Trigonometry (advanced) : Compound Angles: tan(A+B) and tan(A-B)

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Compound Angles: tan(A+B) and tan(A-B)

what you'll learn...

Compound angles of tan & cot

Quickly derive the identities. No need to memorize.

» $\tan (A \pm B) = \frac{\sin (A \pm B)}{\cos (A \pm B)}$ $tan(A+-B) = (sin(A+-B))/(cos(A+-B))$

» $\cot (A \pm B) = \frac{\cos (A \pm B)}{\sin (A \pm B)}$ $cot(A+-B) = (cos(A+-B))/(sin(A+-B))$

tan

Proof for tan(A+B) and tan(A-B) using previous results and algebra of trigonometric functions.

$\tan (A + B)$ $tan(A+B)$
$= \frac{\sin (A + B)}{\cos (A + B)}$ $quad quad = (sin(A+B))/(cos(A+B))$
$= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}$ $quad quad = (sinA cosB + cosA sinB)/(cosA cosB - sinA sinB)$
dividing numerator and denominator by $\cos A \cos B$ $cosA cosB$
$= \frac{\tan A + \tan B}{1 - \tan A \tan B}$ $quad quad = (tan A + tan B)/(1 - tan A tan B)$

$\tan (A - B)$ $tan(A-B)$
$= \tan (A + (- B))$ $quad quad = tan(A+(-B))$
$= \frac{\tan A - \tan B}{1 + \tan A \tan B}$ $quad quad = (tan A - tan B)/(1 + tan A tan B)$

summary

$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ $tan(A+B) = (tan A + tan B)/(1- tan A tan B)$
$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ $tan(A-B) = (tan A - tan B)/(1+ tan A tan B)$

cot

Proof for cot(A+B) and cot(A-B) using previous results and algebra of trigonometric functions.

$\cot (A + B)$ $cot(A+B)$
$= \frac{\cos (A + B)}{\sin (A + B)}$ $quad quad = (cos(A+B))/(sin(A+B))$
$= \frac{\cos A \cos B - \sin A \sin B}{\sin A \cos B + \cos A \sin B}$ $quad quad = (cosA cosB - sinA sinB)/(sinA cosB + cosA sinB)$
dividing numerator and denominator by $\sin A \sin B$ $sinA sinB$
$= \frac{\cot A \cot B - 1}{\cot A + \cot B}$ $quad quad = (cot A cot B -1)/(cot A + cot B)$

$\cot (A - B)$ $cot(A-B)$
$= \cot (A + (- B))$ $quad quad = cot(A+(-B))$
$= \frac{- \cot A \cot B - 1}{\cot A - \cot B}$ $quad quad = (-cot A cot B -1)/(cot A - cot B)$

summary

$\cot (A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ $cot(A+B) = (cot A cot B -1)/(cot A + cot B)$
$\cot (A - B) = \frac{- \cot A \cot B - 1}{\cot A - \cot B}$ $cot(A-B) = (-cot A cot B -1)/(cot A - cot B)$

summary

Trigonometric Identities for compound Angles

$\sin (A + B) = \sin A \cos B + \cos A \sin B$ $sin(A+B)=sinA cosB + cosA sinB$
$\sin (A - B) = \sin A \cos B - \cos A \sin B$ $sin(A-B)=sinA cosB - cosA sinB$
$\cos (A + B) = \cos A \cos B - \sin A \sin B$ $cos(A+B)=cosA cosB - sinA sinB$
$\cos (A - B) = \cos A \cos B + \sin A \sin B$ $cos(A-B)=cosA cosB + sinA sinB$

$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ $tan(A+B) = (tan A + tan B)/(1- tan A tan B)$
$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ $tan(A-B) = (tan A - tan B)/(1+ tan A tan B)$

$\cot (A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ $cot(A+B) = (cot A cot B -1)/(cot A + cot B)$
$\cot (A - B) = \frac{- \cot A \cot B - 1}{\cot A - \cot B}$ $cot(A-B) = (-cot A cot B -1)/(cot A - cot B)$

Outline

It is advised to do the firmfunda version of "basics of Trigonometry" course before doing this.

The outline of material to learn "Advanced Trigonometry" is as follows.
Note: go to detailed outline of Advanced Trigonometry

→ Unit Circle form of Trigonmetric Values

→ Trigonometric Values in all Quadrants

→ Trigonometric Values or any Angles : First Principles

→ Understanding Trigonometric Values in First Quadrant

→ Trigonometric Values in First Quadrant

→ Trigonometric Values of Compound Angles: Geometrical Proof

→ Trigonometric Values of Compound Angles: Algebraic Proof

→ Trigonometric Values of Compound Angles: tan cot

→ Trigonometric Values of Compound Angles: more results