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dk02266
Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.
@wio one cannot assume if cosx=3/5 the sides are of 3 units and 5 units respectively as sine,cosine etc are ratios..
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)
@matricked |dw:1358194140456:dw| \[\begin{split} 1 &= \sqrt{(a/c)^2+(b/c)^2} \\ &= \sqrt{a^2/c^2+b^2/c^2} \\ &= \sqrt{(a^2+b^2)/c^2} \\ &= \sqrt{(a^2+b^2)}/c \\ c&= \sqrt{(a^2+b^2)} \\ \end{split} \]You can proportion the triangle however you want. Why wouldn't you keep it as integers? My method is mathematically sound.