Thread: arccos(3/5)+arccot(1/7)

how do I solve arccos(3/5)+arccot(1/7) ? am i supposed to do cos(arccos(3/5)+arccot(1/7)) making it (3/5)+cos(arccot(1/7)) ? but if so how do i continue?

Originally Posted by weasley74

how do I solve arccos(3/5)+arccot(1/7) ? am i supposed to do cos(arccos(3/5)+arccot(1/7)) making it (3/5)+cos(arccot(1/7)) ? but if so how do i continue?

when you say "solve" what do you mean. because there is no nice way to "simplify" this without a calculator, which is what you seem to be asking us to do. (what you did was totally wrong by the way. cosine is not a linear operator, so you can't apply it as you did. $\displaystyle \cos (A + B) = \cos A \cos B - \sin A \sin B$, secondly, if you are to simplify, you cannot do it by applying a trig function, you'd be changing the value!)

but we're suppose to solve this, in some wierd way.. it's supposed to be 3pi/4.. but no one seems to know how..

actually.. i think i got it..

cos(arccos(3/5))*cos(arccot(1/7))-sin(arccos(3/5))*sin(arccot(1/7))=

= (3/5)*(1/sqrt(50))-(4/5)*(7/sqrt(50))=

= -5/sqrt(50)

and -5/sqrt(50) = cos 3pi/4


right?

Originally Posted by weasley74

actually.. i think i got it..

cos(arccos(3/5))*cos(arccot(1/7))-sin(arccos(3/5))*sin(arccot(1/7))=

= (3/5)*(1/sqrt(50))-(4/5)*(7/sqrt(50))=

= -5/sqrt(50)

and -5/sqrt(50) = cos 3pi/4


right?

yes, that's nice!

yay me! ^^