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Category: Trigonometry

Given-cos-8-6cos-6-13cos-4-8cos-2-cos-7-5cos-5-8cos-3-1-2-then-what-the-value-of-tan-2-

Question Number 137338 by liberty last updated on 01/Apr/21 $$\mathrm{Given}\:\frac{\mathrm{cos}\:\mathrm{8}\theta+\mathrm{6cos}\:\mathrm{6}\theta+\mathrm{13cos}\:\mathrm{4}\theta+\mathrm{8cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{7}\theta+\mathrm{5cos}\:\mathrm{5}\theta+\mathrm{8cos}\:\mathrm{3}\theta}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\mathrm{2}\theta\:? \\ $$ Answered by EDWIN88 last updated on 01/Apr/21 $$\mathrm{consider}\:\mathrm{numerator} \\ $$$$\Leftrightarrow\:\mathrm{cos}\:\mathrm{8}\theta+\mathrm{cos}\:\mathrm{6}\theta+\mathrm{5}\left(\mathrm{cos}\:\mathrm{6}\theta+\mathrm{cos}\:\mathrm{4}\theta\right)+\mathrm{8}\left(\mathrm{cos}\:\mathrm{4}\theta+\mathrm{cos}\:\mathrm{2}\theta\right) \\…

Show-that-180-2cos-1-2-13-sin-1-12-13-

Question Number 6183 by Rasheed Soomro last updated on 17/Jun/16 $${Show}\:{that} \\ $$$$\mathrm{180}°−\mathrm{2cos}^{−\mathrm{1}} \:\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$ Answered by Yozzii last updated on 18/Jun/16 $$\pi−\mathrm{2}{acos}\left(\mathrm{2}/\sqrt{\mathrm{13}}\right)={asin}\left(\mathrm{12}/\mathrm{13}\right)…

if-and-is-corner-in-abc-indication-that-cos-cos-

Question Number 6148 by saiful last updated on 16/Jun/16 $${if}\:\alpha,\beta\:{and}\gamma\:{is}\:{corner}\:{in}\:\Delta{abc}.\:{indication}\:{that}\:{cos}\left(\beta+\gamma\right)=−{cos}\alpha \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $${cos}\left(\beta+\gamma\right)=−{cos}\alpha \\ $$$$−−−−−−−−−− \\ $$$$\because\alpha,\beta,\gamma\:{are}\:{angles}\:{of}\:{a}\:{triangle}…

Determine-the-value-of-sin-3-p-cos-6-p-cos-3-p-sin-6-p-

Question Number 6146 by Ninik last updated on 16/Jun/16 $${Determine}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right) \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right) \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$=\mathrm{sin}\:\left(\left(\frac{\pi}{\mathrm{3}}+{p}\right)−\left(\frac{\pi}{\mathrm{6}}+{p}\right)\right)…