Answer:
[tex]\displaystyle \frac{118}{125}[/tex]
Step-by-step explanation:
Trigonometry
Calculate
[tex]\sin (3\sin^{-1}(2/5))[/tex]
We use the formula for the sine of triple angle:
[tex]\sin 3x=3\sin x-4\sin^3 x[/tex]
And recall:
[tex]\sin \sin^{-1}x=x[/tex]
For this problem, we set:
[tex]x=\sin^{-1}(2/5)[/tex]
Thus:
[tex]\sin 3x=3\sin (\sin^{-1}(2/5))-4\sin^3 (\sin^{-1}(2/5))[/tex]
[tex]\displaystyle \sin 3x=3*\frac{2}{5}-4\left(\frac{2}{5}\right)^3[/tex]
[tex]\displaystyle \sin 3x=\frac{6}{5}-4*\frac{8}{125}[/tex]
[tex]\displaystyle \sin 3x=\frac{6}{5}-\frac{32}{125}[/tex]
[tex]\displaystyle \sin 3x=\frac{6*25}{125}-\frac{32}{125}[/tex]
[tex]\displaystyle \sin 3x=\frac{150-32}{125}[/tex]
[tex]\mathbf{\displaystyle \sin 3x=\frac{118}{125}}[/tex]
