Category: Trigonometry

Question Number 109401 by john santu last updated on 23/Aug/20 $$\frac{JS}{\mathrm{\_}\mathrm{\_}00\mathrm{\_}00\mathrm{\_}\mathrm{\_}00}$$$$solvetheequation4\sin 3x+\frac{1}{3}\cos 3x=3$$ Answered by mr W last updated on 23/Aug/20 $$\mathrm{12}\:\mathrm{sin}\:\mathrm{3}{x}+\mathrm{cos}\:\mathrm{3}{x}=\mathrm{9} \…

Question Number 43810 by gyugfeet last updated on 15/Sep/18 $$\frac{1-cos\theta +co\beta -cos(\theta +\beta )}{1+cos\theta -cos\beta -cos(\theta +\beta )}=tan\frac{\theta }{2}.cot\frac{\beta }{2}$$ Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18 $$=\frac{2{sin}^2\frac{\theta }{2}+2sin(\frac{\theta }{2}+\beta )sin\left(\frac{\theta }{2}\right)}{2{cos}^2\left(\frac{\theta }{2}\right)-2cos(\frac{\theta }{2}+\beta )cos\left(\frac{\theta }{2}\right)}$$$$={tan}\frac{\theta}{\mathrm{2}}×\frac{{sin}\frac{\theta}{\mathrm{2}}+{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)−{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right)} \…

Question Number 109282 by mnjuly1970 last updated on 22/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 174666 by mnjuly1970 last updated on 07/Aug/22 Answered by behi834171 last updated on 08/Aug/22 $$\mathit{c}\mathit{o}\mathit{s}\frac{\mathit{C}}{2}=\sqrt{\frac{\mathit{p}(\mathit{p}-\mathit{c})}{\mathit{a}\mathit{b}}}$$$${\mathit{c}\mathit{o}\mathit{s}}^2\mathbf{\Psi }=1-{\mathit{s}\mathit{i}\mathit{n}}^2\mathbf{\Psi }=1-\frac{4\mathit{a}\mathit{b}}{{(\mathit{a}+\mathit{b})}^2}.\frac{\mathit{p}(\mathit{p}-\mathit{c})}{\mathit{a}\mathit{b}}=$$$$=\mathrm{1}−\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}\right)}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }=…

Question Number 43543 by peter frank last updated on 11/Sep/18 $$provethat\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}$$$$2\cos \theta$$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Sep/18 $$\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta }}}$$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{8}\theta\right)}\:}}…