Trig Fcns part 3: value of sin^2(11pi/6), cos^-1(-sqrt3/2),

[Louise Johnson]

Can anyone check these for me?
Thank you
Louise

#1 The value of sin^2 ( 11pi/6) is ?
answer is (-1/2)^2 or 1/4 sin is the y coordinate which is -1/2 and that squared is 1/4

#2 The value of cos^-1 (-sqrt3/2) , in degree measure is?Answer 165 degrees and 210 degrees because cos is negative in both quadrants 2 and 3.

#3 The value of 16^cos pi/3 isAnswer 256 (this one I am really not sure of as I haven't dealt with an exponent like this.

 

[arthur ohlsten]

11pi/6=11[180 degrees]/6
11pi/6=330 degrees
11pi/6=-30 degrees
sin[-30]= -1/2
sin [11pi/6]=1/2
sin^2[11pi/6]=1/4 answer
2)

the angle whose cos is x, is -sqrt3 / 2
on a xy coordinate system sketch a right triangle whose adjacent side has length sqrt3 in -x direction, and its hypoteneuse = 2 units. this triangle is in the 2nd quadrant
from pythegorian theorem 2^2= y^2+[-sqrt3 ]^2
4=y^2+ 3
y=+/-1


x is 180 degrees +/- 30 degrees
x=210 or 150 answer
not 165

3)
16 ^(cospi/3]
pi/3 = 180/3
pi/3=60 degrees
cos 60 *= 1/2
16^( cospi/3]= 16^1/2 or square root of 16
16^1/2 = +/-4 answer

 

[Louise Johnson]

Thank you
I am going to give this some serious looking over but I just wanted to give a quick thank you!!
Number 3 especially threw me for a loop I had just did 16 exp 1/2 instead of the square of 16....
Thank you
Louise