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Precalculus
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cot[cos^−1(−1/2) + cos^−1(1/2) + tan^−1(1/3)]
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\[\cos^{-1}(-\frac{1}{2})=120\] \[\cos^{-1}(\frac{1}{2})=60\] \[\tan^{-1}(\frac{1}{3})\] is a mystery to me but so far we have \[\cos(180+\tan^{-1}(\frac{1}{3}))\]
now \(\cos(180+x)=-\cos(x)\) so your final job is to find \[-\cos(\tan^{-1}(\frac{1}{3}))\] which is identical to saying "if the tangent is 1/3, what is the cosine? you can do that by drawing a triangle
|dw:1353554940402:dw|
there is a picture of an angle whose tangent is \(\frac{1}{3}\) by pythagoras the hypotenuse is \(\sqrt{10}\) so the cosine of that angle is \[\frac{3}{\sqrt{10}}\] and so your answer is \[-\frac{3}{\sqrt{10}}\]
That makes so much sense, thank you for explaining it to me step by step! :)
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