Menu Close

Show-that-given-4cos-pi-3-2-3-sin-pi-3-5-then-pi-3cos-1-5-28-3tan-1-3-2-

Tinku Tara 0 Comments
Pin




Question Number 1830 by 112358 last updated on 10/Oct/15
Show that given 4cos(π/3)+2(√3)sin(π/3)=5 then π=3cos^(−1) ((5/( (√(28)))))+3tan^(−1) (((√3)/2)) .
$${Show}\:{that}\:{given} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{cos}\frac{\pi}{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{3}}{sin}\frac{\pi}{\mathrm{3}}=\mathrm{5} \\ $$$${then} \\ $$$$\:\:\:\:\:\:\:\pi=\mathrm{3}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{28}}}\right)+\mathrm{3}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:. \\ $$
Answered by Rasheed Soomro last updated on 10/Oct/15
′′4cos(π/3)+2(√3)sin(π/3) ′′ is a constant equal to 5 so to say ′′if 4cos(π/3)+2(√3)sin(π/3)=5′′ is meaningless! Similarly ′′3cos^(−1) ((5/( (√(28)))))+3tan^(−1) (((√3)/2)) ′′ is also a constant and it is not dependant on ′′4cos(π/3)+2(√3)sin(π/3)=5′′ I think that the Question is not meaningful!
$$''\mathrm{4}{cos}\frac{\pi}{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{3}}{sin}\frac{\pi}{\mathrm{3}}\:''\:{is}\:{a}\:{constant}\:{equal}\:{to}\:\mathrm{5} \\ $$$${so}\:{to}\:{say}\:''{if}\:\mathrm{4}{cos}\frac{\pi}{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{3}}{sin}\frac{\pi}{\mathrm{3}}=\mathrm{5}'' \\ $$$${is}\:{meaningless}! \\ $$$${Similarly}\:''\mathrm{3}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{28}}}\right)+\mathrm{3}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:''\:{is} \\ $$$${also}\:{a}\:{constant}\:{and}\:{it}\:{is}\:{not}\:{dependant}\:{on} \\ $$$$''\mathrm{4}{cos}\frac{\pi}{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{3}}{sin}\frac{\pi}{\mathrm{3}}=\mathrm{5}'' \\ $$$${I}\:{think}\:{that}\:{the}\:\:{Question}\:{is}\:{not}\:{meaningful}! \\ $$
Commented by 112358 last updated on 10/Oct/15
The original question is this. Write down a value of θ in the interval (π/4)<θ<(π/2) that satisfies the equation 4cosθ+2(√3)sinθ=5. Hence, or otherwise, show that π=3cos^(−1) ((5/( (√(28)))))+3tan^(−1) (((√3)/2)). Show that π=4sin^(−1) (((7(√2))/(10)))−4tan^(−1) ((3/4)). I just thought that from setting θ=(π/3) the question required (though not strictly) that you work from 4cos(π/3)+2(√3)sin(π/3)=5.
$${The}\:{original}\:{question}\:{is}\:{this}. \\ $$$${Write}\:{down}\:{a}\:{value}\:{of}\:\theta\:{in}\:{the} \\ $$$${interval}\:\:\frac{\pi}{\mathrm{4}}<\theta<\frac{\pi}{\mathrm{2}}\:{that}\:{satisfies} \\ $$$${the}\:{equation} \\ $$$$\mathrm{4}{cos}\theta+\mathrm{2}\sqrt{\mathrm{3}}{sin}\theta=\mathrm{5}.\: \\ $$$${Hence},\:{or}\:{otherwise},\:{show}\:{that} \\ $$$$\pi=\mathrm{3}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{28}}}\right)+\mathrm{3}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right). \\ $$$${Show}\:{that}\: \\ $$$$\pi=\mathrm{4}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{10}}\right)−\mathrm{4}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right).\: \\ $$$${I}\:{just}\:{thought}\:{that}\:{from}\:{setting} \\ $$$$\theta=\frac{\pi}{\mathrm{3}}\:{the}\:{question}\:{required}\:\left({though}\right. \\ $$$$\left.{not}\:{strictly}\right)\:{that}\:{you}\:{work}\:{from} \\ $$$$\mathrm{4}{cos}\frac{\pi}{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{3}}{sin}\frac{\pi}{\mathrm{3}}=\mathrm{5}.\: \\ $$
Commented by 112358 last updated on 10/Oct/15
I′ve corrected the statement and I think I′ve solved it.
$${I}'{ve}\:{corrected}\:{the}\:{statement}\:{and} \\ $$$${I}\:{think}\:{I}'{ve}\:{solved}\:{it}.\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *