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Question Number 152175 by john_santu last updated on 26/Aug/21

what exact value sin 2x if given cos 3x = (2/( (√5)))

$$\mathrm{what}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{if}\:\mathrm{given} \\ $$$$\:\mathrm{cos}\:\mathrm{3x}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$

Commented by MJS_new last updated on 26/Aug/21

cos 3x =4cos^3 x −3cos x sin 2x =±2cos x (√(1−cos^2 x)) cos^3 x −(3/4)cos x −((√5)/(10))=0 ⇒ cos x = { ((−cos ((π/6)+(1/3)arcsin (2/( (√5)))))),((−sin ((1/3)arcsin (2/( (√5)))))),((sin ((π/3)+(1/3)arcsin (2/( (√5)))))) :} ⇒ ±2cos x (√(1−cos^2 x))= { ((∓sin ((π/3)+(2/3)arcsin (2/( (√5)))))),((∓sin ((2/3)arcsin (2/( (√5)))))),((±cos ((π/6)+(2/3)arcsin (2/( (√5)))))) :}

$$\mathrm{cos}\:\mathrm{3}{x}\:=\mathrm{4cos}^{\mathrm{3}} \:{x}\:−\mathrm{3cos}\:{x} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=\pm\mathrm{2cos}\:{x}\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{3}} \:{x}\:−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{cos}\:{x}\:−\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{cos}\:{x}\:=\begin{cases}{−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)}\\{−\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)}\\{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)}\end{cases} \\ $$$$\Rightarrow \\ $$$$\pm\mathrm{2cos}\:{x}\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}}=\begin{cases}{\mp\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)}\\{\mp\mathrm{sin}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)}\\{\pm\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)}\end{cases} \\ $$

Commented by john_santu last updated on 26/Aug/21

i have try by cardano but not solved

$$\mathrm{i}\:\mathrm{have}\:\mathrm{try}\:\mathrm{by}\:\mathrm{cardano}\:\mathrm{but} \\ $$$$\mathrm{not}\:\mathrm{solved} \\ $$

Commented by MJS_new last updated on 26/Aug/21

in this case we need the Trigonometric Method

$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{Trigonometric}\:\mathrm{Method} \\ $$

Commented by MJS_new last updated on 26/Aug/21

I found no “nice” expressions...

$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:``\mathrm{nice}''\:\mathrm{expressions}... \\ $$

Commented by john_santu last updated on 26/Aug/21

what the formula Trigonometry method?

$$\mathrm{what}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{Trigonometry}\:\mathrm{method}? \\ $$

Commented by MJS_new last updated on 26/Aug/21

I posted it in comments to question 114263

$$\mathrm{I}\:\mathrm{posted}\:\mathrm{it}\:\mathrm{in}\:\mathrm{comments}\:\mathrm{to}\:\mathrm{question}\:\mathrm{114263} \\ $$

Answered by mr W last updated on 26/Aug/21

cos 6x=2×((2/( (√5))))^2 −1=(3/5) sin 6x=±(4/5) 6x=2kπ±sin^(−1) (4/5) ⇒sin 2x=sin (1/3)(2kπ±sin^(−1) (4/5))

$$\mathrm{cos}\:\mathrm{6}{x}=\mathrm{2}×\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\mathrm{6}{x}=\pm\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{6}{x}=\mathrm{2}{k}\pi\pm\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}=\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}{k}\pi\pm\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$

Commented by otchereabdullai@gmail.com last updated on 26/Aug/21

nice!

$$\mathrm{nice}! \\ $$

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