well you know the angle is in the 2nd quadrant... so draw a diagram and use pythagoras' theorem |dw:1429478380246:dw| then use the expansions sin(2a) = 2 sin(a)cos(a) cos(2a) = cos^(a) - sin^2(a) etc...
could I just say the hypotenuse is 2? 30-60-90 triangle
@campbell_st
that's correct.... the hypotenuse is 2... so now you can find sin(a) and cos(a) and then answer your questions... be careful with the signs of the terms... in the 2nd quadrant only sin is positive,.
sin2a=2sinAcosA right?
that's correct but be careful cos(A) is negative...
so \[(\frac{ \sqrt{3} }{ 2 })(2)(-\frac{ 1 }{ 2 })=-\frac{ \sqrt{3} }{ 2 }\]
but isnt sin positive?
sin is positive... and cos is negative... so multiplying a positive by a negative gives a negative
okay, so is my answer correct?
@campbell_st
yes... its fine.... because when you think about doubling the size of the angle A to 2A it puts it in the 4th quadrant where sin is negative...
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