can i say that sin(2β-90)= cos(180-2β) ?
O oded244 New member Joined Oct 4, 2007 Messages 38 Oct 17, 2007 #1 can i say that sin(2β-90)= cos(180-2β) ?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Oct 17, 2007 #2 Re: quick trig q Hello, oded244! Can i say that: \(\displaystyle \:\sin(2\beta\,-\,90) \:=\:\cos(180\,-\,2\beta)\) ? . . . . Yes! Click to expand... \(\displaystyle \sin(2\beta\,-\,90)\;=\;\sin(2\beta)\cos(90)\,-\,\sin(90)\cos(2\beta)\) . . . . . . . . . \(\displaystyle = \;\;\;\;\sin(2\beta)\,\cdot\,0\;-\;1\,\cdot\,\cos(2\beta)\) . . . . . . . . . \(\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)\) \(\displaystyle \cos(180 \,-\,2\beta)\;=\;\cos(180)\cos(2\beta)\,+\,\sin(180)\sin(2\beta)\) . . . . . . . . . \(\displaystyle = \;\;\;(-1)\,\cdot\,\cos(2\beta)\;+\;0\,\cdot\,\sin(2\beta)\) . . . . . . . . . \(\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)\)
Re: quick trig q Hello, oded244! Can i say that: \(\displaystyle \:\sin(2\beta\,-\,90) \:=\:\cos(180\,-\,2\beta)\) ? . . . . Yes! Click to expand... \(\displaystyle \sin(2\beta\,-\,90)\;=\;\sin(2\beta)\cos(90)\,-\,\sin(90)\cos(2\beta)\) . . . . . . . . . \(\displaystyle = \;\;\;\;\sin(2\beta)\,\cdot\,0\;-\;1\,\cdot\,\cos(2\beta)\) . . . . . . . . . \(\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)\) \(\displaystyle \cos(180 \,-\,2\beta)\;=\;\cos(180)\cos(2\beta)\,+\,\sin(180)\sin(2\beta)\) . . . . . . . . . \(\displaystyle = \;\;\;(-1)\,\cdot\,\cos(2\beta)\;+\;0\,\cdot\,\sin(2\beta)\) . . . . . . . . . \(\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)\)
