Question

if x is in quadrant ii and cos x = -3/4, find an exact value for sin2x

Answer 1

\[\sin(2x)=2\sin(x)\cos(x)\] you already know that
\[\cos(x)=-\frac{3}{4}\] so you only need
\[\sin(x)\] do you know how to find it?

Answer 2

do you have to use the pythagorean theorem?

Answer 3

yes, to find the other side. and don't forget that you are in quadrant ii so sine is positive

Answer 4

i got sqrt 7?

Answer 5

you should get the other side of the triangle as
\[\sqrt{4^2-3^2}=\sqrt{7}\]so
\[\sin(x)=\frac{\sqrt{7}}{4}\]

Answer 6

oh i see

Answer 7

but now i have sin and cos, im not sure what to do

Answer 8

put it in the formula i wrote above

Answer 9

multiply them together, and mulitply the result by 2

Answer 10

Sin (2x)= 2sin(x) * cos(x) = 2 (√ (7)/4 times (-3/4) =[ (√ (7)/2] [-3/4] =
-3(√ (7)/ 8

Answer 11

I think we need to consider where the angle with measure 2x lies so that the correct sign can be determined. It could be in Quadrant 3 or 4, perhaps.

Answer 12

x is roughly 138 degrees.Twice that is 276 which lies in the fourth quadrant where cosine is positive and sine is negative. So, sin (2x) will be negative as reflected in the answer above.