Category: Trigonometry

Question Number 194315 by cortano12 last updated on 03/Jul/23 $$\:\:\:\:\: \\ $$ Answered by Frix last updated on 03/Jul/23 $$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{transformed}\:\mathrm{to}\:\left[{s}=\mathrm{sin}\:{x}\right]: \\ $$$${s}^{\mathrm{4}} +\frac{{s}^{\mathrm{3}} }{\mathrm{2}}−{s}^{\mathrm{2}} −\frac{{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0}…

Question Number 194304 by 281981 last updated on 02/Jul/23 Answered by cortano12 last updated on 03/Jul/23 $$\:\:\:\:\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)\:\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{sin}\:\left(\mathrm{2A}+\mathrm{2B}\right)\:\mathrm{sin}\:\left(\mathrm{2A}−\mathrm{2B}\right)=\:\mathrm{2} \\ $$$$\:\:\:\underbrace{\boldsymbol{{A}}} \\ $$ Commented by…

Question Number 194252 by horsebrand11 last updated on 01/Jul/23 $$\:\:\mathrm{Find}\:\mathrm{V}\:=\:\mathrm{tan}\:\mathrm{9}°−\mathrm{tan}\:\mathrm{27}°−\mathrm{tan}\:\mathrm{63}°+\mathrm{tan}\:\mathrm{81}° \\ $$ Answered by peter frank last updated on 01/Jul/23 $$\mathrm{Qn}\:\mathrm{177542} \\ $$ Answered by…

Question Number 193749 by Mingma last updated on 19/Jun/23 Answered by som(math1967) last updated on 19/Jun/23 Commented by som(math1967) last updated on 19/Jun/23 $${AD}={DC}={AC}=\mathrm{12}{cm} \\…

Question Number 193595 by Rupesh123 last updated on 17/Jun/23 Answered by MM42 last updated on 17/Jun/23 $${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta−{sin}\alpha{cos}\beta−{sim}\beta{cos}\alpha=\mathrm{0} \\ $$$$\Rightarrow{sin}\alpha\left({sin}\alpha−{cos}\beta\right)+{sin}\beta\left({sin}\beta−{cos}\alpha\right)=\mathrm{0} \\ $$$${sin}\alpha\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)+{sin}\beta\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\left({sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0}…