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
2 comments :
Cool, I wasn't sure about the first formula and google led me here. thanks :D
help me to solve
cot x - tan x=sin2x for 0<=x<=180
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