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Category: Trigonometry

Given-sin-cos-4-3-where-0-lt-lt-pi-4-Find-the-value-of-cos-sin-

Question Number 123160 by liberty last updated on 23/Nov/20 $$\:{Given}\:\mathrm{sin}\:\rho\:+\:\mathrm{cos}\:\rho\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:,\:{where}\: \\ $$$$\:\mathrm{0}\:<\:\rho\:<\:\frac{\pi}{\mathrm{4}}.\:{Find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\rho−\mathrm{sin}\:\rho.\: \\ $$ Answered by benjo_mathlover last updated on 23/Nov/20 $$\Rightarrow\left(\mathrm{sin}\:\rho+\mathrm{cos}\:\rho\right)^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2sin}\:\rho\mathrm{cos}\:\rho=\frac{\mathrm{16}}{\mathrm{9}}…

Question-188645

Question Number 188645 by Rupesh123 last updated on 04/Mar/23 Answered by witcher3 last updated on 05/Mar/23 $$\mathrm{let}\:\mathrm{a}=\mathrm{cos}\left(\frac{\pi}{\mathrm{7}}\right),\mathrm{b}=−\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right);\mathrm{c}=\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{1}+\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}}…

prove-sin18-cos36-1-4-

Question Number 57474 by malwaan last updated on 05/Apr/19 $$\mathrm{prove}\: \\ $$$$\boldsymbol{{sin}}\mathrm{18}×\boldsymbol{{cos}}\mathrm{36}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by Abdo msup. last updated on 05/Apr/19 $$\mathrm{18}\rightarrow{x} \\ $$$$\mathrm{180}\rightarrow\pi\:\Rightarrow\mathrm{180}{x}\:=\mathrm{18}\pi\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{10}}\:\Rightarrow\mathrm{36}\:=\frac{\pi}{\mathrm{5}}…

sin-pi-2-4x-x-cos-pi-x-7-x-1-x-

Question Number 188475 by cortano12 last updated on 02/Mar/23 $$\:\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{4x}+\sqrt{\mathrm{x}}\:\right)\right)\mathrm{cos}\:\left(\pi\left(\mathrm{x}+\mathrm{7}\sqrt{\mathrm{x}}\right)\right)=\mathrm{1} \\ $$$$\:\mathrm{x}=? \\ $$ Answered by Frix last updated on 02/Mar/23 $$\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:=\mathrm{1} \\ $$$$\Leftrightarrow \\…

Question-188376

Question Number 188376 by Rupesh123 last updated on 28/Feb/23 Answered by Frix last updated on 28/Feb/23 $$\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right) \\ $$$$\left(\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{7}}{\mathrm{256}}\left(\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right) \\…