cosine rule : cos(a)=-1/sqrt(2).. how do i evaluate this ? can a = 3/4pi and 5/4pi ?
have you tried taking the arc cosine of both sides?
yes...
or are you supposed to use fancy complicated trigonometric formulas?
3/4pi and 5/4pi both answers?
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yes...do u know this?
YES
how can an angle in triangle = 5/4pi ?
see in 2nd and 3rd quadrant cos will be negative....and you have not been asked for a triangle
no i was using the cosine rule to find angle ..
sorry my bad.. i was finding angle in triangle using cosine rule and got what istated in queston
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and u r aware with the identities cos(pi-theta) = -cos(theta) cos(pi+theta) = - cos(theta)
YES i solved it already .. 3pi/4 and 5pi/4
good...:) then u know...how to solve this type of question?
im not asking how to solve sine equations.. just when using cosine rule to find angle in a triangle as i did.. are both 3pi/4 and 5pi/4 possible angles ? for say alpha
okay so it was originally \[(\sqrt{10})^{2} = (\sqrt{2})^{2} + 2^2 -2(2)(\sqrt{2})\cos(x) \] which simplified to \[\cos(x)=-\frac{1}{\sqrt{2}} \]
oh ok...then rule out that angle...it can't be....sorry
so x=3pi/4,5pi/4 but angles in triangle<pi .: 5pi/4 not soln?
yes...u r right
if u r being asked in a triangle then x=3 pi/4
hm ok..
and uh.. how to solve tan equations? like tan(x)=-1 not quite sure
use the identity tan(pi - theta) = -tan(theta)
do i get x=3pi/4 and 7pi/4 and keep adding 2pi to each one ?
or tan( 2 pi -theta) = -tan(theta)
yes...actually when tan theta= x then x= n(pi) + theta, where n = ....-1,-2,-1,0,1,2,3....
yeah im just wondeirng do u keep adding pi or get the 2 angles and add 2 pi.. or same thing
aksh??
actually tan has a period pi....
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