Question Number 196886 by Amidip last updated on 02/Sep/23 Answered by MM42 last updated on 02/Sep/23 $${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{{p}}{{q}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{p}}{{q}−\mathrm{1}}=\frac{{sin}^{\mathrm{2}} \left(\alpha+\beta\right)}{{sin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)} \\ $$$$\left.\Rightarrow{psin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)=\left({q}−\mathrm{1}\right){sin}^{\mathrm{2}} \left(\alpha+\beta\right)\right) \\ $$$$\Rightarrow{sin}^{\mathrm{2}}…
Question Number 196881 by HeferH last updated on 02/Sep/23 Commented by MathematicalUser2357 last updated on 10/Sep/23 $$\ast\mathrm{Tears}\:\mathrm{paper}\ast \\ $$$$\mathrm{Now}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{question} \\ $$ Answered by HeferH last…
Question Number 196883 by Amidip last updated on 02/Sep/23 Answered by som(math1967) last updated on 02/Sep/23 $$\:\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} {y}=\pi−\mathrm{sin}^{−\mathrm{1}} {z} \\ $$$$\Rightarrow{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }+{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\mathrm{sin}^{−\mathrm{1}}…
Question Number 196816 by ERLY last updated on 01/Sep/23 Answered by MM42 last updated on 01/Sep/23 $$\left.{a}\right){S}=\frac{\mathrm{1}}{\mathrm{2}}+{cost}+{cos}\mathrm{2}{t}+…+{cosnt} \\ $$$$\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{S}={sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}−{sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{5}{t}}{\mathrm{2}}−{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}+…+{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{t}−{sin}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}{t} \\ $$$${S}=\frac{{sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right){t}}{\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}}\:\:\checkmark \\ $$$$\left.{b}\right){S}={sint}+{sin}\mathrm{3}{t}+…+{sin}\left(\mathrm{2}{n}+\mathrm{1}\right){t} \\ $$$$\mathrm{2}{sintS}=\mathrm{1}−{cos}\mathrm{2}{t}+{cos}\mathrm{2}{t}−{cos}\mathrm{4}{t}+{cos}\mathrm{4}{t}−{cos}\mathrm{6}{t}+…+{cos}\mathrm{2}\left({n}+\mathrm{1}\right){t}−{cos}\mathrm{2}{nt}…
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Question Number 196625 by sniper237 last updated on 28/Aug/23 $${Let}\:\:{a}={BC}\:\:{b}={CA}\:\:{c}={AB} \\ $$$${ABC}\:\:{is}\:{rectangle}\:{iff}\:\:{tan}\left(\frac{\hat {{A}}}{\mathrm{2}}\right)=\frac{{a}}{{b}+{c}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196611 by ERLY last updated on 27/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196523 by ERLY last updated on 27/Aug/23 $${resoudre}\:{dans}\:{c}\:{l}\:{equation}\:{sinx}=\mathrm{2}\:\:\:\:\:\bigstar{erly}\:{rolvinst}\bigstar\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<{erly}\:{rolvinst}> \\ $$ Answered by Frix last updated on 26/Aug/23 $$\mathrm{sin}\:{x}\:=\frac{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2i}}=\mathrm{2} \\ $$$$\mathrm{e}^{\mathrm{i}{x}} −\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{i}{x}}…
Question Number 196517 by ERLY last updated on 26/Aug/23 Answered by mr W last updated on 27/Aug/23 $$\mathrm{cos}\:{x}+{i}\:\mathrm{sin}\:{x}={e}^{{ix}} \\ $$$$\mathrm{cos}\:{x}−{i}\:\mathrm{sin}\:{x}={e}^{−{ix}} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}=\mathrm{2} \\…
Question Number 196518 by ERLY last updated on 26/Aug/23 $${resouxre}\:{cosx}=\mathrm{2} \\ $$ Answered by Frix last updated on 26/Aug/23 $$\mathrm{Similar}\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{question} \\ $$$${x}=\mathrm{2}{n}\pi\pm\mathrm{i}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$ Terms…