Question Number 86168 by jagoll last updated on 27/Mar/20 $$\mathrm{given}\: \\ $$$$\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{y}\:=\:\mathrm{2sin}\:\left(\mathrm{x}+\mathrm{y}\right)\: \\ $$$$\mathrm{x}+\mathrm{y}\:\neq\:\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\:\mathrm{tan}\:\frac{\mathrm{y}}{\mathrm{2}}\:=\: \\ $$ Answered by john santu last…
Question Number 20639 by oyshi last updated on 30/Aug/17 $${if}\:\mathrm{cos}\:\left({A}+{B}\right)\mathrm{sin}\:\left({C}+{D}\right)=\mathrm{cos}\:\left({A}−{B}\right) \\ $$$$\mathrm{sin}\:\left({C}−{D}\right) \\ $$$${so}\:{proof}\:\mathrm{cot}\:{A}\mathrm{cot}\:{B}\mathrm{cot}\:{C}=\mathrm{cot}\:{D} \\ $$ Answered by Tinkutara last updated on 30/Aug/17 $$\frac{\mathrm{sin}\:\left({C}+{D}\right)}{\mathrm{sin}\:\left({C}−{D}\right)}=\frac{\mathrm{cos}\:\left({A}−{B}\right)}{\mathrm{cos}\:\left({A}+{B}\right)} \\…
Question Number 20640 by oyshi last updated on 30/Aug/17 $$\mathrm{sin}\:\theta={K}\mathrm{cos}\:\left(\theta−\alpha\right) \\ $$$${so}\:{proof}\:{it},\mathrm{cot}\:\theta=\frac{\mathrm{1}−{K}\mathrm{sin}\:\alpha}{{K}\mathrm{cos}\:\alpha} \\ $$ Answered by mrW1 last updated on 30/Aug/17 $$\mathrm{sin}\:\theta=\mathrm{K}\left(\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha\right) \\ $$$$\left(\mathrm{1}−\mathrm{Ksin}\:\alpha\right)\mathrm{sin}\:\theta=\mathrm{Kcos}\:\alpha\:\mathrm{cos}\:\theta \\…
Question Number 20616 by virus last updated on 29/Aug/17 $${x}={r}\mathrm{cos}\:\theta\mathrm{cos}\:\emptyset,{y}={r}\mathrm{sin}\:\theta\mathrm{sin}\:\emptyset,{z}=\mathrm{sin}\:\theta \\ $$$${then}\:{prove}\:{that}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$ Commented by virus last updated on 01/Sep/17 $${please}\:{solve}\:{this}…
Question Number 86111 by john santu last updated on 27/Mar/20 $$\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{cos}\:{x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by jagoll last updated on 27/Mar/20 $$\Leftrightarrow\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}\:=\:−\frac{\pi}{\mathrm{6}}\:+\:\mathrm{2k}\pi \\ $$$$\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}\pi}\:\left\{−\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\right\}…
Question Number 20576 by gopikrishnan005@gmail.com last updated on 28/Aug/17 $${sec}\left({A}−\mathrm{3}\Pi/\mathrm{2}\right) \\ $$ Answered by Tinkutara last updated on 28/Aug/17 $$\mathrm{sec}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−{A}\right)=−\mathrm{cosec}\:{A} \\ $$ Terms of Service…
Question Number 20506 by Tinkutara last updated on 27/Aug/17 $${Simplify}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2}}}\right),\:\frac{\pi}{\mathrm{4}}\:<\:{x}\:<\:\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$ Answered by ajfour last updated on 27/Aug/17 $$\:\theta=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{cos}\:\left(\pi/\mathrm{4}\right)\mathrm{cos}\:{x}+\mathrm{sin}\:\left(\pi/\mathrm{4}\right)\mathrm{sin}\:{x}\right] \\…
Question Number 20509 by Tinkutara last updated on 27/Aug/17 $${Simplify}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2}}}\right),\:\frac{\mathrm{5}\pi}{\mathrm{4}}\:<\:{x}\:<\:\frac{\mathrm{9}\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20505 by Tinkutara last updated on 27/Aug/17 $${Simplify}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{cos}\:{x}\:+\:\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{sin}\:{x}\right),\:{where} \\ $$$$−\frac{\mathrm{3}\pi}{\mathrm{4}}\:\leqslant\:{x}\:\leqslant\:\frac{\pi}{\mathrm{4}} \\ $$ Answered by ajfour last updated on 27/Aug/17 $${let}\:\mathrm{tan}\:\alpha=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{5}},…
Question Number 151493 by iloveisrael last updated on 21/Aug/21 $$\:\:\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{2}}−\mathrm{6}{x}\right)+\mathrm{sin}\:\left(\pi+\mathrm{4}{x}\right)+\mathrm{sin}\:\left(\mathrm{3}\pi−{x}\right)}{\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{2}}+\mathrm{6}{x}\right)+\mathrm{cos}\:\left(\mathrm{4}{x}−\mathrm{2}\pi\right)+\mathrm{cos}\:\left({x}+\mathrm{2}\pi\right)}\:=? \\ $$ Commented by MJS_new last updated on 21/Aug/21 $$\mathrm{tan}\:{x} \\ $$ Commented by liberty…