Answer:
[tex]cos3x=\frac{3\sqrt{15}}{16}[/tex]
Step-by-step explanation:
Notes that:
[tex]sinx=\frac{2}{8}=\frac{1}{4}[/tex].
We use sinx to find cosx:
[tex]cos^2x=1-sin^2x=1-(\frac{1}{4})^2=\frac{15}{16}[/tex], then
[tex]cosx=\frac{\sqrt{15}}{4}[/tex].
Now we use formule for double angle of sin and cos, to find cos2x and sin2x:
[tex]sin2x=2sinxcosx=2*\frac{1}{4}*\frac{\sqrt{15}}{4}=\frac{\sqrt{15}}{8}[/tex]
[tex]cos2x=cos^2x-sin^2x=\frac{15}{16}-\frac{1}{16}=\frac{14}{16}=\frac{7}{8}[/tex]
Now, we can find cos3x:
[tex]cos3x=cos(2x+x)=cos2x*cosx-sin2x*sinx=[/tex]
[tex]\frac{7}{8}*\frac{\sqrt{15}}{4}-\frac{\sqrt{15}}{8}*\frac{1}{4}[/tex]
[tex]cos3x=\frac{7\sqrt{15}{32}-\frac{\sqrt{15}}{32}=[/tex]
[tex]\frac{6\sqrt{15}}{32}=\frac{3\sqrt{15}}{16}[/tex]
