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Question Number 21296 by NECx last updated on 19/Sep/17

Commented by NECx last updated on 19/Sep/17

please help me solve this. It was solved before but i cant assess the answer again. please help

$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this}.\:\mathrm{It}\:\mathrm{was} \\ $$$$\mathrm{solved}\:\mathrm{before}\:\mathrm{but}\:\mathrm{i}\:\mathrm{cant}\:\mathrm{assess}\:\mathrm{the} \\ $$$$\mathrm{answer}\:\mathrm{again}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$

Commented by NECx last updated on 20/Sep/17

i think then it was solved by smalso334 .... something like that

$$\mathrm{i}\:\mathrm{think}\:\mathrm{then}\:\mathrm{it}\:\mathrm{was}\:\mathrm{solved}\:\mathrm{by}\: \\ $$$$\mathrm{smalso334}\:\:\:....\:\mathrm{something}\:\mathrm{like}\:\mathrm{that} \\ $$

Commented by NECx last updated on 20/Sep/17

please help

$$\mathrm{please}\:\mathrm{help} \\ $$

Answered by Joel577 last updated on 20/Sep/17

2sin^(−1) (x(√6)) + sin^(−1) (4x) = (π/2) sin^(−1) x(√6) = α → sin α = x(√6) sin^(−1) 4x = β → sin β = 4x sin (2sin^(−1) x(√6) + sin^(−1) 4x) = sin (π/2) sin (2α + β) = 1 sin 2α cos β + cos 2α sin β = 1 2sin α cos α cos β + (1 − 2sin^2 α)sin β = 1 2sin α cos α cos β = 2 . x(√6) . (√(1 − 6x^2 )) . (√(1 − 16x^2 )) = 2x(√(6(1 − 6x^2 )(1 − 16x^2 ))) (1 − 2sin^2 α)sin β = (1 − 12x^2 )4x = 4x − 48x^3 2x(√(6(1 − 6x^2 )(1 − 16x^2 ))) + 4x − 48x^3 = 1 4x^2 (√(576x^4 − 132x^2 + 6)) = 48x^3 − 4x + 1 4x^2 (576x^4 − 132x^2 + 6) = 2304x^6 − 384x^4 + 96x^3 + 16x^2 − 8x + 1 2304x^6 − 528x^4 + 24x^2 = 2304x^6 − 384x^4 + 96x^3 + 16x^2 − 8x + 1 144x^4 + 96x^3 − 8x^2 − 8x + 1 = 0 If we plot it into Cartecian plane, it shows x = −(1/2) or x = (1/6)

$$\mathrm{2sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\:+\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:{x}\sqrt{\mathrm{6}}\:=\:\alpha\:\:\rightarrow\:\mathrm{sin}\:\alpha\:=\:{x}\sqrt{\mathrm{6}} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:\mathrm{4}{x}\:=\:\beta\:\:\:\:\rightarrow\:\:\mathrm{sin}\:\beta\:=\:\mathrm{4}{x} \\ $$$$ \\ $$$$\mathrm{sin}\:\left(\mathrm{2sin}^{−\mathrm{1}} {x}\sqrt{\mathrm{6}}\:+\:\mathrm{sin}^{−\mathrm{1}} \:\mathrm{4}{x}\right)\:=\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(\mathrm{2}\alpha\:+\:\beta\right)\:=\:\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{cos}\:\beta\:+\:\mathrm{cos}\:\mathrm{2}\alpha\:\mathrm{sin}\:\beta\:=\:\mathrm{1} \\ $$$$\mathrm{2sin}\:\alpha\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:+\:\left(\mathrm{1}\:−\:\mathrm{2sin}^{\mathrm{2}} \:\alpha\right)\mathrm{sin}\:\beta\:=\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{2sin}\:\alpha\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:=\:\mathrm{2}\:.\:{x}\sqrt{\mathrm{6}}\:.\:\sqrt{\mathrm{1}\:−\:\mathrm{6}{x}^{\mathrm{2}} }\:.\:\sqrt{\mathrm{1}\:−\:\mathrm{16}{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{x}\sqrt{\mathrm{6}\left(\mathrm{1}\:−\:\mathrm{6}{x}^{\mathrm{2}} \right)\left(\mathrm{1}\:−\:\mathrm{16}{x}^{\mathrm{2}} \right)} \\ $$$$\left(\mathrm{1}\:−\:\mathrm{2sin}^{\mathrm{2}} \:\alpha\right)\mathrm{sin}\:\beta\:=\:\left(\mathrm{1}\:−\:\mathrm{12}{x}^{\mathrm{2}} \right)\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}{x}\:−\:\mathrm{48}{x}^{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{2}{x}\sqrt{\mathrm{6}\left(\mathrm{1}\:−\:\mathrm{6}{x}^{\mathrm{2}} \right)\left(\mathrm{1}\:−\:\mathrm{16}{x}^{\mathrm{2}} \right)}\:+\:\mathrm{4}{x}\:−\:\mathrm{48}{x}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \sqrt{\mathrm{576}{x}^{\mathrm{4}} \:−\:\mathrm{132}{x}^{\mathrm{2}} \:+\:\mathrm{6}}\:\:=\:\mathrm{48}{x}^{\mathrm{3}} \:−\:\mathrm{4}{x}\:+\:\mathrm{1} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{576}{x}^{\mathrm{4}} \:−\:\mathrm{132}{x}^{\mathrm{2}} \:+\:\mathrm{6}\right)\:=\:\mathrm{2304}{x}^{\mathrm{6}} \:−\:\mathrm{384}{x}^{\mathrm{4}} \:+\:\mathrm{96}{x}^{\mathrm{3}} \:+\:\mathrm{16}{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\:\mathrm{1} \\ $$$$\mathrm{2304}{x}^{\mathrm{6}} \:−\:\mathrm{528}{x}^{\mathrm{4}} \:+\:\mathrm{24}{x}^{\mathrm{2}} \:\:=\:\mathrm{2304}{x}^{\mathrm{6}} \:−\:\mathrm{384}{x}^{\mathrm{4}} \:+\:\mathrm{96}{x}^{\mathrm{3}} \:+\:\mathrm{16}{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\:\mathrm{1} \\ $$$$\mathrm{144}{x}^{\mathrm{4}} \:+\:\mathrm{96}{x}^{\mathrm{3}} \:−\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{plot}\:\mathrm{it}\:\mathrm{into}\:\mathrm{Cartecian}\:\mathrm{plane},\:\mathrm{it}\:\mathrm{shows} \\ $$$${x}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by Joel577 last updated on 20/Sep/17

If someone have shorter ways to solve this, please explain it to us. Thanks

$$\mathrm{If}\:\mathrm{someone}\:\mathrm{have}\:\mathrm{shorter}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}, \\ $$$$\mathrm{please}\:\mathrm{explain}\:\mathrm{it}\:\mathrm{to}\:\mathrm{us}.\:\mathrm{Thanks} \\ $$

Commented by NECx last updated on 20/Sep/17

thanks boss

$$\mathrm{thanks}\:\mathrm{boss} \\ $$

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