Question Number 176514 by CrispyXYZ last updated on 20/Sep/22
$$\bigtriangleup{ABC},\:\mathrm{sin}{A}=\mathrm{cos}{B}=\mathrm{tan}{C} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}^{\mathrm{3}} {A}+\mathrm{cos}^{\mathrm{2}} {A}−\mathrm{cos}{A}. \\ $$
Answered by mr W last updated on 21/Sep/22
![sin A=cos B ⇒B=A−(π/2) C=π−A−B=((3π)/2)−2A tan C=(1/(tan 2A))=((1−tan^2 A)/(2 tan A))=sin A 1−((sin^2 A)/(cos^2 A))=((2 sin^2 A)/(cos A)) ((2 cos^2 A−1)/(cos A))=2(1−cos^2 A) ⇒cos^3 A+cos^2 A−cos A=(1/2) ✓ cos A=(1/3)[4 sin (−(1/3)sin^(−1) (5/(32)) +((2kπ)/3))−1] since A>(π/2) and −1<cos A<0, the only solution is cos A=−(1/3)[4 sin ((1/3)sin^(−1) (5/(32)) )+1] ⇒A=π−cos^(−1) {(1/3)[4 sin ((1/3)sin^(−1) (5/(32)) )+1]} ≈113.767 843 060 258° A≈113.767843° ⇒sin A≈0.915186 B≈23.767843° ⇒cos B≈0.915186 C≈42.464314° ⇒tan C≈0.915186](https://www.tinkutara.com/question/Q176527.png)
$$\mathrm{sin}\:{A}=\mathrm{cos}\:{B}\:\Rightarrow{B}={A}−\frac{\pi}{\mathrm{2}} \\ $$$${C}=\pi−{A}−{B}=\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{2}{A} \\ $$$$\mathrm{tan}\:{C}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}{A}}=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{A}}{\mathrm{2}\:\mathrm{tan}\:{A}}=\mathrm{sin}\:{A} \\ $$$$\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:{A}}{\mathrm{cos}^{\mathrm{2}} \:{A}}=\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{A}}{\mathrm{cos}\:{A}} \\ $$$$\frac{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{A}−\mathrm{1}}{\mathrm{cos}\:{A}}=\mathrm{2}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{A}\right) \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{3}} \:{A}+\mathrm{cos}^{\mathrm{2}} \:{A}−\mathrm{cos}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$$$\mathrm{cos}\:{A}=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{4}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{32}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)−\mathrm{1}\right] \\ $$$${since}\:{A}>\frac{\pi}{\mathrm{2}}\:{and}\:−\mathrm{1}<\mathrm{cos}\:{A}<\mathrm{0},\: \\ $$$${the}\:{only}\:{solution}\:{is} \\ $$$$\mathrm{cos}\:{A}=−\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{4}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{32}}\:\right)+\mathrm{1}\right] \\ $$$$\Rightarrow{A}=\pi−\mathrm{cos}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{4}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{32}}\:\right)+\mathrm{1}\right]\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\mathrm{113}.\mathrm{767}\:\mathrm{843}\:\mathrm{060}\:\mathrm{258}° \\ $$$$ \\ $$$${A}\approx\mathrm{113}.\mathrm{767843}°\:\Rightarrow\mathrm{sin}\:{A}\approx\mathrm{0}.\mathrm{915186} \\ $$$${B}\approx\mathrm{23}.\mathrm{767843}°\:\Rightarrow\mathrm{cos}\:{B}\approx\mathrm{0}.\mathrm{915186} \\ $$$${C}\approx\mathrm{42}.\mathrm{464314}°\:\Rightarrow\mathrm{tan}\:{C}\approx\mathrm{0}.\mathrm{915186} \\ $$
Commented by Tawa11 last updated on 22/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$