Find derivative of sin(-x)
Answers
y' = - cos(-x)
Step-by-step explanation:
THE DERIVATIVE of sin x is cos x. To prove that, we will use the following identity:
sin A − sin B = 2 cos ½(A + B) sin ½(A − B).
(Topic 20 of Trigonometry.)
Problem 1. Use that identity to show:
sin (x + h) − sin x = Derivative of sin x
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sin (x + h) − sin x = 2 cos ½(x + h + x) sin ½(x + h − x)
= 2 cos ½(2x + h) sin ½h
= Derivative of sin x
Before going on to the derivative of sin x, however, we must prove a lemma; which is a preliminary, subsidiary theorem needed to prove a principle theorem. That lemma requires the following identity:
Problem 2. Show that tan θ divided by sin θ is equal to 1-cos.
tan θ
sin θ = 1
cos θ .
(See Topic 20 of Trigonometry.)
tan θ
sin θ = tan θ · 1
sin θ = sin θ
cos θ · 1
sin θ = 1
cos θ
The lemma we have to prove is discussed in Topic 14 of Trigonometry. (Take a look at it.) Here it is:
LEMMA. When θ is measured in radians, then
Derivative of sin x
Proof. It is not possible to prove that by applying the usual theorems on limits (Lesson 2). We have to go to geometry, and to the meanings of sin θ and radian measure.
Derivative of sin x
Let O be the center of a unit circle, that is, a circle of radius 1;
and let θ be the first quadrant central angle BOA, measured in radians.
Then, since arc length s = rθ, and r = 1, arc BA is equal to θ. (Topic 14 of Trigonometry.)
Draw angle B'OA equal to angle θ, thus making arc AB' equal to arc BA;
draw the straight line BB', cutting AO at P;
and draw the straight lines BC, B'C tangent to the circle.
Then
BB' < arc BAB' < BC + CB'.
Derivative of sin x
Now, in that unit circle, BP = PB' = sin θ, (Topic 17 of Trigonometry),
so that BB' = 2 sin θ;
and BC = CB' = tan θ. (For, tan θ = BC
OB = BC
1 = BC.)
The continued inequality above therefore becomes:
2 sin θ < 2θ < 2 tan θ.
On dividing each term by 2 sin θ:
1 < θ
sin θ < 1
cos θ .
(Problem 2.) And on taking reciprocals, thus changing the sense:
1 > sin θ
θ > cos θ.
(Lesson 11 of Algebra, Theorem 5.)
On changing the signs, the sense changes again :
−1 < − sin θ
θ < −cos θ,
(Lesson 11 of Algebra, Theorem 4),
and if we add 1 to each term:
0 < 1 − sin θ
θ < 1 − cos θ.
Now, as θ becomes very close to 0 (θ Right arrow0), cos θ becomes very close to 1; therefore, 1 − cos θ becomes very close to 0. The expression in the middle, being less than 1 − cos θ, becomes even closer to 0 (and on the left is bounded by 0), therefore the expression in the middle will definitely approach 0. This means:
Derivative of sin x
Which is what we wanted to prove.
The student should keep in mind that for a variable to "approach" 0 or any limit (Definition 2.1), does not mean that the variable ever equals that limit.
The derivative of sin x
d
dx sin x = cos x
To prove that, we will apply the definition of the derivative (Lesson 5). First, we will calculate the difference quotient.
Derivative of sin x = Derivative of sin x , Problem 1,
= Derivative of sin x , on dividing numerator
and denominator by 2,
= Derivative of sin x
We will now take the limit as h Right arrow0. But the limit of a product is equal to the product of the limits. (Lesson 2.) And the factor on the right has the form sin θ/θ. Therefore, according to the Lemma, as h Right arrow0 its limit is 1. Therefore,
d
dx sin x = cos x.
We have established the formula.
The derivative of cos x
d
dx cos x = −sin x
To establish that, we will use the following identity:
cos x = sin ( π
2 − x).
A function of any angle is equal to the cofunction of its complement.
(Topic 3 of Trigonometry).
Therefore, on applying the chain rule:
Derivative of cos x
We have established the formula.
The derivative of tan x
d
dx tan x = sec2x
Now, tan x = sin x
cos x . (Topic 20 of Trigonometry.)
Therefore according to the quotient rule:
d
dx tan x = d
dx sin x
cos x = cos x· cos x − sin x(−sin x)
cos2x
= cos2x + sin2x
cos2x
= 1
cos2x
= sec2x.
We have established the formula.
Problem 3. The derivative of cot x. Prove:
d
dx cot x = −csc2x
d
dx cot x = d
dx cos x
sin x
= sin x(−sin x) − cosx· cos x
sin2x
= −(sin2x + cos2x)
sin2x
= − 1
sin2x
= −csc2x.
The derivative of sec x
d
dx sec x = sec x tan x
Since sec x = 1
cos x = (cos x)−1 ,
then on using the chain rule and the general power rule:
Derivative of sec x
We have established the formula.