5.4.4

Manipulating the Angle Addition Formulae

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Manipulating the Angle Addition Formulae

We can prove other angle addition formulae by manipulating what we have already for sin(A+B)\sin(A+B) and cos(A+B)\cos(A+B).

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Sine subtraction formula

  • We can find the identity for sin(AB)\sin(A-B) by substituting BBB \rightarrow-B into the identity for sin(A+B)\sin(A+B):
    • sin(A+(B))sinAcos(B)+cosAsin(B)\sin(A+(-B)) \equiv \sin A\cos(- B) + \cos A\sin (-B)
    • sin(AB)sinAcosBcosAsinB\sin(A-B) \equiv \sin A\cos B - \cos A\sin B
  • Where we have used cos(B)=cosB\cos(-B) = \cos B and sin(B)=sinB\sin(-B) = -\sin B.
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Cosine subtraction formula

  • We can find the identity for cos(AB)\cos(A-B) by substituting BBB \rightarrow-B into the identity for cos(A+B)\cos(A+B):
    • cos(A+(B))cosAcos(B)sinAsin(B)\cos(A+(-B)) \equiv \cos A\cos(- B) - \sin A\sin (-B)
    • cos(AB)cosAcosB+sinAsinB\cos(A-B) \equiv \cos A\cos B +\sin A\sin B
  • Where we have used cos(B)=cosB\cos(-B) = \cos B and sin(B)=sinB\sin(-B) = -\sin B.
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Tangent addition formula

  • To find the addition formula for tangent, we
    • tan(A+B)sin(A+B)cos(A+B)\tan(A+B) \equiv \frac{\sin(A+B)}{\cos(A+B)}
  • Substituting in our angle addition formulas, we get:
    • tan(A+B)sinAcosB+cosAsinBcosAcosBsinAsinB\tan(A+B) \equiv \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B -\sin A\sin B}
  • Dividing both the numerator and denominator by a factor of cosAcosB\cos A\cos B gives:
    • tan(A+B)sinAcosBcosAcosB+cosAsinBcosAcosBcosAcosBcosAcosBsinAsinBcosAcosB\tan(A+B) \equiv \frac{\small \frac{\sin A\cos B}{\cos A\cos B} + \frac{\cos A\sin B}{\cos A\cos B}}{\small \frac{\cos A\cos B}{\cos A\cos B} - \frac{\sin A \sin B}{\cos A \cos B}}
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Tangent addition formula cont.

  • Cancelling out common factors gives:
    • tan(A+B)sinAcosA+sinBcosB1sinAsinBcosAcosB\tan(A+B) \equiv \frac{\small \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\small 1 - \frac{\sin A \sin B}{\cos A \cos B}}
  • Finally, we can simplify this as:
    • tan(A+B)tanA+tanB1tanAtanB\tan(A+B)\equiv \frac{\tan A+\tan B}{1 - \tan A \tan B}
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Negative tangent

  • We can find the identity for tan(AB)\tan(A-B) by substituting BBB \rightarrow-B into the identity for tan(A+B)\tan(A+B):
    • tan(A+(B))tanA+tan(B)1tanAtan(B)\tan(A+(-B))\equiv \frac{\tan A+\tan (-B)}{1 - \tan A \tan(- B)}
    • tan(AB)tanAtanB1+tanAtanB\tan(A-B) \equiv \frac{\tan A-\tan B}{1 + \tan A\tan B}
  • Where we have used tan(B)=tanB\tan(-B) = -\tan B.

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