Question

Simplify x = cos[2 arcsin(3/5)].

-7/25
7/25
4/5

Answer 1

Answer:

7/25

Step-by-step explanation:

Let $\theta= \arcsin(\frac{3}{5})$ so we have $x=\cos(2\theta)$

As $\cos(2\theta)=\cos^2\theta-\sin^2\theta$, we'll have $\cos[2\arcsin(\frac{3}{5})]=\bigr[\cos(\arcsin(\frac{3}{5}))\bigr]^2-\bigr[(\sin(\arcsin(\frac{3}{5}))\bigr]^2$

To determine $\cos(\arcsin(\frac{3}{5}))$, construct a right triangle with an opposite side of 3 and a hypotenuse of 5. This is because since $\theta=\arcsin(\frac{3}{5})$, then $\sin\theta=\frac{3}{5}=\frac{\text{Opposite}}{\text{Hypotenuse}}$. If you recognize the Pythagorean Triple 3-4-5, you can figure out that the adjacent side is 4, and thus, $\cos\theta=\frac{4}{5}=\frac{\text{Adjacent}}{\text{Hypotenuse}}$. This means that $\cos(\arcsin(\frac{3}{5}))=\frac{4}{5}$.

Hence, $\cos[2\arcsin(\frac{3}{5})]=(\frac{4}{5})^2-(\frac{3}{5})^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$