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Category: Trigonometry

The-maximum-value-of-the-expression-sin-2-x-2a-2-2a-2-3-cos-2-x-where-a-and-x-are-real-numbers-is-1-4-2-2-3-2-4-0-

Question Number 16090 by Tinkutara last updated on 17/Jun/17 $$\mathrm{The}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mid\sqrt{\mathrm{sin}^{\mathrm{2}} \:{x}\:+\:\mathrm{2}{a}^{\mathrm{2}} }\:−\:\sqrt{\mathrm{2}{a}^{\mathrm{2}} \:−\:\mathrm{3}\:−\:\mathrm{cos}^{\mathrm{2}} \:{x}}\mid; \\ $$$$\mathrm{where}\:'{a}'\:\mathrm{and}\:'{x}'\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers},\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{2}} \\…

Find-the-set-of-values-of-x-0-2pi-which-satisfy-sin-x-gt-cos-x-1-pi-4-3pi-4-5pi-4-2pi-2-0-pi-4-5pi-4-2pi-3-pi-4-5pi-4-4-0-3pi-4-5pi-4-2pi-

Question Number 16092 by Tinkutara last updated on 17/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:{x}\:\in\:\left[\mathrm{0},\:\mathrm{2}\pi\right] \\ $$$$\mathrm{which}\:\mathrm{satisfy}\:\mathrm{sin}\:{x}\:>\:\mathrm{cos}\:{x}. \\ $$$$\left(\mathrm{1}\right)\:\left(\frac{\pi}{\mathrm{4}},\:\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\:\cup\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\:\mathrm{2}\pi\right) \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{4}}\right)\:\cup\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\:\mathrm{2}\pi\right) \\ $$$$\left(\mathrm{3}\right)\:\left(\frac{\pi}{\mathrm{4}},\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{0},\:\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\:\cup\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\:\mathrm{2}\pi\right) \\ $$ Commented by Tinkutara…

The-range-of-function-f-sin-2-1-1-sin-2-is-1-1-2-2-3-1-3-2-4-3-2-

Question Number 16089 by Tinkutara last updated on 17/Jun/17 $$\mathrm{The}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function} \\ $$$${f}\left(\theta\right)\:=\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \:\theta}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\left[\mathrm{1},\:\infty\right) \\ $$$$\left(\mathrm{2}\right)\:\left[\mathrm{2},\:\infty\right) \\ $$$$\left(\mathrm{3}\right)\:\left[\mathrm{1},\:\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$$$\left(\mathrm{4}\right)\:\left[\frac{\mathrm{3}}{\mathrm{2}},\:\infty\right) \\ $$ Commented…

Question-147118

Question Number 147118 by bramlexs22 last updated on 18/Jul/21 Answered by Olaf_Thorendsen last updated on 20/Jul/21 $$\mathrm{As}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{is}\:\mathrm{45}°\:\mathrm{the}\:\mathrm{cylinder}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{developped}\:\mathrm{as}\:\mathrm{a}\:\mathrm{square}. \\ $$$$\mathrm{The}\:\mathrm{walk}\:\mathrm{is}\:\mathrm{the}\:\mathrm{diagonal}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}. \\ $$$${l}\:=\:\sqrt{\mathrm{2}}{h}\:=\:\mathrm{9}\sqrt{\mathrm{2}} \\ $$…

If-sides-a-b-c-of-ABC-are-in-H-P-prove-that-sin-2-A-2-sin-2-B-2-sin-2-C-2-are-in-H-P-

Question Number 16041 by Tinkutara last updated on 17/Jun/17 $$\mathrm{If}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{of}\:\Delta{ABC}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}., \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{A}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{B}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$ Answered by Tinkutara last updated on…

Question-16039

Question Number 16039 by Tinkutara last updated on 17/Jun/17 Commented by RasheedSoomro last updated on 17/Jun/17 $$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2ab}\:\mathrm{cos}\:\mathrm{60}° \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{a}^{\mathrm{2}}…

Solve-sin-2x-cos-2x-0-for-pi-lt-x-lt-pi-

Question Number 15975 by myintkhaing last updated on 16/Jun/17 $${Solve}\:{sin}\:\mathrm{2}{x}\:+\:{cos}\:\mathrm{2}{x}\:=\mathrm{0}, \\ $$$${for}\:−\pi<{x}<\pi. \\ $$ Commented by tawa tawa last updated on 16/Jun/17 $$\mathrm{sin2x}\:+\:\mathrm{cos2x}\:=\:\mathrm{0} \\ $$$$\mathrm{Note}:\:\mathrm{sin2x}\:=\:\mathrm{2sinxcosx}\:\:\mathrm{and}\:\:\mathrm{cos2x}\:=\:\mathrm{cos}^{\mathrm{2}}…