arctan(63/16)= arcsin(5/13)+arccos(3/5)

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Show that `arctan(63/16)=arcsin(5/13)+arccos(3/5)` .

(1) If `alpha = arcsin(5/13)` , then `alpha` is an acute angle of a right triangle in the first quadrant with hypotenuse 13, and side opposite `alpha` is 5. Then the third side is 12 (Apply the pythagorean theorem or recognize the pythagorean triple). So `alpha = arctan(5/12)` .

(2) If `beta=arccos(3/5)` , then `beta` is an acute angle of a right triangle in the first quadrant with hypotenuse 5, and side adjacent to `beta` is 3. Then the third side is 4, and `beta = arctan(4/3) ` .

(3) Thus `arctan(63/16)=arcsin(5/13)+arccos(3/5)=arctan(5/12)+arctan(4/3)` Take the tangent of both sides -- since all angles are in the first quadrant, `tan(arctan theta)=theta` .

Now `tan(arctan(63/16))=tan(arctan(5/12)+arctan(4/3))`

`63/16=(5/12+4/3)/(1-5/12*4/3)` (`tan(a+b)=(tana+tanb)/(1-tanatanb)` )

`63/16=63/16` as required.

Approved by eNotes Editorial Team
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