{\cos h-1}{h}\right)1+\cos x\left(\lim_{h\to0}\frac{\sin h }{h}\right)\]And then apply a couple of limit identities:\[\large\rm \lim_{h\to0}\frac{1-\cos h}{h}=0,\qquad\qquad\qquad \lim_{h\to0}\frac{\sin h}{h}=1\]

Answer 7

Once you've remembered that sine and cosine's derivatives are all related, you might forget which one is which. If you can imagine the graphs of sine and cosine however, it makes it a lot easier.

So let's look at the graph of sine and figure out if its derivative is +cosine or -cosine.


So look, from left to right we see that the sine curve has positive slope up, then gets flat (zero slope), then curves down, and then is flat again. Then it all repeats.

So right at x=0 the graph of sine looks like the graph of \(y=x\) which we know has derivative 1. Similarly, which trig function is 1 at x=0? \(+\cos(x)\). I think this geometric interpretation is one of the most important so I hope you spend some time thinking about this and making sure you see why it must be true!