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Question Number 147492 by Kunal12588 last updated on 21/Jul/21

2(√(19)) cos [(1/3)tan^(−1) (((45(√3))/(28)))] it is equal to 8. How?

$$\mathrm{2}\sqrt{\mathrm{19}}\:\mathrm{cos}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\right)\right] \\ $$$${it}\:{is}\:{equal}\:{to}\:\mathrm{8}.\:{How}? \\ $$

Commented by MJS_new last updated on 21/Jul/21

just an idea x^3 −57x−56=0 x_1 =−7 x_2 =−1 x_3 =8 using the trigonometric solution formula (because p=−57∧q=−56 ⇒ (p^3 /(27))+(q^2 /4)<0 ⇒ Cardano′s solution method is not possible) we get x_1 =−2(√(19))cos ((π/6)+(1/3)arcsin ((28(√(19)))/(361))) x_2 =−2(√(19))sin ((1/3)arcsin ((28(√(19)))/(361))) x_3 =2(√(19))sin ((π/3)+(1/3)arcsin ((28(√(19)))/(361))) now we must show cos ((1/3)arctan ((45(√3))/(28))) =sin ((π/3)+(1/3)arcsin ((28(√(19)))/(361)))

$$\mathrm{just}\:\mathrm{an}\:\mathrm{idea} \\ $$$${x}^{\mathrm{3}} −\mathrm{57}{x}−\mathrm{56}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\mathrm{7} \\ $$$${x}_{\mathrm{2}} =−\mathrm{1} \\ $$$${x}_{\mathrm{3}} =\mathrm{8} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{solution}\:\mathrm{formula} \\ $$$$\left(\mathrm{because}\:{p}=−\mathrm{57}\wedge{q}=−\mathrm{56}\:\Rightarrow\:\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\Rightarrow\right. \\ $$$$\left.\mathrm{Cardano}'\mathrm{s}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}\right) \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}\sqrt{\mathrm{19}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{28}\sqrt{\mathrm{19}}}{\mathrm{361}}\right) \\ $$$${x}_{\mathrm{2}} =−\mathrm{2}\sqrt{\mathrm{19}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{28}\sqrt{\mathrm{19}}}{\mathrm{361}}\right) \\ $$$${x}_{\mathrm{3}} =\mathrm{2}\sqrt{\mathrm{19}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{28}\sqrt{\mathrm{19}}}{\mathrm{361}}\right) \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{must}\:\mathrm{show} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:\frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\right)\:=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{28}\sqrt{\mathrm{19}}}{\mathrm{361}}\right) \\ $$

Commented by Kunal12588 last updated on 22/Jul/21

thank you sir is there any other method than cardano when discriminant<0

$${thank}\:{you}\:{sir} \\ $$$${is}\:{there}\:{any}\:{other}\:{method}\:{than}\:{cardano}\:{when} \\ $$$${discriminant}<\mathrm{0} \\ $$

Commented by MJS_new last updated on 22/Jul/21

Cardano works only for D≥0 for D<0 we need the Trigonometric Method see my comments to question 114263

$$\mathrm{Cardano}\:\mathrm{works}\:\boldsymbol{{only}}\:\mathrm{for}\:{D}\geqslant\mathrm{0} \\ $$$$\mathrm{for}\:{D}<\mathrm{0}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{Trigonometric}\:\mathrm{Method} \\ $$$$\mathrm{see}\:\mathrm{my}\:\mathrm{comments}\:\mathrm{to}\:\mathrm{question}\:\mathrm{114263} \\ $$

Answered by EDWIN88 last updated on 21/Jul/21

Commented by Kunal12588 last updated on 21/Jul/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by mr W last updated on 21/Jul/21

s=2(√(19)) cos [(1/3)tan^(−1) (((45(√3))/(28)))] =2(√(19)) cos [(1/3)cos^(−1) (((28)/( 19(√(19)))))] =2(√((−((−57)/3)))) cos [(1/3)cos^(−1) (−((−56)/(2(√((−((−57)/3))^3 )))))]>0 which is root of x^2 −57x−56=0 (x+1)(x+7)(x−8)=0 8 is the only positive root, therefore 2(√(19)) cos [(1/3)tan^(−1) (((45(√3))/(28)))]=8

$${s}=\mathrm{2}\sqrt{\mathrm{19}}\:\mathrm{cos}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{19}}\:\mathrm{cos}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{28}}{\:\mathrm{19}\sqrt{\mathrm{19}}}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\left(−\frac{−\mathrm{57}}{\mathrm{3}}\right)}\:\mathrm{cos}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}^{−\mathrm{1}} \left(−\frac{−\mathrm{56}}{\mathrm{2}\sqrt{\left(−\frac{−\mathrm{57}}{\mathrm{3}}\right)^{\mathrm{3}} }}\right)\right]>\mathrm{0} \\ $$$${which}\:{is}\:{root}\:{of} \\ $$$${x}^{\mathrm{2}} −\mathrm{57}{x}−\mathrm{56}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{7}\right)\left({x}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\mathrm{8}\:{is}\:{the}\:{only}\:{positive}\:{root},\:{therefore} \\ $$$$\mathrm{2}\sqrt{\mathrm{19}}\:\mathrm{cos}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{45}\sqrt{\mathrm{3}}}{\mathrm{28}}\right)\right]=\mathrm{8} \\ $$

Commented by Kunal12588 last updated on 22/Jul/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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