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: sin(2β − 90) = cos(180 − 2β)\displaystyle \:\sin(2\beta\,-\,90) \:=\:\cos(180\,-\,2\beta)sin(2β−90)=cos(180−2β) ? . . . . Yes! Click to expand... sin(2β − 90) = sin(2β)cos(90) − sin(90)cos(2β)\displaystyle \sin(2\beta\,-\,90)\;=\;\sin(2\beta)\cos(90)\,-\,\sin(90)\cos(2\beta)sin(2β−90)=sin(2β)cos(90)−sin(90)cos(2β) . . . . . . . . . = sin(2β) ⋅ 0 − 1 ⋅ cos(2β)\displaystyle = \;\;\;\;\sin(2\beta)\,\cdot\,0\;-\;1\,\cdot\,\cos(2\beta)=sin(2β)⋅0−1⋅cos(2β) . . . . . . . . . = −cos(2β)\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)=−cos(2β) cos(180 − 2β) = cos(180)cos(2β) + sin(180)sin(2β)\displaystyle \cos(180 \,-\,2\beta)\;=\;\cos(180)\cos(2\beta)\,+\,\sin(180)\sin(2\beta)cos(180−2β)=cos(180)cos(2β)+sin(180)sin(2β) . . . . . . . . . = (−1) ⋅ cos(2β) + 0 ⋅ sin(2β)\displaystyle = \;\;\;(-1)\,\cdot\,\cos(2\beta)\;+\;0\,\cdot\,\sin(2\beta)=(−1)⋅cos(2β)+0⋅sin(2β) . . . . . . . . . = −cos(2β)\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)=−cos(2β)
Re: quick trig q Hello, oded244! Can i say that: sin(2β − 90) = cos(180 − 2β)\displaystyle \:\sin(2\beta\,-\,90) \:=\:\cos(180\,-\,2\beta)sin(2β−90)=cos(180−2β) ? . . . . Yes! Click to expand... sin(2β − 90) = sin(2β)cos(90) − sin(90)cos(2β)\displaystyle \sin(2\beta\,-\,90)\;=\;\sin(2\beta)\cos(90)\,-\,\sin(90)\cos(2\beta)sin(2β−90)=sin(2β)cos(90)−sin(90)cos(2β) . . . . . . . . . = sin(2β) ⋅ 0 − 1 ⋅ cos(2β)\displaystyle = \;\;\;\;\sin(2\beta)\,\cdot\,0\;-\;1\,\cdot\,\cos(2\beta)=sin(2β)⋅0−1⋅cos(2β) . . . . . . . . . = −cos(2β)\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)=−cos(2β) cos(180 − 2β) = cos(180)cos(2β) + sin(180)sin(2β)\displaystyle \cos(180 \,-\,2\beta)\;=\;\cos(180)\cos(2\beta)\,+\,\sin(180)\sin(2\beta)cos(180−2β)=cos(180)cos(2β)+sin(180)sin(2β) . . . . . . . . . = (−1) ⋅ cos(2β) + 0 ⋅ sin(2β)\displaystyle = \;\;\;(-1)\,\cdot\,\cos(2\beta)\;+\;0\,\cdot\,\sin(2\beta)=(−1)⋅cos(2β)+0⋅sin(2β) . . . . . . . . . = −cos(2β)\displaystyle =\;\;\;\;\;\;\;\;\;-\cos(2\beta)=−cos(2β)
