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Question Number 139710 by ajfour last updated on 30/Apr/21

If 3sin x+5cos x=5 find value(s) of 5sin x−3cos x.

$${If}\:\:\mathrm{3sin}\:{x}+\mathrm{5cos}\:{x}=\mathrm{5} \\ $$$${find}\:{value}\left({s}\right)\:{of}\:\:\mathrm{5sin}\:{x}−\mathrm{3cos}\:{x}. \\ $$

Answered by MJS_new last updated on 30/Apr/21

3sin x +5cos x =5 ((3tan x+5)/( (√(1+tan^2 x))))=5 ⇒ tan x =0∨tan x =((15)/8) 5sin x −3cos x =−((3−5tan x)/( (√(1+tan^2 x))))= { ((−3)),(3) :}

$$\mathrm{3sin}\:{x}\:+\mathrm{5cos}\:{x}\:=\mathrm{5} \\ $$$$\frac{\mathrm{3tan}\:{x}+\mathrm{5}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}=\mathrm{5}\:\Rightarrow\:\mathrm{tan}\:{x}\:=\mathrm{0}\vee\mathrm{tan}\:{x}\:=\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\mathrm{5sin}\:{x}\:−\mathrm{3cos}\:{x}\:=−\frac{\mathrm{3}−\mathrm{5tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}=\begin{cases}{−\mathrm{3}}\\{\mathrm{3}}\end{cases} \\ $$

Answered by john_santu last updated on 30/Apr/21

let 5sin x−3cos x=p ⇒9sin^2 x+30sin xcos x+25cos^2 x=25 ⇒9cos^2 x−30sin xcos x+25sin^2 x=p^2 ⇒9+25=p^2 +25 ⇒p^2 =9 ; p=±3 (∗) (√(25+9)) cos (x−α) = p ⇒−1≤ (p/( (√(34)) )) ≤ 1

$${let}\:\mathrm{5sin}\:{x}−\mathrm{3cos}\:{x}={p} \\ $$$$\Rightarrow\mathrm{9sin}\:^{\mathrm{2}} {x}+\mathrm{30sin}\:{x}\mathrm{cos}\:{x}+\mathrm{25cos}\:^{\mathrm{2}} {x}=\mathrm{25} \\ $$$$\Rightarrow\mathrm{9cos}\:^{\mathrm{2}} {x}−\mathrm{30sin}\:{x}\mathrm{cos}\:{x}+\mathrm{25sin}\:^{\mathrm{2}} {x}={p}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}+\mathrm{25}={p}^{\mathrm{2}} +\mathrm{25} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\mathrm{9}\:;\:{p}=\pm\mathrm{3} \\ $$$$\left(\ast\right)\:\sqrt{\mathrm{25}+\mathrm{9}}\:\mathrm{cos}\:\left({x}−\alpha\right)\:=\:{p} \\ $$$$\Rightarrow−\mathrm{1}\leqslant\:\frac{{p}}{\:\sqrt{\mathrm{34}}\:}\:\leqslant\:\mathrm{1}\: \\ $$

Answered by mr W last updated on 30/Apr/21

3 sin x+5 cos x=5 ...(i) 5 sin x−3 cos x=k ...(ii) (i)^2 +(ii)^2 : 3^2 +5^2 =5^2 +k^2 ⇒k=±3 or (√(3^2 +5^2 ))sin (x+α)=5 ⇒sin (x+α)=(5/( (√(3^2 +5^2 )))) ⇒cos (x+α)=±(3/( (√(3^2 +5^2 )))) k=(√(3^2 +5^2 )) cos (x+α)=±3

$$\mathrm{3}\:\mathrm{sin}\:{x}+\mathrm{5}\:\mathrm{cos}\:{x}=\mathrm{5}\:\:\:...\left({i}\right) \\ $$$$\mathrm{5}\:\mathrm{sin}\:{x}−\mathrm{3}\:\mathrm{cos}\:{x}={k}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({ii}\right)^{\mathrm{2}} : \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$$$\Rightarrow{k}=\pm\mathrm{3} \\ $$$$ \\ $$$${or} \\ $$$$\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\mathrm{sin}\:\left({x}+\alpha\right)=\mathrm{5} \\ $$$$\Rightarrow\mathrm{sin}\:\left({x}+\alpha\right)=\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{cos}\:\left({x}+\alpha\right)=\pm\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }} \\ $$$${k}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:\mathrm{cos}\:\left({x}+\alpha\right)=\pm\mathrm{3} \\ $$

Answered by ajfour last updated on 30/Apr/21

3sin x+5cos x=5 5sin x−3cos x=t sin x=((15+5t)/(34)) cos x=((25−3t)/(34)) 25(3+t)^2 +(25−3t)^2 =(34)^2 34t^2 +25×34=34×34 ⇒ t^2 =9 ⇒ t=±3

$$\mathrm{3sin}\:{x}+\mathrm{5cos}\:{x}=\mathrm{5} \\ $$$$\mathrm{5sin}\:{x}−\mathrm{3cos}\:{x}={t} \\ $$$$\mathrm{sin}\:{x}=\frac{\mathrm{15}+\mathrm{5}{t}}{\mathrm{34}} \\ $$$$\mathrm{cos}\:{x}=\frac{\mathrm{25}−\mathrm{3}{t}}{\mathrm{34}} \\ $$$$\mathrm{25}\left(\mathrm{3}+{t}\right)^{\mathrm{2}} +\left(\mathrm{25}−\mathrm{3}{t}\right)^{\mathrm{2}} =\left(\mathrm{34}\right)^{\mathrm{2}} \\ $$$$\mathrm{34}{t}^{\mathrm{2}} +\mathrm{25}×\mathrm{34}=\mathrm{34}×\mathrm{34} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} =\mathrm{9}\:\:\:\:\Rightarrow\:\:{t}=\pm\mathrm{3} \\ $$

Answered by ajfour last updated on 30/Apr/21

Thanks both Sirs.

$${Thanks}\:{both}\:{Sirs}. \\ $$

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