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Category: Trigonometry

sin-12-13-and-is-in-the-2nd-quadrent-prove-cos-5-13-

Question Number 71146 by sadimuhmud 136 last updated on 12/Oct/19 $$\mathrm{sin}\:\alpha=\frac{\mathrm{12}}{\mathrm{13}}\:\:\mathrm{and}\:\alpha\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{2nd}\:\mathrm{quadrent}. \\ $$$$\mathrm{prove}\:\mathrm{cos}\:\alpha=−\frac{\mathrm{5}}{\mathrm{13}} \\ $$ Commented by Rio Michael last updated on 12/Oct/19 $${sin}\alpha\:=\:\frac{\mathrm{12}}{\mathrm{13}} \\…

arcsin-x-arcsin-1-x-t-sin-x-sin-1-x-t-sin-cos-cos-sin-sin-t-x-1-1-x-1-x-1-x-x-1-x-1-t-pi-2-

Question Number 136679 by mnjuly1970 last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:{arcsin}\sqrt{{x}}\:+{arcsin}\left(\sqrt{\mathrm{1}−{x}}\:\right)={t} \\ $$$$\:\:\:\:\:\:{sin}\left(\alpha\right)=\sqrt{{x}}\:\:\:\:{sin}\left(\beta\right)=\sqrt{\mathrm{1}−{x}}\: \\ $$$$\:\:\:\:\alpha+\beta={t} \\ $$$$\:\:\:\:{sin}\left(\alpha\right){cos}\left(\beta\right)+{cos}\left(\alpha\right){sin}\left(\beta\right)={sin}\left({t}\right) \\ $$$$\:\:\sqrt{{x}\:}\:.\sqrt{\mathrm{1}−\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}\:.\sqrt{\mathrm{1}−{x}}\: \\ $$$$={x}+\mathrm{1}−{x}=\mathrm{1} \\ $$$$\:\:\:{t}=\frac{\pi}{\mathrm{2}}… \\ $$$$\:\:\: \\…

Solve-the-following-equation-x-sin-x-2-3-pi-

Question Number 5590 by Rasheed Soomro last updated on 21/May/16 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation} \\ $$$$\mathrm{x}−\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2}}{\mathrm{3}}\pi\:. \\ $$ Commented by Yozzii last updated on 24/May/16 $${Numerical}\:{methods}\:{work},\:{but}\:{one}\: \\ $$$${may}\:{wonder}\:{what}\:{the}\:{exact}\:{value}\left({s}\right)…

cos180-tan-x-

Question Number 5479 by keagan j last updated on 15/May/16 $${cos}\mathrm{180}.{tan}\left(−{x}\right) \\ $$ Answered by Rasheed Soomro last updated on 15/May/16 $${cos}\:\mathrm{180}=−\mathrm{1},{tan}\left(−{x}\right)=−{tan}\left({x}\right)\:\left[{tan}\:\:{is}\:{odd}\:{function}\right] \\ $$$$\therefore\:{cos}\mathrm{180}.{tan}\left(−{x}\right)=\left(−\mathrm{1}\right)\left(−{tan}\:{x}\right) \\…

The-angle-of-elevation-of-point-A-and-B-from-P-are-and-respectively-The-bearing-of-A-and-B-from-P-are-S20-W-and-S40-E-and-their-distance-from-P-measured-on-the-map-are-3cm-and-1cm-respetively-

Question Number 5405 by sanusihammed last updated on 14/May/16 $${The}\:{angle}\:{of}\:{elevation}\:{of}\:{point}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:\alpha\:{and}\:\beta\: \\ $$$${respectively}.\:{The}\:{bearing}\:{of}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:{S}\mathrm{20}°{W}\:{and} \\ $$$${S}\mathrm{40}°{E}\:{and}\:{their}\:{distance}\:{from}\:{P}\:\:{measured}\:{on}\:{the}\:{map}\:{are} \\ $$$$\mathrm{3}{cm}\:{and}\:\mathrm{1}{cm}\:{respetively}.\:{A}\:{is}\:{higher}\:{than}\:{B}.\:{Prove}\:{that}\:{the}\: \\ $$$${elevation}\:{of}\:{A}\:{from}\:{B}\:{is} \\ $$$$ \\ $$$$\frac{{tan}^{−\mathrm{1}} \left[\mathrm{3}{tan}\alpha\:−\:{tan}\beta\right]}{\:\sqrt{\mathrm{7}}\:} \\ $$$$…