How do you simplify the expression $sin (arctan (\frac{1}{4}) + arccos (\frac{3}{4}))$ $sin (arctan (1/4) + arccos (3/4) )$ ?

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Answer 1 · 2016-06-26T14:25:47

$sin (arctan (\frac{1}{4}) + arccos (\frac{3}{4})) = \frac{3 \sqrt{17} + 4 \sqrt{119}}{68}$ $sin (arctan(1/4)+arccos(3/4))=(3sqrt17+4sqrt119)/68$

Let $A = arctan (\frac{1}{4})$ $A=arctan(1/4)$
Let $B = arccos (\frac{3}{4})$ $B=arccos(3/4)$

$sin (A + B) = sin A cos B + cos A sin B$ $sin (A+B)=sin A cos B + cos A sin B$

$sin (A + B) = (\frac{1}{\sqrt{17}}) (\frac{3}{4}) + (\frac{4}{\sqrt{17}}) (\frac{\sqrt{7}}{4})$ $sin (A+B)=(1/sqrt17)(3/4) + (4/sqrt17) (sqrt7/4)$

$sin (A + B) = \frac{3 + 4 \sqrt{7}}{4 \sqrt{17}}$ $sin (A+B)=(3+4sqrt7)/(4sqrt17)$

$sin (A + B) = \frac{3 \sqrt{17} + 4 \sqrt{119}}{68}$ $sin (A+B)=(3sqrt17+4sqrt119)/68$

$sin (arctan (\frac{1}{4}) + arccos (\frac{3}{4})) = \frac{3 \sqrt{17} + 4 \sqrt{119}}{68}$ $sin (arctan(1/4)+arccos(3/4))=(3sqrt17+4sqrt119)/68$

God bless....I hope the explanation is useful.

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