Graph of sine function with tangent at (0,0). 
Graph of cosine function with tangent at (0,1). 

Examining Figure 1, for the sine function, we see the slope is positive and the tangent line appears to be making an angle of about 45° with the Xaxis. This suggests that its slope is close to or equal to $1$. 
Examining Figure 2, for the cosine function, we see the tangent line appears to be horizontal. This suggests that its slope is close to or equal to 0. 

In another informal approach, we can make a table of estimates for the values of the derivatives using the definition. 

Recall that $\sin(0) = 0$, so the difference quotient simplifies to
$\frac {\sin(h)} h$.
Table I.F.3.1 gives some values for this difference
quotient when $h$ is close to $0$. These estimates suggest that the
limit is equal to $1$.
Table I.F.3.1 
Recall that $\cos(0) = 1$, so the difference quotient simplifies to
$\frac {\cos(h)  1} h$.Table I.F.3.2 gives some values for this difference
quotient when $h$ is close to $0$. These estimates suggest that the
limit is equal to $0$.
Table I.F.3.2 
Proof: i) First let's consider the case when $h > 0$ and think about $\sin(h)$ as visualized in a unit circle. See Figure 5. The
sine value is represented by the vertical line segment from the Xaxis
to the point on the circle determined by the arc on the circle of length
$h$. From this figure, at least, it seems to make sense that $\sin(h)
< h$. In Figure 5, $h$ is the length of the arc with one endpoint $Q$ and the other endpoint $P$ with coordinates $(\cos(h),\sin(h))$. Looking further at Figure 5 you can change $h$ by moving the point $P$ closer to $Q$ on the arc to see at what happens near the point $Q (1,0)$. Check the box to add the chord $PQ$ to the figure. The length of the chord connecting $Q$ to $P$ certainly must be less than h because the shortest distance between two points is determined by the line segment connecting the points. But the chord $PQ$ is the hypotenuse of the right triangle with its right angle at the point $R (\cos(h),0)$. Therefore the chord $PQ$ is longer than the vertical line segment $PR$, which has length $\sin(h)$. Thus, $\sin(h) < h$ when $h > 0$, so when we divide by $h$ we have $\frac {\sin(h)} h < 1$. Now check the box "Show Tangent to Circle" in Figure 5 and turn your attention to the right triangle formed with right angle at $Q (1,0)$, and an angle of measure $h$ at $(0,0)$. The altitude of this triangle is $\tan(h)$. [Can you explain why?$^1$] Therefore the area of this triangle is $\frac 1 2 \cdot 1 \cdot \tan(h) = \frac 1 2 \cdot \frac {\sin(h)} {\cos(h)}$. 
Figure 5

But this triangle contains the sector of the circle formed by the line
segment $QO$ and making an angle $h$ at $(0,0)$. This sector has an area
of $\frac 1 2 \cdot h$. Since the area of the sector is less than the area of the triangle, we have $\frac 1 2 \cdot h <\frac 1 2 \cdot \frac {\sin(h)} {\cos(h)}$. (2)
Since $h > 0$ and $\cos(h) > 0$ when $h$ is close to $0$, we can multiply both sides of the inequality (2) by $2\cos(h)$ and divide by $h$ to obtain the new inequality visualized in Figure 6:
$\cos(h) < \frac {\sin(h)} h < 1$. (4)
At this point it is time to think! (Step IV.) $\lim_{h \to 0^+} \frac {\sin(h)} h = 1$. (5) Now what about the case when $h < 0$ and $h \to 0$?
In this case the data suggest that $\frac {\sin(h)} h = \frac {\sin(h)}{h}$. To see why this is true, let $k = h$. Recall that $\sin(h) = \sin(h)$, so $\sin(k) = \sin(h)$. When $h<0, k = h > 0$, and if $h \to 0$, then $k = h \to 0$ as well. Thus we see that $\frac {\sin(h)} h = \frac {\sin(h)}{h} = \frac {sin(k)} k$.
Our previous analysis (5) showed that when $k > 0$ and $k\to 0$, $\frac {\sin(k)} k \to 1$. So when $h < 0$ and $h \to 0$, $\frac {\sin(h)} h= \frac {\sin(k)} k \to 1$, i.e., $\lim_{h \to 0^} \frac {\sin(h)} h = 1$. (6)
Check this as well visually with Figure 6.

Figure 6

Here are two useful identities from trigonometry, based on the addition formulae for sine
and cosine. These will be used in one of our proofs when we get to Step II (subtract) in the four
step method for finding the derivative.
Lemma I.F.8:
i) $\sin(x)  \sin(a)=2 \sin(\frac 12 (x  a)) \cos(\frac12 (x + a))$.
ii) $\cos(x)  \cos(a)=  2\sin(\frac12(x  a)) \sin(\frac12(x+a))$.
Proof 1 of i) Remark: Using Lemma I.F.8 it is not too hard to find the general derivatives of the sine and cosine. We will give the argument for the derivative of the sine, leaving the proof for the cosine using the lemma as an exercise for the reader. In finding the derivative we'll use the
four step method based on the definition
with $f (x) = \sin(x)$. For the purpose of the Step IV analysis (Think), we first consider the factor $\frac { \sin(\frac 12 (x  a))} {\frac 12 (xa)}$. Notice that when $x$ is close to $a$, the expression $\frac 12 (x a)$ is close to $0$. Think of $\frac 12 (x a)$ as a number which we can call $h$. Now the factor is $\frac { \sin(h)} h$. It's time to think! When $x \to a$ , $h \to 0$. Our work
in finding $\sin'(0)$ showed that $\frac { \sin(h)} h \to 1 $, or
equivalently, $\frac { \sin(\frac 12 (x  a))} {\frac 12 (xa)} \to 1$. Using the definition of the cosine function on the unit circle should make it sensible to see that when $x \to a$ , $\frac 12 (x+ a) \to a$, and thus $\cos( \frac 12 (x + a)) \to \cos(a)$. Putting this all together, we have $f'(a) = \lim_{x \to a}\frac {f(x)f(a)} {xa}$ 
Proof 2 of i) Remark: In this proof we make use of the addition identity: $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$ In finding the derivative this time we'll use the
four step method based on the definition
with $f (x) = \sin(x)$. Thus the difference quotient in Step III (divide) is
For the purpose of the Step IV analysis (Think),
we first consider the term $ \cos(a)\frac { \sin(h) }h$. Now consider the term $\sin(a) \frac {[\cos(h) 1]} h$ . Our recent work in finding $\cos'(0)$ showed that $\frac {\cos(h)1} h \to 0$, so it should make sense that $\sin(a) \frac {[\cos(h) 1]} h \to 0$ . Putting this all together, we have $f'(a) = \lim_{h \to 0}\frac {f(a+h)f(a)} {h}$
So replacing a with x we have the general result: $D_x\sin(x) = \sin'(x) =\cos(x)$. EOP. 
Proof of ii) using $h\to 0$: We'll use the four step method based on the definition with $f (x) = \cos(x)$.
For the purpose of the Step IV analysis (Think),
we first consider the term $ \sin(a)\frac { \sin(h) }h$. Now consider the term $\cos(a) \frac {[\cos(h) 1]} h$ . Our recent work in finding $\cos'(0)$ showed that $\frac {\cos(h)1} h \to 0$, so it should make sense that $\cos(a) \frac {[\cos(h) 1]} h \to 0$ . Putting this all together, we have $f'(a) = \lim_{h \to
0}\frac {f(a+h)f(a)}
{h}$

Summary of derivatives for sine and cosine function 



Graphical Interpretation of the Sine and Its Derivative: Note that the value of the sine’s derivative is $0$ when $x$ is $ \pm \frac {\pi} 2$. For the graph of the sine at these same $x$’s, it seems the corresponding tangent lines would be horizontal. 
Graphical Interpretation of the Cosine and Its Derivative: Note that the value of the cosine’s derivative is $0$ when $x$ is $0$ and $\pm \pi$. For the graph of the cosine at these same $x$’s, it seems the corresponding tangent lines would be horizontal.. 
Solution: For the tangent line at $x = a$ to be horizontal, it must have a slope of $0$. So we are looking for numbers $a$ between $0$ and $2\pi$ where the derivative of $y$ is $0$ at $x = a$. Now from the derivative calculus we have that $\frac{dy}{dx} = \frac d{dx}\sin(x)  \frac d{dx} \cos(x) = \cos(x) +\sin(x)$ . Setting this derivative to $0$ at $x = a$, we see that $\cos(a) + \sin(a) = 0$. Solving this trigonometric equation we find that for this value of $a$, $\cos(a) =  \sin(a)$, and thus $\tan (a) = 1$. So for the tangent line at $x = a$ to be horizontal with $a$ between $0$ and $2\pi$, it must be that $a$ is either $\frac 34 \pi$ or $\frac 74 \pi$. See Figure 11.
Figure 11

I.F.3.a [Optional] The Derivative of The Sine: A More Geometric View.
Here is an alternative demonstration that $\sin’(a) = \cos(a)$.
Consider sin(x) and cos(x) as defined on the unit circle. That is, an arc on the unit circle
starting at (1,0) and having length x has its endpoint with coordinates (cos(x), sin(x)). Similarly
the arc of length a on the unit circle has its endpoint with coordinates (cos(a), sin(a)) .
When x is close to a , the value of sin(x)  sin(a) represents the length of the vertical side S of a
right triangle PQR with vertices P(cos(x),sin(x)), Q (cos(a),sin(a)) and R (cos(x),sin(a)). [See
Figure 12.] The difference xa represents the difference in the lengths between arc x and arc a.
When x is very close to a, x  a is approximately equal to the length of the hypotenuse H of
triangle PQR, i.e, the length of the chord PQ.
Thus $\frac {\sin(x)\sin(a)} {xa}$ is approximated by S/H which is the cosine of the angle RPQ (formed at P) which we'll call θ, i.e.,$\frac {\sin(x)\sin(a)} {xa} \approx \frac SH= \cos(\Theta)$.
Now θ depends on the choice of x in a very useful fashion. When x is very close to a, the angle PQO (formed at Q) is almost a right angle. θ is the complement of the angle PQR labeled φ in Figure 12. Since RQ is parallel to the horizontal axis, the angle RQO has the same measure as angle QOU. So these two angles are both labeled a in Figure 12.
Now we notice that when x is very close to a, angle RQO is close to being the complement of the angle PQR labeled φ, and thus the angle θ has a measure that is close to a. So, when x is close to a, cos (θ) is close to the cos(a).
Thus as x→a and we see that sin'(a) = cos(a). EOP
Exercises I.F.3.
1. Find $\frac{dy}{dx}$ when
a. $y = 5 \sin(x)  3
\cos(x)$; $y =
\ln(x^2) + \sin (x)$
b. $ y = \cos(x)  5x  3$ ; $ y = 3 \sin (x) + \frac 1x$
2. Find all $x$ where $f'(x) = 0$ when
a. $f(x) = \sin(x)$
d.
$f(x) = \sin(x) + x$
b. $f(x) = \cos(x)$
e. $f(x) =
\cos(x)  x$
c. $f(x) = \sin(x) + \cos(x)$
3. Consider $F(x) = \sin(x)$ is a probability
distribution for the random variable $X$ on the interval $[0, π/2]$.
Find the probability density at $x=π/4, π/6,$ and $π/3$. Explain briefly
why the values for $X$ are more likely to fall closer to $π/6$ then to
$π/4$ or $π/3$.
4. Prove Lemma ii). Use this to prove Theorem I.F.9.ii).
5. Give a geometric analysis for showing $\cos'(x) = \sin(x)$.
[Hint: Discuss why $\frac{\cos(x)\cos(a)}{xa} \approx
\sin(\Theta)$. How do you explain the "$$" in the geometry?] Now
complete the argument showing that $\frac{\cos(x)\cos(a)} {xa}\approx
\sin(θ) \to \sin(a)$.
6. Challenge: Give a geometric analysis for showing $\tan'(x) = sec^2(x)$.
[Hint: Discuss why $\frac{tan(x)tan(a)} {xa} \approx
\sec^2(a)$. How do you explain the term $\tan(a)$ in the geometry?] Now
complete the argument.
7. Challenge: Give a geometric analysis for showing $\sec'(x) = \sec(x)\tan(x)$.
[Hint: Discuss why $\frac{\sec(x)\sec(a)}{xa} \approx
\sec(a)\tan(a)$. How do you explain the term $\sec(a)$ in the geometry?]
Now complete the argument.
8. Find a function P where $ P'(x) = f(x)$ for the following:
a. $f(x) = 5 \cos(x)  \frac 3{x^2}
+ 2x^{ 3/5}$ b. $ f(x) = 2 \sin(x)  3x^2 +
2x^{ 5/7}$ .
9. Explain why $\frac {d\sin(x)}{dx} = \cos(x)$ but
$\sin'(x)=\cos(x)$ and $\frac {d\cos(x)}{dx} = \sin(x)$ while
$\cos'(x)=\sin(x)$..
Discuss these results with regard to the graphical interpretation of these derivative.
10. Use the addition identity to prove that
a. $\sin(x+π/2) = \cos(x); \sin(x+π) = \sin(x)$;
b. $\cos(x+π/2) =\sin(x); \cos(x+π) =\cos(x)$.
Use the identities in parts a) and b) to explain why
c. $ \frac {d\sin(x+π/2)}{dx} = \sin(x); \frac {d\sin(x+π)}{dx}= \cos(x)$;
d. $\frac{d\cos(x+π/2)}{dx} =\cos(x); \frac{d\cos(x+π)}{dx} = \sin(x)$.
Discuss these results with regard to the graphical interpretation of these derivative.