What does $cos (arctan (2)) + sin (a r c sec (1))$ $cos(arctan(2))+sin(arcsec(1))$ equal?

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Answer 1 · 2016-09-30T23:58:54

$cos (arctan (2)) + sin (a r c sec (1)) = \frac{\sqrt{5}}{5}$ $cos(arctan(2)) + sin(arcsec(1)) = sqrt(5)/5$

Consider the right angled triangle with sides $1, 2, \sqrt{5}$ $1, 2, sqrt(5)$ ...

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Then:

$tan α = \frac{opposite}{adjacent} = \frac{2}{1} = 2$ $tan alpha = "opposite"/"adjacent" = 2/1 = 2$

and

$cos α = \frac{adjacent}{hypotenuse} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$ $cos alpha = "adjacent"/"hypotenuse" = 1/sqrt(5) = sqrt(5)/5$

So:

$cos (arctan (2)) = \frac{\sqrt{5}}{5}$ $cos(arctan(2)) = sqrt(5)/5$

Note also that:

$sec (0) = \frac{1}{cos (0)} = \frac{1}{1} = 1$ $sec(0) = 1/cos(0) = 1/1 = 1$

$sin (0) = 0$ $sin(0) = 0$

So we find:

$sin (a r c sec (1)) = sin (0) = 0$ $sin(arcsec(1)) = sin(0) = 0$

Putting it together:

$cos (arctan (2)) + sin (a r c sec (1)) = \frac{\sqrt{5}}{5}$ $cos(arctan(2)) + sin(arcsec(1)) = sqrt(5)/5$

What does cos(arctan(2))+sin(arcsec(1))cos(arctan(2))+sin(arcsec(1)) equal?