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Category: Trigonometry

find-the-domaine-and-simplify-the-function-f-x-arcos-1-x-2-1-x-2-

Question Number 92438 by Ar Brandon last updated on 06/May/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{domaine}\:\mathrm{and}\:\mathrm{simplify} \\ $$$$\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arcos}\left(\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right) \\ $$ Commented by Joel578 last updated on…

sin-3-x-cos-4-x-0-

Question Number 92399 by jagoll last updated on 06/May/20 $$\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\mathrm{x}\right)\:=\:\mathrm{0} \\ $$ Commented by john santu last updated on 06/May/20 $$\mathrm{sin}\:^{\mathrm{3}} {x}\:+\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}}…

Question-157925

Question Number 157925 by cortano last updated on 29/Oct/21 Answered by FongXD last updated on 30/Oct/21 $$\mathrm{Given}:\:\frac{\mathrm{sin}\alpha\mathrm{cos}\alpha}{\mathrm{sin}\beta\mathrm{cos}\beta}=\frac{\mathrm{8}}{\mathrm{5}}\:\:\:\left(\mathrm{1}\right)\:\mathrm{and}\:\frac{\mathrm{sin}\alpha\mathrm{cos}\beta}{\mathrm{sin}\beta\mathrm{cos}\alpha}=\mathrm{4}\:\:\:\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow\:\left(\mathrm{1}\right)\centerdot\left(\mathrm{2}\right):\:\frac{\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{sin}^{\mathrm{2}} \beta}=\frac{\mathrm{32}}{\mathrm{5}}\:\mathrm{and}\:\left(\mathrm{1}\right)\boldsymbol{\div}\left(\mathrm{2}\right):\:\frac{\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{cos}^{\mathrm{2}} \beta}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Leftrightarrow\:\mathrm{5}−\mathrm{5sin}^{\mathrm{2}}…

If-sin-A-sin-B-p-and-cos-A-cos-B-q-show-that-tan-A-tan-B-8pq-p-2-q-2-2-4q-2-

Question Number 26778 by tawa tawa last updated on 29/Dec/17 $$\mathrm{If}\:\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{A}\right)\:+\:\mathrm{sin}\left(\mathrm{B}\right)\:=\:\mathrm{p}\:\:\:\:\:\mathrm{and}\:\:\:\:\mathrm{cos}\left(\mathrm{A}\right)\:+\:\mathrm{cos}\left(\mathrm{B}\right)\:=\:\mathrm{q} \\ $$$$\mathrm{show}\:\mathrm{that}\:\:\:\:\mathrm{tan}\left(\mathrm{A}\right)\:+\:\mathrm{tan}\left(\mathrm{B}\right)\:=\:\frac{\mathrm{8pq}}{\left(\mathrm{p}^{\mathrm{2}} \:+\:\mathrm{q}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\:\mathrm{4q}^{\mathrm{2}} } \\ $$ Commented by tawa tawa last updated…