SOLUTION: Evaluate exactly: sin[arctan(5/sqrt15) - arcsin(3/sqrt10)]

Algebra.Com
Question 884510: Evaluate exactly:
sin[arctan(5/sqrt15) - arcsin(3/sqrt10)]

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!


Let 

and 

 means the angle whose tangent is .
The tangent is  so let's draw a right triangle
containing angle  by using the numerator of , which
is 5, for the opposite side, and using the denominator of ,
which is  for the opposite side.

 means the angle whose sine is .
The sine is  so let's draw another right triangle
containing angle  by using the numerator of , which is
3, for the opposite side, and using the denominator of , which
is  for the hypotenuse.





We calculate the hypotenuse of the first right triangle and
the adjacent side to  in the second one:
















Put those values on the triangles:



Now



      

Edwin

RELATED QUESTIONS

Evaluate cos(Arcsin (3/5) + Arctan... (answered by jsmallt9)
Evaluate without a calculator: A) sin{ Arctan [(-5/13)]} B)cos{ Arcsin[(-3/4)]}... (answered by stanbon)
Evaluate: (a) Arcsin (-1/2) (b) sin [ Arcsin... (answered by stanbon)
Cos(arcsin(5/13)-arctan(3/4)) (answered by lwsshak3)
if c=arctan(3) + arcsin(5/13) find... (answered by Alan3354)
evaluate log((sqrt10)^3(sqrt10)^5(sqrt10)) without using a... (answered by lwsshak3)
Evaluate without calculator. Variables are positive numbers. I've been trying but can't... (answered by AnlytcPhil)
How do you solve problems like arcsin(sin pi/6) or cos(arctan (-radical... (answered by stanbon)
Please solve this equation exactly. sin(arccos(3/5) - arctan(5/12)) = (answered by lwsshak3,Edwin McCravy)