If Cos theta + Sin theta=√2Cos theta, show that Cos theta-Sin theta = √2Sin theta
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let theta = x
cos x + sin x = root2 cos x
squaring on both side, we get......
cos2x + sin2x + 2cosxsinx = 2cos2x
2sinxcosx = 2cos2x - cos2x - sin2x
2sinxcosx = cos2x - sin2x
2sinxcosx = (cosx+sinx) (cosx - sinx)
2sinxcosx = (root2 cosx) (cosx - sinx)
2sinxcosx/root2 cosx = cosx - sinx
root2 sinx = cosx - sinx
cos x + sin x = root2 cos x
squaring on both side, we get......
cos2x + sin2x + 2cosxsinx = 2cos2x
2sinxcosx = 2cos2x - cos2x - sin2x
2sinxcosx = cos2x - sin2x
2sinxcosx = (cosx+sinx) (cosx - sinx)
2sinxcosx = (root2 cosx) (cosx - sinx)
2sinxcosx/root2 cosx = cosx - sinx
root2 sinx = cosx - sinx
devendrayadu196:
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