Question Number 126562 by mey3nipaba last updated on 21/Dec/20 $${Prove}\:{that}\:\mathrm{2sin}\:\frac{\theta+\phi}{\mathrm{2}}\mathrm{cos}\:\frac{\theta−\phi}{\mathrm{2}}=\mathrm{sin}\:\theta+\mathrm{sin}\:\emptyset \\ $$$${I}\:{need}\:{help}\:{immediately}\:{please} \\ $$ Commented by mr W last updated on 21/Dec/20 $$\theta=\frac{\theta+\varphi}{\mathrm{2}}+\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\varphi=\frac{\theta+\varphi}{\mathrm{2}}−\frac{\theta−\varphi}{\mathrm{2}}…
Question Number 192045 by Safiullah_21 last updated on 06/May/23 $$\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{16}}]{\mathrm{2}}\right)}×\left(\frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}_{} }}\right) \\ $$$$ \\ $$$$\Rightarrow_{} \frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)}\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 126499 by liberty last updated on 21/Dec/20 $$\:\:\:\:\mathrm{4}\:\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=? \\ $$ Answered by benjo_mathlover last updated on 21/Dec/20 $$\:{let}\:{x}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)\:\Rightarrow\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\mathrm{tan}\:\mathrm{2}{x}=\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{2}/\mathrm{5}}{\mathrm{1}−\mathrm{1}/\mathrm{25}}\: \\…
Question Number 191935 by Rupesh123 last updated on 04/May/23 Answered by som(math1967) last updated on 04/May/23 $${Direction}\:{cosine}\:{of}\:{L} \\ $$$${cos}\alpha,{cos}\beta,{cos}\gamma \\ $$$$\:\therefore{cos}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \beta+{cos}^{\mathrm{2}} \gamma=\mathrm{1} \\…
Question Number 126381 by liberty last updated on 20/Dec/20 $$\:\mathrm{arctan}\:\mathrm{1}\:+\:\mathrm{arctan}\:\mathrm{2}\:+\:\mathrm{arctan}\:\mathrm{3}\:=? \\ $$ Answered by benjo_mathlover last updated on 20/Dec/20 $$\left(\bullet\right)\:\mathrm{arctan}\:\mathrm{1}+\mathrm{arctan}\:\mathrm{2}=\mathrm{arctan}\:\left(\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}−\mathrm{1}.\mathrm{2}}\right) \\ $$$$\:\mathrm{arctan}\:\left(−\mathrm{3}\right) \\ $$$$\left(\bullet\bullet\right)\:\mathrm{arctan}\:\left(−\mathrm{3}\right)+\mathrm{arctan}\:\left(\mathrm{3}\right)=\mathrm{arctan}\:\left(\frac{−\mathrm{3}+\mathrm{3}}{\mathrm{1}−\left(−\mathrm{3}\right)\left(\mathrm{3}\right)}\right) \\…
Question Number 191901 by mehdee42 last updated on 03/May/23 $${if}\:\:\mathrm{0}<\theta<\mathrm{45}^{\mathrm{0}} \:\:{which}\:{is}\:{bigger}\:? \\ $$$$\mathrm{2}{tan}\theta\:\:{or}\:\:{tan}\mathrm{2}\theta \\ $$ Answered by mr W last updated on 03/May/23 Commented by…
Question Number 60814 by Forkum Michael Choungong last updated on 26/May/19 $${find}\:{x}\:{given}\:{that} \\ $$$$\mathrm{9}^{{sin}^{\mathrm{2}} {x}} +\mathrm{9}^{{cos}^{\mathrm{2}} {x}} =\mathrm{2}\: \\ $$$$ \\ $$ Answered by $@ty@m…
Question Number 191804 by Shlock last updated on 30/Apr/23 Commented by Shlock last updated on 30/Apr/23 x=? Commented by AST last updated on 30/Apr/23 $${It}\:{follows}\:{from}\:{sin}^{\mathrm{2}}…
Question Number 126148 by Study last updated on 17/Dec/20 $${A},{B},{C}\:{and}\:{D}\:{are}\:{angles}\:{of}\:{tetragon} \\ $$$${if}\:{A}+{D}=\mathrm{240}^{°\:} \:,\:\:\:{C}+{D}=\mathrm{200}{g} \\ $$$${B}+{D}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\:\:,\:{find}\:{D}=??? \\ $$ Answered by mahdipoor last updated on 17/Dec/20 $${A}+{B}+{C}+{D}=\mathrm{360}^{°}…
Question Number 60536 by hovea cw last updated on 21/May/19 $$\mathrm{cosx}=\mathrm{sin3x} \\ $$$$\mathrm{find}\:\mathrm{x}\:\mathrm{with}\:\mathrm{solution}\:\: \\ $$$$\mathrm{pllllllz} \\ $$ Answered by mr W last updated on 21/May/19…