Trig

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Find the exact value of sinx/2 if cosx = 2/3 and 270 < x < 360.
A)1/3
B)-1/3
C)sqrt 6/6
D)-sqrt 6/6

C, since I KNOW cosx is always positive but I don't know the work involved. I know the half angle formula

  • Trig -

    First of all, x/2 will be in the second quadrant, since x is in the fourth quadrant. The sine of x/2 will therefore be positive.

    Use the formula for sin (x/2) in terms of cos x.

    sin(x/2) = sqrt([1-cos(x)]/2) = sqrt (1/6) = sqrt6/6
    You got the right answer, but you it ssmes to have been a lucky guess.

    Cos x is NOT always positive, but it is in this case.

  • Trig -

    Thank you. It wasn't really a guess it was either C or D and then I just knew it was positive so that just leaves C.

  • Trig -

    Jon,
    Perhaps it would help if you drew an x-y axis system with a unit radius vector in each of the four quadrants.
    then in quadrant 1
    sin T = y/1 so +
    cos T = x/1 so -
    tan T = y/x so +
    then in quadrant 2
    sin T = y/1 so +
    cos T = x/1 so - because x is - in q 2
    tan T = y/x so -
    then in quadrant 3
    sin T = y/1 so -
    cos T = x/1 so -
    tan T = y/x so + because top and bottom both -
    then in quadrant 4
    sin T = y/1 so -
    cos T = x/1 so +
    tan T = y/x so -

    sin has same sign as its inverse csc
    cos has same sign as its inverse sec
    tan has same sign as its inverse ctan

  • Trig TYPO -

    then in quadrant 1
    sin T = y/1 so +
    cos T = x/1 so +
    tan T = y/x so +

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