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

maths > advanced-trigonometry

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

what you'll learn...

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

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

» $\sin (A - B) = \sin (A + (- B))$ $sin(A-B) = sin(A+(-B))$

» $\cos (A + B)$ $cos(A+B)$ $= \sin ((90 - A) - B)$ $= sin((90-A) -B)$

» $\cos (A - B)$ $cos(A-B)$ $= \sin ((90 - A) + B)$ $= sin((90-A)+B)$

Geometrical Proof

geometrical proof » $\sin (A - B)$ $sin(A-B)$ from $\bar{R T} = \bar{P Q 1}$ $bar(RT)=bar(PQ1)$
» $\cos (A - B)$ $cos(A-B)$ from $\bar{R S} = \bar{P Q}$ $bar(RS)=bar(PQ)$

algebraic equivalent

It was geometrically proven that $\cos (A + B) = \cos A \cos B - \sin A \sin B$ $cos(A+B)=cosA cosB - sinA sinB$ . For the same result, we can work out a proof using algebra of trigonometric functions.

$\cos (A + B)$ $cos(A+B)$
$= \sin (90 - A - B)$ $quad quad = sin(90-A-B)$
$= \sin [(90 - A) + (- B)]$ $quad quad = sin[(90-A) + (-B)]$
$= \sin (90 - A) \cos (- B) + \cos (90 - A) \sin (- B)$ $quad quad = sin(90-A)cos(-B)+cos(90-A)sin(-B)$
$= \cos A \cos B - \sin A \sin B$ $quad quad = cosA cosB - sinA sinB$

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.

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

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

Geometrical proof for $\sin (A - B)$ $sin(A-B)$ and $\cos (A - B)$ $cos(A-B)$ is outlined below.

equate square of chord lengths to derive the result given.

${\bar{R T}}^{2} = {\bar{P Q 1}}^{2}$ $bar(RT)^2 = bar(PQ1)^2$
$\Rightarrow \sin (A - B) = \sin A \cos B - \cos A \sin B$ $=> sin(A-B) = sinA cosB - cosA sinB$

${\bar{R S}}^{2} = {\bar{P Q}}^{2}$ $bar(RS)^2 = bar(PQ)^2$
$\Rightarrow \cos (A - B) = \cos A \cos B + \sin A \sin B$ $=> cos(A-B) = cosA cosB + sinA sinB$

example

Compute $\sin 75$ $sin75$ .

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

$\sin 75$ $sin75$
$= \sin (45 + 30)$ $quad = sin(45+30)$
$= \sin 45 \cos 30 + \cos 45 \sin 30$ $quad = sin45cos30 + cos45sin30$
$= \frac{1 + \sqrt{3}}{2 \sqrt{2}}$ $quad = (1+ sqrt(3))/(2sqrt(2))$

summary

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

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