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sin (Arc sec 4)
Please walk me through how you would approach solving this
Anyone?
so basically you want the answer to this: \[\sin(\sec ^{-1} (4))\]?
Well, there are no calculators allowed, so it has to be a special triangle....
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sec A = hypotenuse/ base which means sin(arcsec(4)) => sin(arcsec(4/1)) => h =4, b =1 now sin of same anlge = p/h => sqrt(4^2-1)/4 => sqrt(15)/4
Thanks - that makes sense!
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