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    find correct to 1dp, two smallest positive values for theta which satisfy:

    1) a) sin theta = 0.1 b ) costheta =0.8


    2) Find all values of theta in the interval -180<equalto theta <equalto 180:
    sin theta = 0.8


    I don't know how to start these off .
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    Use your inverse trig buttons on your calculator to find the first solutions?
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    (Original post by Gelato)
    find correct to 1dp, two smallest positive values for theta which satisfy:

    1) a) sin theta = 0.1 b ) costheta =0.8


    2) Find all values of theta in the interval -180<equalto theta <equalto 180:
    sin theta = 0.8


    I don't know how to start these off .


    for
    a) on your calc:

    inverse sin(0.1) = theta
    inverse cos(0.8)= theta

    if it looks wrong, maybe try in it radians?


    2)
    inverse sin(0.8) = theta

    then the answers are theta +180
    theta+ 2(180)
    theta+3(180)...... (see where i'm going?)
    theta-180
    theta-2(180)......

    because -180<=theta<=180
    if keep adding or subtracting multiples of pi until you are outside of the limits.
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    Sin(theta) =\= sin(theta+180) Try it with any value of theta other than 0 and see if they're equal.

    sin(theta) = sin(180-theta), cos(theta)=cos(360-theta), tan(theta)=tan(180+theta) and of course adding or subtracting 360 degrees to anything won't change it (so sin(theta)=sin(540-theta)=sin(theta-360), etc)

    But those overly simple rules won't always give all of the values of theta, you have to use the method that involves seeing in which "quadrant" (ie when the angle is acute, obtuse, 180<angle<270 or 270<angle<360) each of the functions is positive, that I can't really explain clearly enough to be of any use.
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    oh yeah i forgot about the quadrant bit.

    look into the CAST diagram for that bit
 
 
 
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