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Question Number 127169 by bramlexs22 last updated on 27/Dec/20

If (1+cos x)(1+sin x) = (3/2)   then (1−cos x)(1−sin x) =?    nice trigonometry

$${If}\:\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:{then}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)\:=? \\ $$$$ \\ $$$${nice}\:{trigonometry}\: \\ $$

Commented by Dwaipayan Shikari last updated on 27/Dec/20

(1−cosx)(1−sinx)(1+cosx)(1+sinx)=(3/2)(1−cosx)(1−sinx)  ⇒sin^2 x cos^2 x=(3/2)(1−cosx)(1−sinx)=(3/2)a  1+cosx+sinx+sinxcosx=(3/2)→(a)  1−cosx−sinx+sinxcosx=a→(b)     (a+b)→   2sinxcosx=a−(1/2)  (3/2)a=(1/4)(a−(1/2))^2 ⇒a^2 +(1/4)−a=6a⇒a^2 −7a+(1/4)⇒a=(7/2)±2(√3)  Here a=(7/2)−2(√3)

$$\left(\mathrm{1}−{cosx}\right)\left(\mathrm{1}−{sinx}\right)\left(\mathrm{1}+{cosx}\right)\left(\mathrm{1}+{sinx}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}−{cosx}\right)\left(\mathrm{1}−{sinx}\right) \\ $$$$\Rightarrow{sin}^{\mathrm{2}} {x}\:{cos}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}−{cosx}\right)\left(\mathrm{1}−{sinx}\right)=\frac{\mathrm{3}}{\mathrm{2}}{a} \\ $$$$\mathrm{1}+{cosx}+{sinx}+{sinxcosx}=\frac{\mathrm{3}}{\mathrm{2}}\rightarrow\left({a}\right) \\ $$$$\mathrm{1}−{cosx}−{sinx}+{sinxcosx}={a}\rightarrow\left({b}\right) \\ $$$$\:\:\:\left({a}+{b}\right)\rightarrow\:\:\:\mathrm{2}{sinxcosx}={a}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{a}=\frac{\mathrm{1}}{\mathrm{4}}\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}−{a}=\mathrm{6}{a}\Rightarrow{a}^{\mathrm{2}} −\mathrm{7}{a}+\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{a}=\frac{\mathrm{7}}{\mathrm{2}}\pm\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${Here}\:{a}=\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by bramlexs22 last updated on 27/Dec/20

Commented by liberty last updated on 27/Dec/20

(1+sin x)(1+cos x)=(3/2)  (sin (x/2)+cos (x/2))^2 (2cos^2 (x/2))=(3/2)  (sin (x/2)+cos (x/2)).cos (x/2) =± ((√3)/2)  sin (x/2).cos (x/2)+cos^2 (x/2) =±((√3)/2)  ((sin x)/2)+((1+cos x)/2)= ±((√3)/2)  ⇒sin x+cos x = (√3)−1 ; other value is  outside the range .  note: range of (sin x+cos x)= [−(√2) ,(√2) ]   and−1−(√3) <−(√2)   ∴ (sin x+cos x)^2 =((√3)−1)^2       1+2sin xcos x = 4−2(√3)  then f(x)=(1−sin x)(1−cos x)                      = 1−(sin x+cos x)+sin xcos x                      = 1−((√3)−1)+((3−2(√3))/2)                      = ((7−4(√3))/2) ≈ 0.035898

$$\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{2cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right).\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:=\pm\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{{x}}{\mathrm{2}}.\mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{2}}+\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{2}}=\:\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\sqrt{\mathrm{3}}−\mathrm{1}\:;\:{other}\:{value}\:{is} \\ $$$${outside}\:{the}\:{range}\:. \\ $$$${note}:\:{range}\:{of}\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)=\:\left[−\sqrt{\mathrm{2}}\:,\sqrt{\mathrm{2}}\:\right]\: \\ $$$${and}−\mathrm{1}−\sqrt{\mathrm{3}}\:<−\sqrt{\mathrm{2}}\: \\ $$$$\therefore\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{1}+\mathrm{2sin}\:{x}\mathrm{cos}\:{x}\:=\:\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${then}\:{f}\left({x}\right)=\left(\mathrm{1}−\mathrm{sin}\:{x}\right)\left(\mathrm{1}−\mathrm{cos}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}−\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)+\mathrm{sin}\:{x}\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)+\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\approx\:\mathrm{0}.\mathrm{035898} \\ $$$$ \\ $$

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