cos-x-sin-x-4-5-5sin-x-

Question Number 89047 by jagoll last updated on 15/Apr/20

$$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{5sin}\:{x}\:=\:? \\ $$

Commented by john santu last updated on 15/Apr/20

± (√(1−sin^2 x)) = (4/5)−sin x 1−sin^2 x = ((16)/(25)) −(8/5)sin x +sin^2 x 2sin^2 x−(8/5)sin x−(9/(25)) = 0 ⇒sin x = ((((8/5)−((4(√(34)))/5)))/4) 5 sin x = 2−(√(34)) ≈ −3.831

$$\pm\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{4}}{\mathrm{5}}−\mathrm{sin}\:{x} \\ $$$$\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\:=\:\frac{\mathrm{16}}{\mathrm{25}}\:−\frac{\mathrm{8}}{\mathrm{5}}\mathrm{sin}\:{x}\:+\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{2sin}\:^{\mathrm{2}} {x}−\frac{\mathrm{8}}{\mathrm{5}}\mathrm{sin}\:{x}−\frac{\mathrm{9}}{\mathrm{25}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:{x}\:=\:\frac{\left(\frac{\mathrm{8}}{\mathrm{5}}−\frac{\mathrm{4}\sqrt{\mathrm{34}}}{\mathrm{5}}\right)}{\mathrm{4}} \\ $$$$\mathrm{5}\:\mathrm{sin}\:{x}\:=\:\mathrm{2}−\sqrt{\mathrm{34}}\:\approx\:−\mathrm{3}.\mathrm{831} \\ $$

Answered by MJS last updated on 15/Apr/20

t=tan (x/2) ⇔ x=2arctan t ((1−t^2 )/(t^2 +1))+((2t)/(t^2 +1))=(4/5) −5t^2 +10t+5=4t^2 +4 t^2 −((10)/9)t−(1/9)=0 t=(5/9)±((√(34))/9) ⇒ 5sin x =((10t)/(t^2 +1))=2±((√(34))/2)

$${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Leftrightarrow\:{x}=\mathrm{2arctan}\:{t} \\ $$$$\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$−\mathrm{5}{t}^{\mathrm{2}} +\mathrm{10}{t}+\mathrm{5}=\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4} \\ $$$${t}^{\mathrm{2}} −\frac{\mathrm{10}}{\mathrm{9}}{t}−\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{5}}{\mathrm{9}}\pm\frac{\sqrt{\mathrm{34}}}{\mathrm{9}} \\ $$$$\Rightarrow\:\mathrm{5sin}\:{x}\:=\frac{\mathrm{10}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2}\pm\frac{\sqrt{\mathrm{34}}}{\mathrm{2}} \\ $$

Commented by john santu last updated on 15/Apr/20

t^2 = ((59 ± 10(√(34)))/(81)) ⇒ t^2 +1 = ((140 ±10(√(34)))/(81)) 10t = ((50±10(√(34)))/9) ((10t)/(t^2 +1)) = ((50±10(√(34)))/9) ×((81)/(140±10(√(34)))) = (((5±(√(34)))×9)/(14±(√(34)))) ?

$${t}^{\mathrm{2}} \:=\:\frac{\mathrm{59}\:\pm\:\mathrm{10}\sqrt{\mathrm{34}}}{\mathrm{81}}\:\Rightarrow\:{t}^{\mathrm{2}} +\mathrm{1}\:=\:\frac{\mathrm{140}\:\pm\mathrm{10}\sqrt{\mathrm{34}}}{\mathrm{81}} \\ $$$$\mathrm{10}{t}\:=\:\frac{\mathrm{50}\pm\mathrm{10}\sqrt{\mathrm{34}}}{\mathrm{9}} \\ $$$$\frac{\mathrm{10}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\mathrm{50}\pm\mathrm{10}\sqrt{\mathrm{34}}}{\mathrm{9}}\:×\frac{\mathrm{81}}{\mathrm{140}\pm\mathrm{10}\sqrt{\mathrm{34}}} \\ $$$$=\:\frac{\left(\mathrm{5}\pm\sqrt{\mathrm{34}}\right)×\mathrm{9}}{\mathrm{14}\pm\sqrt{\mathrm{34}}}\:?\: \\ $$

Commented by MJS last updated on 15/Apr/20

=9(((5±(√(34)))(14∓(√(34))))/((14±(√(34)))(14∓(√(34)))))=9((36±9(√(34)))/(162))=2±((√(34))/2)

$$=\mathrm{9}\frac{\left(\mathrm{5}\pm\sqrt{\mathrm{34}}\right)\left(\mathrm{14}\mp\sqrt{\mathrm{34}}\right)}{\left(\mathrm{14}\pm\sqrt{\mathrm{34}}\right)\left(\mathrm{14}\mp\sqrt{\mathrm{34}}\right)}=\mathrm{9}\frac{\mathrm{36}\pm\mathrm{9}\sqrt{\mathrm{34}}}{\mathrm{162}}=\mathrm{2}\pm\frac{\sqrt{\mathrm{34}}}{\mathrm{2}} \\ $$

Commented by jagoll last updated on 15/Apr/20

what the correct answer? i see has 2 answer

$${what}\:{the}\:{correct}\:{answer}? \\ $$$${i}\:{see}\:{has}\:\mathrm{2}\:{answer}\: \\ $$

Commented by MJS last updated on 15/Apr/20

sin x +cos x =(4/5) has 2 solutions in [0; 2π[ ⇒ 2 correct answers

$$\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\:=\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{has}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{in}\:\left[\mathrm{0};\:\mathrm{2}\pi\left[\right.\right. \\ $$$$\Rightarrow\:\mathrm{2}\:\mathrm{correct}\:\mathrm{answers} \\ $$

Commented by jagoll last updated on 15/Apr/20

$${o}.\:{thanks}\:{very}\:{much}\:{sir}.\: \\ $$

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