Category: Trigonometry

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

Question Number 194252 by horsebrand11 last updated on 01/Jul/23 $$\mathrm{Find}\mathrm{V}=\tan 9°-\tan 27°-\tan 63°+\tan 81°$$ Answered by peter frank last updated on 01/Jul/23 $$\mathrm{Qn}177542$$ 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 $${sin}^2\alpha +{sin}^2\beta -sin\alpha cos\beta -sim\beta cos\alpha =0$$$$\Rightarrow sin\alpha (sin\alpha -cos\beta )+sin\beta (sin\beta -cos\alpha )=0$$$$sin\alpha \left(2sin(\frac{\alpha +\beta }{2}-\frac{\pi }{4})cos(\frac{\alpha -\beta }{2}+\frac{\pi }{4})\right)+sin\beta \left(2sin(\frac{\alpha +\beta }{2}-\frac{\pi }{4})cos(\frac{\beta -\alpha }{2}+\frac{\pi }{4})\right)=0$$$$\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}…

Question Number 193622 by a.lgnaoui last updated on 17/Jun/23 $$\mathrm{determiner}\mathrm{l}\mathrm{angle}\mathbf{x}$$ Commented by a.lgnaoui last updated on 17/Jun/23 Answered by cherokeesay last updated on…