Question

Find the exact value of sin(2arccos(-3/5)) .


ASAP

Answer 1

`\sin(2\arccos(-(3)/(5)))=-(24)/(25)`

Explanation:

Let
`\sin(2\arccos(-(3)/(5)))=\sin(2\theta)=2\sin\theta\cos\theta` and
`\theta=\arccos(-(3)/(5))` so that
`\cos\theta=-(3)/(5)`. Now we'll need to find
`\sin\theta` having known
`\cos\theta`:

`\displaystyle \cos\theta=\frac{\text{Adjacent}}{\text{Hypotenuse}}=(-3)/(5)\\\\\sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}}=(√(5^2-(-3)^2))/(5)=(√(25-9))/(5)=(√(16))/(5)=(4)/(5)`

Therefore,
`2\sin\theta\cos\theta=2((4)/(5))(-(3)/(5))=2((-12)/(25))=-(24)/(25)`, which makes
`\sin(2\arccos(-(3)/(5)))=-(24)/(25)`