sum and difference formulas

[alyren]

use the sum and difference formulas to find the exact value of the expression

sin(-11pi/12)

what do i do with the negative sign?

i try
sin[(-3pi/12) - (8pi/12)]
sin[(-pi/4) - (2pi/3)]
sin(-pi/4)cos(2pi/3) - sin(2pi/3)cos(-pi/4)

somehow it don't feel right..

 

[Deleted member 4993]

use the sum and difference formulas to find the exact value of the expression

sin(-11pi/12)

what do i do with the negative sign?

i try
sin[(-3pi/12) - (8pi/12)]
sin[(-pi/4) - (2pi/3)]
sin(-pi/4)cos(2pi/3) - sin(2pi/3)cos(-pi/4)

somehow it don't feel right..

\(\displaystyle sin(-\theta) \ = \ - sin(\theta)\)

 

[iambryanatkinson]

http://texasinstrumentsbaiiplus.floatshelves.com/

 

[BigGlenntheHeavy]

\(\displaystyle sin\bigg(\frac{-11\pi}{12}\bigg) \ = \ sin\bigg(\frac{-3\pi}{12}-\frac{8\pi}{12}\bigg) \ = \ sin\bigg(\frac{-\pi}{4}-\frac{2\pi}{3}\bigg)\)

\(\displaystyle = \ sin\bigg[\bigg(\frac{-\pi}{4}\bigg)-\bigg(\frac{2\pi}{3}\bigg)\bigg]\)

\(\displaystyle = \ sin(-\pi/4)cos(2\pi/3)-cos(-\pi/4)sin(2\pi/3)\)

\(\displaystyle Note: \ sin(-\pi/4) \ = \ -sin(\pi/4), \ odd \ function \ and \ cos(-\pi/4) \ = \cos(\pi/4), \ even \ function.\)

\(\displaystyle Hence, \ we \ have: \ sin\bigg(\frac{-11\pi}{12}\bigg) \ = \ -sin(\pi/4)cos(2\pi/3)-cos(\pi/4)sin(2\pi/3)\)

\(\displaystyle = \ \bigg(\frac{-1}{\sqrt2}\bigg)\bigg(\frac{-1}{2}\bigg)-\bigg(\frac{1}{\sqrt2}\bigg)\bigg(\frac{\sqrt3}{2}\bigg) \ = \ \bigg(\frac{1}{2\sqrt2}\bigg)-\bigg(\frac{\sqrt3}{2\sqrt2}\bigg) \ = \ \frac{1-\sqrt3}{2\sqrt2} \ = \ \frac{\sqrt2-\sqrt6}{4}.\)