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Question Number 119295 by bobhans last updated on 23/Oct/20

Let a and b non negative real numbers If sin x+acos x=b , express ∣ asin x−cos x∣ in terms of a and b.

$$\:{Let}\:{a}\:{and}\:{b}\:\:{non}\:{negative}\:{real}\:{numbers}\: \\ $$$${If}\:\mathrm{sin}\:{x}+{a}\mathrm{cos}\:{x}={b}\:,\:{express}\: \\ $$$$\mid\:{a}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\mid\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$$$ \\ $$

Answered by bemath last updated on 23/Oct/20

consider relations a^2 +1 = (sin^2 x+cos^2 x)(a^2 +1) ⇒ a^2 +1=a^2 sin^2 x+sin^2 x+a^2 cos^2 x+cos^2 x = (a^2 sin^2 x−2asin xcos x+cos^2 x)+ (a^2 cos^2 x+2acos xsin x+sin^2 x) = (asin x−cos x)^2 +(acos x+sin x)^2 ⇒(asin x−cos x)^2 =a^2 +1−b^2 ⇔ asin x−cos x =± (√(a^2 −b^2 +1)) ⇔ ∣asin x−cos x∣ = (√(a^2 −b^2 +1))

$${consider}\:{relations}\:{a}^{\mathrm{2}} +\mathrm{1}\:=\:\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)\left({a}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} +\mathrm{1}={a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{2}} {x}+{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$$\:\:\:\:\:\:\:\:=\:\left({a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{2}{a}\mathrm{sin}\:{x}\mathrm{cos}\:{x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{2}{a}\mathrm{cos}\:{x}\mathrm{sin}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x}\right) \\ $$$$\:\:\:\:\:\:\:=\:\left({a}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} +\left({a}\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{1}−{b}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:{a}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\:=\pm\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Leftrightarrow\:\mid{a}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\mid\:=\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{1}}\: \\ $$$$ \\ $$

Commented by bobhans last updated on 23/Oct/20

great!

$${great}! \\ $$

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