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Prove : \(\sin (90 - \theta) = \cos \theta\)
Actually I am confused that how can I : prove that sin in 2nd quadrant is +ve ? That is my basic confusion ...
can u use sin (A-B) formula?
No idea. Actually that was just in start and the formula arises later, so I think that there must be an easier method?
draw a right triangle
:( not so good but OK ? :)
|dw:1350046739720:dw| nothing is there without naming , OK now I think.
|dw:1350046691525:dw| now find sin 90-theta and cos theta using basic defination
OK wait please !
In case of sin 90 - theta , will perpendicular be : adjacent (base) and base will be opposite?
know the basic defination of sin and cos ? |dw:1350046897401:dw| apply it here, in terms of a and b,c
OH IT IS soh cah toa and not soa cah toa , sorry :(
sin = opposite / hypotenuse cos = adjacent / hypotenuse
sin (90-theta) = b/c cos (theta) = b/c and hence sin(90-theta) = cos(theta) , got it now? lol jk. Well is this right @hartnn
yep, hence proved.
next identity please.
prove that sin in 2nd quadrant is +ve ? That is my basic confusion ...
here, put theta = -theta
and use cos(-theta) = cos theta
how to prove that cos(-theta)=theta ?
again with the use of right triangle @hartnn ?
how will u find -theta in triangle ?
oh yes .. silly me . So, a cartesian plane ?
no, cartesian plane will not work also.
ccw : +ve angle cw : -ve angle
because lengths will be positive , so ratio will always come out to be positive, hence, maybe cartesian plane won't work
Some how agreed .
searched net now, http://www.mathsisfun.com/algebra/trig-four-quadrants.html is that the correct one ?
yes, its correct....we could prove it using co-ordinates.... (cos t,sin t)= (x,y) so sin t is +ve in 2nd quadrant
doubts ended :) thanks a lo.
mathisfun is always a nice reference....
more we knw more confuse we become
more we practice, more the confusion vanishes....
more we practice, more ways we look at things n hence more confusions which ofcourse gives kick ;p