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# How do you evaluate cos(tan^-1 (1/4) + cos^-1 (1/2))?

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How do you evaluate cos(tan^-1 (1/4) + cos^-1 (1/2))?

Guest Mar 19, 2017
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#2
+77081
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arctan (1/4) =  14.04°

arccos (1/2)   = 60°

So

cos ( arctan (1/4)  + arccos (1/2) )   =

cos (14.04 + 60)  =

cos (74.04)   ≈ 0.275

CPhill  Mar 20, 2017
#3
+1

Find the third side of the right-angled triangles and read off the other two ratios,

Angle A = $$\tan^{-1}(1/4)=\cos^{-1}(4/\sqrt{17})=\sin^{-1}(1/\sqrt{17}),$$

Angle B = $$\cos^{-1}(1/2)=\sin^{-1}(\sqrt{3}/2)=\tan^{-1}\sqrt{3}$$.

$$\displaystyle \cos(A+B)= \cos(A)\cos(B)-\sin(A)\sin(B)\\=\frac{4}{\sqrt{17}}.\frac{1}{2}-\frac{1}{\sqrt{17}}.\frac{\sqrt{3}}{2}=\frac{4-\sqrt{3}}{2\sqrt{17}}\approx0.275029.$$

Guest Mar 20, 2017

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