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    Hey guys just started the trig chapter in C2 and I really have no idea how to go about these question, any help would really be appreciated

    a) Given that angle A is obtuse and that sin A = 5/14 root 3, find the exact value of cos A.
    b) Given that 180 <B< 360 and that tan B = -21/20, find the exact value of cos B.
    c) Find all possible values of sin C for which cos C = 1/2.
    d) Find the values of D for which -180 degrees <D< 180 degrees and tan D = 5 sin D.

    Many thanks
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    (Original post by adil12)
    Hey guys just started the trig chapter in C2 and I really have no idea how to go about these question, any help would really be appreciated

    a) Given that angle A is obtuse and that sin A = 5/14 root 3, find the exact value of cos A.
    b) Given that 180 <B< 360 and that tan B = -21/20, find the exact value of cos B.
    c) Find all possible values of sin C for which cos C = 1/2.
    d) Find the values of D for which -180 degrees <D< 180 degrees and tan D = 5 sin D.

    Many thanks
    a) Consider B=180-A. As sinB=sinA use the cosB=\sqrt{1-sin^2B}
    consider that cosA=cos(180-B)=-cosB
    b) AS 270<B<360 (tanB<0) cosB>0
    the identity: cosB=\frac{1}{\sqrt{1-tan^2B}}
    c) cosC=1/2 for C1=60 and C2=-60=(360-60)=300
    This means two sinC value a positive and a negative
    d) sinD/cosD=5 sinD -> sinD=5sinD cosD -> sinD(1-5cosD)=0
    this wil give the solutions
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    a) Use \sin^2A+\cos^2A=1 to get cosA in terms of sinA. Bear in mind that through doing this, you would need to consider both the positive and negative square root to get cosA. To determine which value of cosA you want, look at the graph of y=cosx (or use a CAST diagram, whichever you were taught) to tell you whether cosA is positive or negative.

    b) Use 1+\tan^2B=\sec^2B=1/\cos^2B to get cosB in terms of tanB. As before, this would give you a positive and negative solotion for cosB so use 180 <B< 360 as well as the fact that tanB is negative to determine roughly what angle B is and hence whether cosB is positive or negative.

    c) This is a similar method to part A except that you just write down the positive and negative solution

    d) You can rewrite this as \frac{\sin D}{\cos D}=5\sin D. Now be careful here. You can't just divide both sides by sinD because you'll lose a solution. Instead, write this as \frac{\sin D}{\cos D}-5\sin D=0 so \sin D(1/\cos D -5)=0 and solve that.
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    Cheers guys
 
 
 
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