If sinØ= -4/5 and Ø terminates in Quadrant III, find the exact value of sin 2Ø

Question

A. 16/25

B. 24/-25

C.24/25

D. -9/25

Answer 1

Okay, this one is tough to draw using this software, so I will ask you to draw alongside my calculations.

We have SinΘ = -4/5. Ignore the negative for a moment and remember that:

SinΘ = Opposite                      That makes the opposite 4, and the hypotenuse 5. Using the Pythagorean theorem

           Hypotenuse                  will get you this equation:

Opp² + Adj² = Hyp²

            or...

4² + Adj² = 5²                        Solve for the adjacent and you will get 3. Now for the fun part!

Remember your trig formulas? There are only 60+, so of course you do. In this case, you need the double-angle rule:

SIN(2Θ) = 2SINΘ*COSΘ          From here it is fairly easy. You already know SINΘ, and you can easily get Cosine, since

COSΘ = Adjacent =    3           But what sign? In Quadrant III, Cos is always negative. That also works for Sin being

             Hypotenuse   5            a negative value.

Simply multiply:

SIN(2Θ) =2*  -4  * -3    = 24

                      5      5       25