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Question Number 86490 by ram roop sharma last updated on 29/Mar/20
If sin θ + cos θ=(√2) cos θ then cos θ−sin θ is equal to
$$\mathrm{If}\:\:\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta=\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta\:\mathrm{then} \\ $$$$\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by john santu last updated on 29/Mar/20
sin θ = (√2) cos θ−cos θ tan θ = (√2) −1 let cos θ−sin θ = k sin θ−cos θ = −k (sin θ−cos θ)(sin θ+cos θ) = −k(√2) cos θ −cos 2θ = −k(√2) cos θ 2cos^2 θ −1 = k(√2) cos θ k = ((2cos^2 θ−1)/( (√2) cos θ)) = (√2) cos θ−(1/( (√2))) cos θ = (√2) ((1/( (√2) ((√(2−(√2)))))))−(1/(2(√(2−(√2))))) = (1/( (√(2−(√2))))) − (1/(2(√(2−(√2))))) = (1/(2(√(2−(√2)))))
$$\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta−\mathrm{cos}\:\theta \\ $$$$\mathrm{tan}\:\theta\:=\:\sqrt{\mathrm{2}}\:−\mathrm{1}\: \\ $$$$\mathrm{let}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\:=\:\mathrm{k}\: \\ $$$$\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\:=\:−\mathrm{k} \\ $$$$\left(\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right)\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)\:=\:−\mathrm{k}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta \\ $$$$−\mathrm{cos}\:\mathrm{2}\theta\:=\:−\mathrm{k}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2cos}\:^{\mathrm{2}} \theta\:−\mathrm{1}\:=\:\mathrm{k}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{k}\:=\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}}{\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta}\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\theta \\ $$$$=\:\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\:−\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}} \\ $$
Answered by som(math1967) last updated on 29/Mar/20
sinθ=((√2)−1)cosθ ((√2)+1)sinθ=((√2)+1)((√2)−1)cosθ (√2)sinθ +sinθ=1.cosθ (√2)sinθ=cosθ−sinθ ∴cosθ−sinθ=(√2)sinθ ans
$${sin}\theta=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){cos}\theta \\ $$$$\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){sin}\theta=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){cos}\theta \\ $$$$\sqrt{\mathrm{2}}{sin}\theta\:+{sin}\theta=\mathrm{1}.{cos}\theta \\ $$$$\sqrt{\mathrm{2}}{sin}\theta={cos}\theta−{sin}\theta \\ $$$$\therefore{cos}\theta−{sin}\theta=\sqrt{\mathrm{2}}{sin}\theta\:{ans} \\ $$
Commented by peter frank last updated on 29/Mar/20
thank you
$${thank}\:{you} \\ $$

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