tan(π/4) = 1

Lemma: tan π/4 = 1

Proof:

(1) Using definition:

tan(π/4) = sin(π/4)/cos(π/4)

(2) Using the triangle definition of sin and cosine (see here), we know that:

sin (x) = cos (π/2 - x)

(3) This then gives us that:

sin (π/4) = cos(π/2 - π/4) = cos(2π/4 - π/4) = cos(π/4)

(4) So that:

tan(π/4) = cos(π/4)/cos(π/4) = 1

QED

cot 2x = (1/2) [cot x - tan x]

Lemma 1: cot 2x = (1/2)[cot x - tan x]

Proof:

(1) cot 2x = 1/tan(2x) = cos(2x)/sin(2x) [See here for definition of tan]

(2) cos(2x) = cos²(x) - sin²(x) [See Lemma 3, here]

(3) sin(2x) = 2(sin x)(cos x) [See Lemma 2, here]

(4) cos(2x)/sin(2x) = [cos²(x) - sin²(x)]/2(sin x)(cos x) =

= (1/2)[cos(x)/sin(x) - sin(x)/cos(x)] = (1/2)[cot(x) - tan(x)]

QED

Lemma 2: tan(-b) = -tan b

Proof:

(1) tan(-b) = sin(-b)/cos(-b) [See here for definition of tangent]

(2) sin(-b) = -sin b [See Property 4, here]

(3) cos(-b) = cos b [See Property 9, here]

(4) So, tan(-b) = (-sin b)/(cos b) = (-1)(sin b)/(cos b) = (-1)tan b = -tan b.

QED

Lemma 3: tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)]

Proof:

(1) tan(a + b) = sin(a + b)/cos(a + b) [See here for definition of tangent]

(2) sin(a + b) = (sin a)(cos b) + (cos a)(sin b) [See Theorem 1, here]

(3) cos(a + b) = (cos a)(cos b) - (sin a)(sin b) [See Theorem 2, here]

(4) So that:

sin(a + b)/cos(a + b) = [(sin a)(cos b) + (cos a)(sin b)]/[(cos a)(cos b) - (sin a)(sin b)] =

(sin a)(cos b)/[(cos a)(cos b) - (sin a)(sin b)] + (cos a)(sin b)/[(cos a)(cos b) - (sin a)(sin b)]

(5) Multiplying both sides of the fractions by 1/(cos a)(cos b) gives us:

(tan a)/[1 - (tan a)(tan b)] + (tan b)/[1 - (tan a)(tan b)] =

= [tan a + tan b ]/[1 - (tan a)(tan b)]

QED

Corollary 3.1: tan(a - b) = [tan a - tan b]/[1 + (tan a)(tan b)]

Proof:

(1) tan(a - b) = tan (a + (-b)) = [tan a + tan (-b)]/[1 - (tan a)(tan -b)]

(2) Using Theorem 2 above, we have:

[tan a + tan (-b)]/[1 - (tan a)(tan -b)] = [tan a - tan b]/[1 + (tan a)(tan b)]

QED

Lemma 4: tan (2x) = (2 tan x)/(1 - tan² x)

Proof:

(1) tan(2x) = tan(x + x)

(2) Using Theorem 3 above:

tan(x + x) = [tan x + tan x]/[1 - (tan x)(tan x)] =

= (2 tan x)/(1 - tan² x)

QED

Lemma 5: 2 cos mx cos nx = cos(m + n)x + cos(m -n)x

Proof:

(1) Using cos(a + b) = cos(a)cos(b) - sin(a)sin(b) [See Theorem 2, here]

cos(m+n)x = cos(mx + nx) = cos(mx)cos(nx) - sin(mx)sin(nx)

cos(m-n)x = cos(mx - nx) = cos(mx)cos(-nx) - sin(mx)sin(-nx)

(2) Using cos(-x) = cos(x) [see Property 9, here] and sin(-x) = -sin(x) [see Property 4, here], we have:

cos(m - n)x = cos(mx)cos(nx) + sin(mx)sin(nx)

(3) So:

cos(m+n)x + cos(m-n)x = cos(mx)cos(nx) - sin(mx)sin(nx) + cos(mx)cos(nx) + sin(mx)sin(nx) =
2 cos(mx)cos(nx)

QED