Question Number 137864 by n0y0n last updated on 07/Apr/21 $$\:\mathrm{if}\:, \\ $$$$\:\:\:\:\mathrm{acos}\alpha+\mathrm{bcos}\beta=\mathrm{Acos}\emptyset \\ $$$$\:\mathrm{proof}\:\mathrm{that}\: \\ $$$$\:\:\:\:\mathrm{asin}\alpha+\mathrm{bsin}\beta=\mathrm{Asin}\emptyset \\ $$ Commented by mr W last updated on…
Question Number 137861 by SLVR last updated on 07/Apr/21 Commented by SLVR last updated on 07/Apr/21 $${kindly}\:{help}\:{me}…{please} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 137851 by liberty last updated on 07/Apr/21 $${If}\:\mathrm{2}^{\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{3sin}\:\theta+\mathrm{1}} \:+\:\mathrm{2}^{\mathrm{2}−\mathrm{2sin}\:^{\mathrm{2}} \theta+\mathrm{3sin}\:\theta} \:=\:\mathrm{9} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:. \\ $$ Answered by bobhans last updated on 07/Apr/21…
Question Number 6750 by Tawakalitu. last updated on 22/Jul/16 Answered by Yozzii last updated on 22/Jul/16 $${Let}\:\angle{QPR}=\angle{RPS}=\alpha>\mathrm{0},\:\angle{PQR}=\beta>\mathrm{0}\:{and}\:\angle{PSR}=\delta>\mathrm{0}. \\ $$$${By}\:{the}\:{sine}\:{rule},\:{in}\:\bigtriangleup{QPR},\: \\ $$$$\frac{{QR}}{{sin}\alpha}=\frac{{QP}}{{sin}\angle{PRQ}}=\frac{\mathrm{3}{a}}{{sin}\left(\pi−\left(\alpha+\beta\right)\right)} \\ $$$$\Rightarrow{QR}=\frac{\mathrm{3}{asin}\alpha}{{sin}\left(\alpha+\beta\right)}. \\ $$$${In}\:\bigtriangleup{PRS},\:{by}\:{the}\:{sine}\:{rule}\:{we}\:{have}…
Question Number 137813 by bramlexs22 last updated on 07/Apr/21 $$\mathrm{cos}\:\left(\mathrm{arctan}\:\left(\frac{\mathrm{21}}{\mathrm{60}}\right)\right)=? \\ $$ Answered by mr W last updated on 07/Apr/21 $${say}\:{t}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{21}}{\mathrm{60}} \\ $$$$\mathrm{tan}\:{t}=\frac{\mathrm{21}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{20}} \\…
Question Number 137811 by bramlexs22 last updated on 07/Apr/21 $${Given}\:\begin{cases}{\mathrm{cos}\:\left({x}−{y}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{x}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:{x}}\end{cases} \\ $$$${where}\:\mathrm{270}°\:<\:{y}<\:\mathrm{360}°\:.\:{find} \\ $$$$\mathrm{sin}\:\mathrm{2}{y}\:. \\ $$ Answered by EDWIN88 last updated on 07/Apr/21 $$\Leftrightarrow\:\mathrm{cos}\:\left({x}−{y}\right)+\mathrm{cos}\:\left({x}+{y}\right)=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{cos}\:{x} \\…
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Question Number 6723 by Tawakalitu. last updated on 16/Jul/16 $${An}\:{MTN}\:{mask}\:{is}\:{erected}\:{at}\:{a}\:{point}\:{P}\:\:{in}\:{ilaro}\:{town}.\:{At}\:{a}\:{point} \\ $$$${B}\:{due}\:{west},\:{The}\:{angle}\:{of}\:{elevation}\:{of}\:{its}\:{top}\:{is}\:\beta\:{and}\:{at}\:{point}\: \\ $$$${C}\:{due}\:{south},\:{the}\:{angle}\:{of}\:{elevation}\:{is}\:\alpha.\:{With}\:{the}\:{aid}\:{of}\:{an}\: \\ $$$${appropriate}\:{diagam}.\:{show}\:{that}\:{the}\:{angle}\:{of}\:{elevation}\:{of}\:{the}\:{top} \\ $$$${from}\:{a}\:{point}\:{due}\:{south}\:{of}\:{B}\:{and}\:{due}\:{west}\:{of}\:{C}\:{is}\: \\ $$$$ \\ $$$${cot}^{−\mathrm{1}} \left[{cot}^{\mathrm{2}} \left(\beta\right)\:+\:{cot}^{\mathrm{2}} \left(\alpha\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 6715 by Tawakalitu. last updated on 15/Jul/16 $${Solve}\:{the}\:{following}\:{equation}\:{for}\:\mathrm{0}\:<\:\Theta\:<\:\mathrm{360}^{{o}} \\ $$$${cosx}\:+\:{cos}\mathrm{3}{x}\:+\:{cos}\mathrm{5}{x}\:+\:{cos}\mathrm{7}{x}\:=\:\mathrm{0} \\ $$ Answered by Yozzii last updated on 16/Jul/16 $${We}\:{know}\:{the}\:{identity}\:{cosa}+{cosb}=\mathrm{2}{cos}\frac{{a}+{b}}{\mathrm{2}}{cos}\frac{{a}−{b}}{\mathrm{2}}\:{for}\:{a},{b}\in\mathbb{R}….\left(\mathrm{1}\right) \\ $$$${Let}\:{u}={cosx}+{cos}\mathrm{3}{x}+{cos}\mathrm{5}{x}+{cos}\mathrm{7}{x}. \\…
Question Number 137771 by bemath last updated on 06/Apr/21 Commented by bemath last updated on 06/Apr/21 $${any}\:{one}\:{solve}\:{the}\:{question}\:{number} \\ $$$$\mathrm{16}? \\ $$ Commented by Ñï= last…