Question

if cos ∅ - sin ∅ = √2 sin ∅, prove that cos ∅ + sin ∅
= √2 cos∅
$\large{✰} \bf \underline\color{red}{Now Prove\ \textless \ br /\ \textgreater \ - }$
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Answer 1

☘Aɴsᴡᴇʀ☘

cos θ + sin θ = √2cos θ

cos θ + sin θ = √2cos θSquaring both side, we get

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θcos2θ – 2 × cosθ × sinθ = sin2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θcos2θ – 2 × cosθ × sinθ = sin2θNow adding sin2θ both side, we get

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θcos2θ – 2 × cosθ × sinθ = sin2θNow adding sin2θ both side, we getcos2θ -2 × cosθ × sinθ + sin2θ = sin2θ + sin2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θcos2θ – 2 × cosθ × sinθ = sin2θNow adding sin2θ both side, we getcos2θ -2 × cosθ × sinθ + sin2θ = sin2θ + sin2θcosθ–sinθ2 = 2sin2θ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θcos2θ – 2 × cosθ × sinθ = sin2θNow adding sin2θ both side, we getcos2θ -2 × cosθ × sinθ + sin2θ = sin2θ + sin2θcosθ–sinθ2 = 2sin2θcos θ – sin θ = √2sinθ

cos θ + sin θ = √2cos θSquaring both side, we getcosθ+sinθ2 = 2cos2θcos2θ + sin2θ + 2 × cosθ × sinθ = 2cos2θsin2θ + 2 × cosθ × sinθ = 2cos2θ – cos2θsin2θ + 2 × cosθ × sinθ = cos2θcos2θ – 2 × cosθ × sinθ = sin2θNow adding sin2θ both side, we getcos2θ -2 × cosθ × sinθ + sin2θ = sin2θ + sin2θcosθ–sinθ2 = 2sin2θcos θ – sin θ = √2sinθ∴ cos θ – sin θ = √2sinθ

Answer 2

Answer:

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