Given,. cosθ+sinθ=√2cosθ. squaring on both the sides, we get,. cos2θ+sin2θ+2sinθcosθ=2cos2θ. cos2θ−sin2θ=2cosθsinθ. (cosθ+sinθ)(cosθ−sinθ)=2cosθsinθ. √2cosθ(cos ... |
19 мая 2020 г. · We have, cos θ + sin θ = √2 cos θ Squaring both side, (cos θ + sin θ) 2 = 2 cos 2 θ ⇒ cos 2 θ + sin 2 θ + 2 sin θ. |
Cos theta+sin theta =√2cos theta show thatcos theta-sin theta =√2sin theta. Open in App. Solution. cosθ+sinθ = √2cosθ on ... cosθsinθ+cosθ=21−2cos2θ. |
7 авг. 2024 г. · To prove that cosθ−sinθ=√2sinθ given that cosθ+sinθ=√2cosθ, we will follow these steps: Step 1: Start with the given equation · If sinθ+cosθ=√2 ... |
If cosθ+sinθ=√2cosθ, then cosθ−sinθ=_____. √2cosθ+sinθ; √2cosθ; √2tanθ; √2sinθ. A. √2cosθ+sinθ. B. √2tanθ. C. √2cosθ. D. √2sinθ. |
Given: Cos θ + Sin θ = \(√{2}\) Cos θ Calculation: ⇒ Squaring both sides, we get ⇒ \(cos^2\theta+sin^2\theta+2c. |
16 нояб. 2022 г. · Given: cos θ + sin θ = √2 cos θ. Calculation: Taking square both the sides 2 2. ⇒ cos2 θ + sin2 θ + 2.sin θ . cos θ = 2 cos2 θ. |
30 мая 2014 г. · Question :- → If cos θ + sin θ = √2cos θ , then prove that cos θ - sin θ = √2sin θ .Answer :- We have, → cos θ + sin θ = √2cos θ . |
21 мая 2019 г. · Given that: cos x + sin x = √2 cos x, or sin x = cos x (√2–1) Next to prove, cos x - sin x = √2 sin x, take LHS = cos x - sin x = [sin x ... |
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