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![sin[pi/6+cos^(- 1)(-(sqrt(3))/2)]=` - YouTube](https://i.ytimg.com/vi/hAtJRyfcGwo/maxresdefault.jpg "sin[pi/6+cos^(- 1)(-(sqrt(3))/2)]=` - YouTube")
sin[pi/6+cos^(- 1)(-(sqrt(3))/2)]=` - YouTube
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SOLVED: Evaluate arccos a. in radians b. in degrees 2. Evaluate arctan in radians b. in degrees Evaluate arcsin in radians b. in degrees Evaluate arccos in radians b. in degrees
If A+B+C=180 prove that cos^2a + cos^2B + cos^2C = 1 - 2COSACOSBCOSC - Brainly.com
If you get a maths question asking you to convert 3/pi radians into degree measure? Which value of pi will you take 3.142 or 180° In its solution 3/pi ×180°/pi…will the pi^2
the value of arc `cos sqrt(2/3)``-`arc `cos ((sqrt(6)+1)/(2sqrt(3)))` is - YouTube
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Why does arccos (-1/2) equal to pi
