If sin x = -2/3 and x is in the third quadrant, how do you find the value of sin 2x?

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Answer 1 · 2016-11-06T14:18:53

#sin2x = (4sqrt(5))/9#

#sin2x# is equivalent to #2sinxcosx#.

We must hence find cosine of #x# to evaluate.

Letting #x# be the adjacent side, #y# be the opposite side and #r# be the hypotenuse, by pythagorean theorem we have:

#x^2 + y^2 = r^2#

#x^2 + (-2)^2 = 3^2#

#x^2 = 5#

#x =+- sqrt(5)#

Since we're in quadrant #3#, cosine is negative. It is impossible to have a negative hypotenuse, so #x# measures #sqrt(5)# units.

Hence, #cosx = -sqrt(5)/3#

We can now evaluate.

#sin(2x) = 2sinxcosx = 2(-2/3)(-sqrt(5)/3) = (4sqrt(5))/9#

Hopefully this helps!