if cos ∅ - sin ∅ =√2 sin∅, prove that cos ∅ + sin∅ =√2 cos∅
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Here I am writing Ф as A because it is difficult for me to write Ф Always.
![Given : cos A - sin A = (\sqrt{2} sin A) Given : cos A - sin A = (\sqrt{2} sin A)](https://tex.z-dn.net/?f=Given+%3A+cos+A+-+sin+A+%3D++%28%5Csqrt%7B2%7D+sin+A%29)
On Squaring both sides, we get
![(cos A - sin A)^2 = ( \sqrt{2} sin A)^2 (cos A - sin A)^2 = ( \sqrt{2} sin A)^2](https://tex.z-dn.net/?f=%28cos+A+-+sin+A%29%5E2+%3D+%28+%5Csqrt%7B2%7D+sin+A%29%5E2)
cos^2A+ sin^2A - 2sinAcosA = 2sin^2 A
cos^2A = sin^2A + 2sinAcosA
Add cos^2A on both sides, we get
cos^2 A + cos^2A = sin^2A + cos^2A + 2sinAcosA
2cos^2A = (cosA+sinA)^2
![\sqrt{2} cos A = (cos A + sin A) \sqrt{2} cos A = (cos A + sin A)](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+cos+A+%3D+%28cos+A+%2B+sin+A%29)
Hope this helps!
On Squaring both sides, we get
cos^2A+ sin^2A - 2sinAcosA = 2sin^2 A
cos^2A = sin^2A + 2sinAcosA
Add cos^2A on both sides, we get
cos^2 A + cos^2A = sin^2A + cos^2A + 2sinAcosA
2cos^2A = (cosA+sinA)^2
Hope this helps!
siddhartharao77:
:-)
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