Question

Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.

Answer 1

@dk02266 Can you do it?

Answer 2

@wio
one cannot assume if cosx=3/5 the sides are of 3 units and 5 units respectively as sine,cosine etc are ratios..

Answer 3

aliter
let arccos (3/5)= x then cos x=3/5
now we sinx =sart(1-cos^2x)=sqrt(1- 9/25)=sqrt (16/25) =4/5
thus sinx=4/5
or sin( arccos (3/5))=4/5 as x= arccos (3/5)

Answer 4

@matricked

\[\begin{split}
1 &= \sqrt{(a/c)^2+(b/c)^2} \\
&= \sqrt{a^2/c^2+b^2/c^2} \\
&= \sqrt{(a^2+b^2)/c^2} \\
&= \sqrt{(a^2+b^2)}/c \\
c&= \sqrt{(a^2+b^2)} \\
\end{split}
\]You can proportion the triangle however you want.
Why wouldn't you keep it as integers? My method is mathematically sound.