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    'If sinC= 1/ (square root of 5) and C is obtuse, find sin2C, cos2C and tan2C.'

    I know how to when it is acute but how do I solve it when C is obtuse??
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    (Original post by Carokelly1234)
    'If sinC= 1/ (square root of 5) and C is obtuse, find sin2C, cos2C and tan2C.'

    I know how to when it is acute but how do I solve it when C is obtuse??
    \sin C=\frac{1}{\sqrt{5}} for two values of C in the range 0\leq C \leq 2\pi. One acute and one obtuse.
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    if C is obtuse then 2C will be in the 3rd or 4th quadrants ie sin2C will be negative ?
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    (Original post by the bear)
    if C is obtuse then 2C will be in the 3rd or 4th quadrants ie sin2C will be negative ?
    how do you find out what cosC is?
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    Let's do a similar but easier one. If we had an acute angle A such that sin A = 3/5, then you could draw a right angled triangle in the first quadrant with an angle of A next to the origin, an opposite side of length 3, and a hypotenuse of length 5. You could then use Pythagoras to work out the length of the adjacent side, and then work out the cosine of A. Note that there is no need to work out what A is, and that in most cases, you can do this without a calculator.

    Your problem is like this, but when the triangle is not in the first quadrant, one or more of the adjacent and opposite sides will have a negative length, depending on whether they are going the "right" way or the "wrong" way. (The hypotenuse is always considered to have positive length.)
 
 
 
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