Trigonometry

[whitmack]

I need some help finding the exact value of the expressions below:


sin x =3/4 cos y =-1/3 find sin(x - y) in intervals (pie/2, pie)



sin{inv sin 2/3 +inv cos 1/3}

 

[skeeter]

x,y in the interval (pi/2, pi)

use two reference triangles in quadrant II to calculate cosx and siny ...

sinx = 3/4, cosx = -sqrt(7)/4

cosy = -1/3, siny = sqrt(8)/3

now use the difference formula for sine, substituting in the values above ...

sin(x - y) = sinx*cosy - cosx*siny

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sin[arcsin(2/3) + arccos(1/3)]

let a = arcsin(2/3), b = arccos(1/3)

so, sina = 2/3 and cosb = 1/3

two refernce triangles again, this time in quadrant I since arcsin and arccos values are both (+).

sina = 2/3 ... cosa = sqrt(5)/3

cosb = 1/3 ... sina = sqrt(8)/3

same drill ...
sin(a+b) = sina*cosb + cosa*sinb



btw, pie is something you eat ... pi is the greek letter that represents the ratio between the circumference and diameter of a circle.

 

[whitmack]

Thanks for the help and pi info!

WRM