2007-11-19 10:14:26 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
cos-1 for both please forgot the inverse
2007-11-19 10:21:14 · update #1
Simplify the expression
tan[arccos(3/4) + arccos(5/13)]
___________
α = arccos(3/4)
cosα = 3/4
sin²α = 1 - cos²α = 1 - 9/16 = 7/16
sinα = √7 / 4
tanα = sinα / cosα = (√7/4) / (3/4) = √7 / 3
Similarly
β = arccos(5/13)
cosβ = 5/13
sin²β = 1 - cos²β = 1 - 25/169 = 144/169
sinβ = 12/13
tanβ = sinβ / cosβ = (12/13) / (5/13) = 12/5
________
Now use the law of angle addition for tangents.
tan[arccos(3/4) + arccos(5/13)]
= tan[arctan(√7/3) + arctan(12/5)]
= (√7/3 + 12/5) / [1 - (√7/3)(12/5)]
= (5√7 + 36) / (15 - 12√7)
= (36 + 5√7) / (15 - 12√7)
__________
This would have been a simpler expression if the first arccos had been arccos(3/5).
2007-11-19 10:41:22 · answer #1 · answered by Northstar 7 · 1⤊ 0⤋
If x=cos-1(3/4) then cosx=3/4, right?
So sinx=sqrt(1-(3/4)^2)=sqrt(1-(9/16))=
=sqrt(7/16)=sqrt(7)/4
Likewise, if cosy=5/13 then siny=12/13
Now tan(x+y)=sin(x+y)/cos(x+y)=
=(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)=
=(sqrt(7)*5 + 3*12)/(3*5-sqrt(7)*12)=
=-(960+507*sqrt(7))/783
(if I haven't messed up my arithmetic towards the end)
2007-11-19 11:00:14 · answer #2 · answered by Angelos 5 · 0⤊ 1⤋
tan(t) = -7 < 0 ? ?/2 < t < ? or 3?/2 < t < 2? cos(t) > 0 ? -?/2 < t < ?/2 then ? -?/2 < t <0 tan(t) = -7 a / b = 7 or a / b = 7 /a million anticipate: a = 7 b = a million c^2 = a^2 + b^2 or c^2 = 7^2 + a million^2 = 50 c = ?50 sec(t) = c / b = ?50 / a million = ?50
2016-11-12 03:24:00 · answer #3 · answered by ? 4 · 0⤊ 0⤋
are the angles given in radians?
are you sure you want cos() or cos-1()?
2007-11-19 10:18:42 · answer #4 · answered by graham e 2 · 0⤊ 0⤋
*ascud* is wrong
the exact value is
-2.939202959
2007-11-19 10:59:12 · answer #5 · answered by wolf 6 · 0⤊ 0⤋
-11.3556
2007-11-19 10:50:08 · answer #6 · answered by ascud2000 2 · 0⤊ 1⤋