Difference between revisions of "Angle addition identities"

(Created page with "The trigonometric angle addition identities state the following identities: <math>\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)</math> <math>\cos(x + y) = \cos (x) \cos...")
 
(Proofs)
 
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<math>\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)</math>
 
<math>\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)</math>
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<math>\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)</math>
 
<math>\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)</math>
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<math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}</math>
 
<math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}</math>
There are many proofs of these identities. For the sake of brevity, we list only one here.
 
  
Euler's identity states that <math>e^{ix} = \cos (x) + i \sin(x)</math>. We have that\begin{align*} \cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \ &= e^{ix} \cdot e^{iy} \ &= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \ &= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y)) \end{align*}By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.
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{{stub}}
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==Proofs==
  
To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by <math>\cos (x) \cos (y)</math>, and simplify.\begin{align*} \tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \ &= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \ &= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(y)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \ &= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)} \end{align*}as desired.
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<asy>
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unitsize(216);
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real d = 1/cos(radians(35));
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real d1 = d * cos(radians(55));
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real d2 = d * sin(radians(55));
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pair O = (0,0);
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pair A = (cos(radians(20)),0);
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pair B = (cos(radians(20)),sin(radians(20)));
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pair C = (cos(radians(20)),d2);
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pair D = (d1,d2);
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draw(O--A--B--O--D--B--O--D--C--B);
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dot(O);
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dot(B);
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dot(A,red);
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dot(C,green);
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dot(D,blue);
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label("O",O,SW);
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label("$\alpha$",shift(dir(10)/5)*O);
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label("$\beta$",shift(dir(37.5)/5)*O);
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label("A",A,SE,red);
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label("B",B,E);
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label("C",C,NE,green);
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label("D",D,dir(122.5),blue);
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label("$\cos \alpha$",O--A,S);
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label("$\sin \alpha$",A--B,E);
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label("1",O--B,dir(302.5));
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label("$\frac{\cos \alpha \sin \beta}{\cos \beta}$",B--C,E);
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label("$\frac{\sin \alpha \sin \beta}{\cos \beta}$",C--D,N);
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label("$\frac{\sin \beta}{\cos \beta}$",B--D,dir(200));
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label("$\frac{1}{\cos \beta}$",D--O,dir(325));
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</asy>
  
{{stub}}
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<math>\fontsize{18}{27}\selectfont \sin (\alpha + \beta ) = \frac{\left( \sin \alpha + \frac{\cos \alpha \sin \beta}{\cos \beta} \right)}{\frac{1}{\cos \beta}} = \cos \beta \times \left( \sin \alpha + \frac{\cos \alpha \sin \beta}{\cos \beta} \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta</math>
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<math>\fontsize{18}{27}\selectfont \cos (\alpha + \beta ) = \frac{\left( \cos \alpha - \frac{\sin \alpha \sin \beta}{\cos \beta} \right)}{\frac{1}{\cos \beta}} = \cos \beta \times \left( \cos \alpha - \frac{\sin \alpha \sin \beta}{\cos \beta} \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta</math>
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<math>\fontsize{18}{27}\selectfont \tan (\alpha + \beta ) = \frac{\sin (\alpha + \beta )}{\cos (\alpha + \beta )} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}{1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}</math>
  
 
==See Also==
 
==See Also==
 
* [[Trigonometric identities]]
 
* [[Trigonometric identities]]

Latest revision as of 19:46, 13 January 2024

The trigonometric angle addition identities state the following identities:

$\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$

$\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$

$\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$

This article is a stub. Help us out by expanding it.

Proofs

[asy] unitsize(216); real d = 1/cos(radians(35)); real d1 = d * cos(radians(55)); real d2 = d * sin(radians(55)); pair O = (0,0); pair A = (cos(radians(20)),0); pair B = (cos(radians(20)),sin(radians(20))); pair C = (cos(radians(20)),d2); pair D = (d1,d2); draw(O--A--B--O--D--B--O--D--C--B); dot(O); dot(B); dot(A,red); dot(C,green); dot(D,blue); label("O",O,SW); label("$\alpha$",shift(dir(10)/5)*O); label("$\beta$",shift(dir(37.5)/5)*O); label("A",A,SE,red); label("B",B,E); label("C",C,NE,green); label("D",D,dir(122.5),blue); label("$\cos \alpha$",O--A,S); label("$\sin \alpha$",A--B,E); label("1",O--B,dir(302.5)); label("$\frac{\cos \alpha \sin \beta}{\cos \beta}$",B--C,E); label("$\frac{\sin \alpha \sin \beta}{\cos \beta}$",C--D,N); label("$\frac{\sin \beta}{\cos \beta}$",B--D,dir(200)); label("$\frac{1}{\cos \beta}$",D--O,dir(325)); [/asy]

$\fontsize{18}{27}\selectfont \sin (\alpha + \beta ) = \frac{\left( \sin \alpha + \frac{\cos \alpha \sin \beta}{\cos \beta} \right)}{\frac{1}{\cos \beta}} = \cos \beta \times \left( \sin \alpha + \frac{\cos \alpha \sin \beta}{\cos \beta} \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\fontsize{18}{27}\selectfont \cos (\alpha + \beta ) = \frac{\left( \cos \alpha - \frac{\sin \alpha \sin \beta}{\cos \beta} \right)}{\frac{1}{\cos \beta}} = \cos \beta \times \left( \cos \alpha - \frac{\sin \alpha \sin \beta}{\cos \beta} \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\fontsize{18}{27}\selectfont \tan (\alpha + \beta ) = \frac{\sin (\alpha + \beta )}{\cos (\alpha + \beta )} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}{1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

See Also