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Category: Trigonometry

Question-190306

Question Number 190306 by cortano12 last updated on 31/Mar/23 Answered by Frix last updated on 31/Mar/23 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{triangle}\:{u},\:{v},\:{w}\:\mathrm{is} \\ $$$$\frac{\sqrt{\left({u}+{v}+{w}\right)\left({u}+{v}−{w}\right)\left({u}+{w}−{v}\right)\left({v}+{w}−{u}\right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{2}\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{u}^{\mathrm{2}} {w}^{\mathrm{2}} +{v}^{\mathrm{2}}…

In-AB-C-If-sin-A-1-2-2-3-A-

Question Number 190185 by mnjuly1970 last updated on 29/Mar/23 $$ \\ $$$$\:\:\:\:\:\:\mathrm{In}\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\::\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{If}\:,\:\mathrm{sin}\:\left(\hat {\mathrm{A}}\:\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\:\sqrt{\:\mathrm{2}+\:\sqrt{\mathrm{3}}}}\:\:\:\:\:\Rightarrow\:\:\:\:\:\hat {\mathrm{A}}\:=\:?\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$ Commented by…

probar-con-h-0-sin-x-h-sin-x-h-sin-h-2-h-2-cos-x-h-2-

Question Number 59006 by arcana last updated on 03/May/19 $${probar}\:{con}\:{h}\neq\mathrm{0} \\ $$$$\frac{{sin}\left({x}+{h}\right)−{sin}\left({x}\right)}{{h}}=\frac{{sin}\left({h}/\mathrm{2}\right)}{{h}/\mathrm{2}}{cos}\left({x}+\frac{{h}}{\mathrm{2}}\right) \\ $$$$ \\ $$ Commented by arcana last updated on 25/May/19 $${usando}\:\:\mathrm{2}{sin}\left({x}\right){cos}\left({y}\right)={sin}\left({x}+{y}\right)−{sin}\left({y}−{x}\right) \\…

nice-calculus-in-AB-C-sin-2-A-sin-2-B-sin-2-C-2-prove-that-AB-C-is-right-triangle-Good-luck-

Question Number 124501 by mnjuly1970 last updated on 03/Dec/20 $$\:\:\:\:\:\:\:\:\:\:….{nice}\:\:{calculus}.. \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\::\:{sin}^{\mathrm{2}} \left({A}\right)+{sin}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({C}\right)=\mathrm{2} \\ $$$${prove}\:{that}:\:{A}\overset{\Delta} {{B}C}\:{is}\:{right}\:{triangle} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{G}{ood}\:{luck}. \\ $$ Answered by…

Question-189941

Question Number 189941 by Rupesh123 last updated on 24/Mar/23 Answered by Frix last updated on 24/Mar/23 $$\mathrm{2arctan}\:{x}\:+\mathrm{arctan}\:{y}\:=\mathrm{arctan}\:\frac{{x}^{\mathrm{2}} {y}−\mathrm{2}{x}−{y}}{{x}^{\mathrm{2}} +\mathrm{2}{xy}−\mathrm{1}} \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$ Commented by…

Question-189897

Question Number 189897 by Mingma last updated on 23/Mar/23 Answered by mr W last updated on 24/Mar/23 $${a}^{\mathrm{2}} =\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$${b}^{\mathrm{2}} =\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{1}+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}}…