Derivative of sine and cosine

In today's blog, I bring together some of the results that I presented earlier to determine the derivative for sine and cosine.

Lemma 1: sin(A+B) - sin(A - B) = 2*cos(A)sin(B)

Proof:

(1) sin(A+B) = cos(B)*sin(A) + cos(A)*sin(B) [See here for proof]

(2) sin(A-B) = sin(A+(-B)) = cos(-B)*sin(A) + cos(A)*sin(-B) =

(3) since cos(-B) = cos(B) [See here] and sin(-B) = -sin(B) [See here], we get:

sin(A-B) = cos(B)*sin(A) - cos(A)*sin(B)

(4) sin(A+B) - sin(A-B) =

= cos(B)*sin(A) + cos(A)*sin(B) - cos(B)*sin(A) + cos(A)*sin(B) =

= 2*cos(A)*sin(B)

QED

Lemma 2: sin P - sin Q = 2*cos [(P+Q)/2 ] * sin[(P-Q)/2]

Proof:

(1) Let A = (P+Q)/2

(2) Let B = (P-Q)/2

(3) A+B = (P+Q)/2 + (P-Q)/2 = (P+P+Q-Q)/2 = P

(4) A - B = (P+Q)/2 - (P-Q)/2 = (P-P+Q+Q)/2 = Q

(5) sin(P) - sin(Q) = sin(A+B) - sin(A-B) = 2*cos(A)*sin(B) [See Lemma 1 above]

(6) Putting it all together gives us:

sin(P) - sin(Q) = 2*cos(A)*sin(B) = 2*cos[(P+Q)/2]*sin[(P-Q)/2]

QED

Theorem 1: d/dx(sin x) = cos x

Proof:

(1) Let y = sin(x)

(2) dy/dx = lim (Δx → 0) [sin(x + Δx) - sin(x)]/Δx

(3) From Lemma 2 above we know that:
sin(x+Δx) - sin(x) = 2*cos[(x+Δx+x)/2]*sin[(x+Δx-x)/2] =
= 2*cos(x + Δx/2)*sin(Δx/2)

(4) So, substituting (3) into (2) gives us:

dy/dx = lim(Δx → 0) [ 2*cos(x+Δx/2)*sin(Δx/2) ]

(5) Using the Product Law (see here), we know that:

lim(Δx → 0) [ 2*cos(x+Δx/2)*sin(Δx/2) ] =
lim(Δx → 0)[cos(x+Δx/2)] * lim(Δx → 0)[sin(Δx/2)/(Δx/2)]

(6) Now, if we set θ = Δx/2, we know that:
lim (θ → 0) [ sin(θ)/θ ] = 1 (See here for proof)

(7) We also know that
lim(Δx → 0)[cos(x + Δx/2)] = cos x

(8) This then gives us:
dy/dx = cos x * 1 = cos x.

QED

Lemma 3: cos(A+B) - cos(A-B) = -2*sin(A)sin(B)

Proof:

(1) cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) [See here for proof]

(2) cos(A-B) = cos(A+(-B)) = cos(A)*cos(-B)-sin(A)*sin(-B)

(3) Since cos(-B) = cos(B) and sin(-B) = -sin(B) [See here], we have:

cos(A-B) = cos(A)*cos(B)+sin(A)*sin(B)

(4) cos(A+B) - cos(A-B) =

=cos(A)*cos(B) - sin(A)*sin(B) - cos(A)*cos(B) - sin(A)*sin(B) =

= -2*sin(A)*sin(B)

QED

Lemma 4: cos P - cos Q = -2*sin[(P+Q)/2]*sin[(P-Q)/2]

Proof:

(1) Let A = (P+Q)/2

(2) Let B = (P-Q)/2

(3) A + B = (P+Q)/2 + (P-Q)/2 = (P+Q+P-Q)/2 = P

(4) A - B = (P+Q)/2 - (P-Q)/2 = (P + Q - P + Q)/2 = Q

(5) cos(P) - cos(Q) = cos(A+B) - cos(A-B) = -2*sin(A)*sin(B) [From Lemma 3 above]

(6) So,

cos(P) - cos(Q) = -2*sin(A)*sin(B) = -2*sin[(P+Q)/2]*sin[(P-Q)/2]

QED

Theorem 2: d/dx(cos x) = -sin x

Proof:

(1) Let y = cos(x)

(2) dy/dx = lim(Δx → 0) [ cos(x + Δ x) - cos(x) ]/Δx

(3) Now, from Lemma 4 above:

cos(x +Δx) - cos(x) = -2*sin[(x + Δx + x)/2]*sin[(x+Δx-x)/2] =
= -2*sin(x + Δx/2)*sin(Δx/2)

(4) Once again, applying the Product Rule for limits gives us:

dy/dx = lim(Δx → 0) [ cos(x + Δ x) - cos(x) ]/Δx =
lim(Δx → 0)[ -sin(x + Δx/2) ] * lim(Δx → 0) [ sin (Δx/2)/(Δx/2)]

(5) Again, setting θ = Δx/2 gives us:
lim(Δx → 0)[ sin(θ)/θ ] = 1 [See here for proof]

(6) We can also see that:
lim(Δx → 0)[ -sin(x + Δ x/2) ] = -sin(x)

(7) Putting this all together gives us:
dy/dx = -sin(x)*1 = -sin(x)

QED