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dk02266
Group Title
Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.
 one year ago
 one year ago
dk02266 Group Title
Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.
 one year ago
 one year ago

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wio Group TitleBest ResponseYou've already chosen the best response.1
dw:1358142890651:dw
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
@dk02266 Can you do it?
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
@wio one cannot assume if cosx=3/5 the sides are of 3 units and 5 units respectively as sine,cosine etc are ratios..
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
aliter let arccos (3/5)= x then cos x=3/5 now we sinx =sart(1cos^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)
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
@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.
 one year ago
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