SOLUTION: sin[tan^-1(5/12)-sin^-1(-1)] I need to solve this equation for 0<theta<2pi I am familiar with the Product to sum formulas but I do not see where that would play in here becau

Question 749158: sin[tan^-1(5/12)-sin^-1(-1)]
I need to solve this equation for 0 I am familiar with the Product to sum formulas but I do not see where that would play in here because of the arctan(5/12). I know that tan is sin/cos. I tried to see if I can maybe solve it knowing that much, but I cannot seem to figure it out.
Any help or feedback would be great! Thank you!
Mayra
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
sin[tan^-1(5/12)-sin^-1(-1)]
let x=the angle whose tan=5/12
tan(x)=5/12
hypotenuse=13 (5-12-13 right triangle)
sin(x)=5/13
cos(x)=12/13
..
let y=the angle whose sin=-1
sin(y)=-1
cos(y)=√(1-sin^2(y))=0
...
sin[tan^-1(5/12)-sin^-1(-1)]=sin(x-y)=sin(x)cos(y)-cos(x)sin(y)=5/13*0-12/13*(-1)=12/13
..
Check: (with calculator)
tan(x)=5/12
x=22.62�
sin(y) =-1
y=270�
x-y=(22.62-270)=-247.38 (in Q3)
reference angle:67.38�
sin(x+y)=sin(67.38�)≈0.923..
exact value=12/13≈0.923..

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