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proof of angle sum identities∗
rspuzio†
2013-03-21 23:34:02
We will derive the angle sum identities for the various trigonometric functions
here. We begin by deriving the identity for the sine by means of a geometric
argument and then obtain the remaining identities by algebraic manipulation.
Theorem 1.
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
Proof. Let us make the restrictions 0◦ < x < 90◦ and 0◦ < y < 90◦ for the
time being. Then we may draw a triangle ABC such that ∠CAB = x and
∠ABF = y:
, (0, 0); (70, 0)∗∗@−; (40, 30)∗∗@−; (0, 0)∗∗@−, (−2, −2)∗A, (72, −2)∗B, (40, 32)∗C
Since the angles of a triangle add up to 180◦ , we must have ∠BCA = 180◦ −x−y,
so we have sin(∠BCA) = sin(180◦ − x − y) = sin(x + y).
We now draw perpendiculars two different ways in order to derive ratios.
First, we drop a perpendicular AD from C to AB:
, (0, 0); (70, 0)∗∗@−; (40, 30)∗∗@−; (0, 0)∗∗@−, (40, 30); (40, 0)∗∗@−, (−2, −2)∗A, (72, −2)∗B, (40, 32)∗C, (40, −2
Since ACD and BCD are right triangles we have, by definition,
cot(∠CAB) = AD/CD
cot(∠ABC) = BD/CD
sin(∠CAB) = CD/AC.
Second, we draw a perpendicular AE form A to BC. Depending on whether
x + y < 90◦ or x + y < 90◦ the point E will or will not lie between B and C, as
illustrated below. (There is also the case x + y = 90◦ , but it is trivial.)
, (0, 0); (70, 0)∗∗@−; (30, 40)∗∗@−; (0, 0)∗∗@−, (0, 0); (35, 35)∗∗@−, (−2, −2)∗A, (72, −2)∗B, (30, 42)∗C, (36, 36)
, (0, 0); (70, 0)∗∗@−; (35, 35)∗∗@−; (0, 0)∗∗@−, (0, 0); (40, 30)∗∗@−, (−2, −2)∗A, (72, −2)∗B, (41, 31)∗C, (35, 37)
Either way, ABE and ACE are right triangles, and we have, by definition,
sin(∠BCA) = AE/AC
sin(∠ABC) = AE/AB.
∗ hProofOfAngleSumIdentitiesi
created: h2013-03-21i by: hrspuzioi version: h39798i
Privacy setting: h1i hProofi h43-00i h51-00i h42-00i h33B10i
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Combining these ratios, we find that
sin(∠BCA)/ sin(∠ABC) = AB/AC.
To finish deriving the sum identity, we manipulate the ratios derived above
algebraically and use the fact that AD + BD = AB:
sin(x + y) = sin(∠BCA) = AB sin(∠ABC)/AC
= (AD + BD) sin(∠ABC)/AC
= CD (cot(∠CAB) + cot(∠ABC)) / sin(∠ABC)AC
cos(∠CAB) cos(∠ABC)
+
= sin(∠CAB) sin(∠ABC)
sin(∠CAB)
sin(∠ABC)
= sin(∠CAB) cos(∠ABC) + cos(∠CAB) sin(∠ABC)
= sin(x) cos(y) + cos(x) sin(y)
To lift the restriction on the range of x and y, we use the identities for
complements and negatives of angles.
Entry under construction
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