Question #88774

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Question #88774

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Answer 1 · 2016-12-11T08:31:35

We know

${(cos θ + sin θ)}^{2} + {(cos θ - sin θ)}^{2} = 2 ({cos}^{2} θ + {sin}^{2} θ)$ $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)$

$\Rightarrow {(\sqrt{2} cos θ)}^{2} + {(cos θ - sin θ)}^{2} = 2 \times 1$ $=>(sqrt2costheta)^2+(costheta-sintheta)^2=2xx1$

$\Rightarrow {(cos θ - sin θ)}^{2} = 2 - 2 {cos}^{2} θ$ $=>(costheta-sintheta)^2=2-2cos^2theta$

$\Rightarrow {(cos θ - sin θ)}^{2} = 2 (1 - {cos}^{2} θ)$ $=>(costheta-sintheta)^2=2(1-cos^2theta)$

$\Rightarrow {(cos θ - sin θ)}^{2} = 2 {sin}^{2} θ$ $=>(costheta-sintheta)^2=2sin^2theta$

$\Rightarrow cos θ - sin θ = \sqrt{2} sin θ$ $=>costheta-sintheta=sqrt2sintheta$

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Question #88774

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