 Compound Angles: cos(A+B), sin(A-B), cos(A-B)

what you'll learn...

sin(A-B), cos(A+B), cos(A-B)

»  $\mathrm{sin}\left(A+B\right)$$\sin \left(A + B\right)$ Proven result
Quickly derive the identities. No need to memorize.

»  $\mathrm{sin}\left(A-B\right)=\mathrm{sin}\left(A+\left(-B\right)\right)$$\sin \left(A - B\right) = \sin \left(A + \left(- B\right)\right)$

»  $\mathrm{cos}\left(A+B\right)$$\cos \left(A + B\right)$$=\mathrm{sin}\left(\left(90-A\right)-B\right)$$= \sin \left(\left(90 - A\right) - B\right)$

»  $\mathrm{cos}\left(A-B\right)$$\cos \left(A - B\right)$$=\mathrm{sin}\left(\left(90-A\right)+B\right)$$= \sin \left(\left(90 - A\right) + B\right)$

Geometrical Proof »  $\mathrm{sin}\left(A-B\right)$$\sin \left(A - B\right)$ from $\overline{RT}=\overline{PQ1}$$\overline{R T} = \overline{P Q 1}$
»  $\mathrm{cos}\left(A-B\right)$$\cos \left(A - B\right)$ from $\overline{RS}=\overline{PQ}$$\overline{R S} = \overline{P Q}$

algebraic equivalent

It was geometrically proven that $\mathrm{cos}\left(A+B\right)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B$$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$. For the same result, we can work out a proof using algebra of trigonometric functions.

$\mathrm{cos}\left(A+B\right)$$\cos \left(A + B\right)$
$\quad \quad = \sin \left(90 - A - B\right)$
$\quad \quad = \sin \left[\left(90 - A\right) + \left(- B\right)\right]$
$\quad \quad = \sin \left(90 - A\right) \cos \left(- B\right) + \cos \left(90 - A\right) \sin \left(- B\right)$
$\quad \quad = \cos A \cos B - \sin A \sin B$

To calculate trigonometric values for compound angle, we will switch to using the proofs with algebra of trigonometric functions, as it is simpler. But, equivalently a geometrical proof can be worked out.

A-B

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

$\mathrm{sin}\left(A-B\right)$$\sin \left(A - B\right)$
$\quad \quad = \sin \left(A + \left(- B\right)\right)$
$\quad \quad = \sin A \cos \left(- B\right) + \cos A \sin \left(- B\right)$
$\quad \quad = \sin A \cos B - \cos A \sin B$

$\mathrm{cos}\left(A-B\right)$$\cos \left(A - B\right)$
$\quad \quad = \cos \left(A + \left(- B\right)\right)$
$\quad \quad = \cos A \cos \left(- B\right) - \sin A \sin \left(- B\right)$
$\quad \quad = \cos A \cos B + \sin A \sin B$ Geometrical proof for $\mathrm{sin}\left(A-B\right)$$\sin \left(A - B\right)$ and $\mathrm{cos}\left(A-B\right)$$\cos \left(A - B\right)$ is outlined below.

equate square of chord lengths to derive the result given.

${\overline{RT}}^{2}={\overline{PQ1}}^{2}$${\overline{R T}}^{2} = {\overline{P Q 1}}^{2}$
$⇒\mathrm{sin}\left(A-B\right)=\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B$$\implies \sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$

${\overline{RS}}^{2}={\overline{PQ}}^{2}$${\overline{R S}}^{2} = {\overline{P Q}}^{2}$
$⇒\mathrm{cos}\left(A-B\right)=\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B$$\implies \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

example

Compute $\mathrm{sin}75$$\sin 75$.

The answer is consider ${75}^{\circ }$${75}^{\circ}$ as sum of standard angles ${45}^{\circ }$${45}^{\circ}$ and ${30}^{\circ }$${30}^{\circ}$'.

$\mathrm{sin}75$$\sin 75$
$\quad = \sin \left(45 + 30\right)$
$\quad = \sin 45 \cos 30 + \cos 45 \sin 30$
$\quad = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$

summary

$\mathrm{sin}\left(A-B\right)=\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B$$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$
$\mathrm{cos}\left(A-B\right)=\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B$$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin 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