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Category: Trigonometry

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

Solve-sin-2x-2sin-2-x-3-cos-x-for-x-real-and-0-x-2pi-

Question Number 136268 by EDWIN88 last updated on 20/Mar/21 $$\mathrm{Solve}\:\mathrm{sin}\:\left(\mathrm{2x}\right)−\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{for}\: \\ $$$$\mathrm{x}\:\mathrm{real}\:\mathrm{and}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

3-cos-x-13-sin-2x-3-17-cos-x-3-

Question Number 136121 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{3}\:\mathrm{cos}\:{x}\:=\:\mathrm{13}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\:+\:\mathrm{17}\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{3}}\right) \\ $$ Answered by EDWIN88 last updated on 19/Mar/21 $$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{3}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{3cos}\:\mathrm{3t}\:=\:\mathrm{13sin}\:\mathrm{2t}+\mathrm{17cos}\:\mathrm{t} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{4cos}\:^{\mathrm{3}} \mathrm{t}−\mathrm{3cos}\:\mathrm{t}\right)=\mathrm{26sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:+\mathrm{17cos}\:\mathrm{t} \\ $$$$\mathrm{12cos}\:^{\mathrm{3}}…

2-cos-x-cos-2x-sin-2x-1-2cos-x-2sin-x-

Question Number 136122 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{2}\left(\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}\right)\:+\mathrm{sin}\:\mathrm{2}{x}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)=\mathrm{2sin}\:{x}\: \\ $$ Answered by EDWIN88 last updated on 18/Mar/21 $$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{2sin}\:\mathrm{xcos}\:\mathrm{x}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{x}\right)−\mathrm{2sin}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{2sin}\:\mathrm{x}\left[\:\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}\:\right]=\mathrm{0}…