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Category: Trigonometry

An-MTN-mask-is-erected-at-a-point-P-in-ilaro-town-At-a-point-B-due-west-The-angle-of-elevation-of-its-top-is-and-at-point-C-due-south-the-angle-of-elevation-is-With-the-aid-of-an-appropriat

Question Number 6723 by Tawakalitu. last updated on 16/Jul/16 $${An}\:{MTN}\:{mask}\:{is}\:{erected}\:{at}\:{a}\:{point}\:{P}\:\:{in}\:{ilaro}\:{town}.\:{At}\:{a}\:{point} \\ $$$${B}\:{due}\:{west},\:{The}\:{angle}\:{of}\:{elevation}\:{of}\:{its}\:{top}\:{is}\:\beta\:{and}\:{at}\:{point}\: \\ $$$${C}\:{due}\:{south},\:{the}\:{angle}\:{of}\:{elevation}\:{is}\:\alpha.\:{With}\:{the}\:{aid}\:{of}\:{an}\: \\ $$$${appropriate}\:{diagam}.\:{show}\:{that}\:{the}\:{angle}\:{of}\:{elevation}\:{of}\:{the}\:{top} \\ $$$${from}\:{a}\:{point}\:{due}\:{south}\:{of}\:{B}\:{and}\:{due}\:{west}\:{of}\:{C}\:{is}\: \\ $$$$ \\ $$$${cot}^{−\mathrm{1}} \left[{cot}^{\mathrm{2}} \left(\beta\right)\:+\:{cot}^{\mathrm{2}} \left(\alpha\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}}…

Solve-the-following-equation-for-0-lt-lt-360-o-cosx-cos3x-cos5x-cos7x-0-

Question Number 6715 by Tawakalitu. last updated on 15/Jul/16 $${Solve}\:{the}\:{following}\:{equation}\:{for}\:\mathrm{0}\:<\:\Theta\:<\:\mathrm{360}^{{o}} \\ $$$${cosx}\:+\:{cos}\mathrm{3}{x}\:+\:{cos}\mathrm{5}{x}\:+\:{cos}\mathrm{7}{x}\:=\:\mathrm{0} \\ $$ Answered by Yozzii last updated on 16/Jul/16 $${We}\:{know}\:{the}\:{identity}\:{cosa}+{cosb}=\mathrm{2}{cos}\frac{{a}+{b}}{\mathrm{2}}{cos}\frac{{a}−{b}}{\mathrm{2}}\:{for}\:{a},{b}\in\mathbb{R}….\left(\mathrm{1}\right) \\ $$$${Let}\:{u}={cosx}+{cos}\mathrm{3}{x}+{cos}\mathrm{5}{x}+{cos}\mathrm{7}{x}. \\…

If-Cos-x-cos-y-x-y-cos-then-prove-that-tan-2-x-y-x-y-tan-2-

Question Number 71918 by lalitchand last updated on 22/Oct/19 $$\mathrm{If}\:\mathrm{Cos}\theta=\frac{\mathrm{x}\:\mathrm{cos}\beta\:−\:\mathrm{y}}{\mathrm{x}\:−\:\mathrm{y}\:\mathrm{cos}\beta}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{tan}\frac{\theta}{\mathrm{2}}\:=\sqrt{\frac{\mathrm{x}−\mathrm{y}}{\mathrm{x}+\mathrm{y}}\:}\:\mathrm{tan}\frac{\beta}{\mathrm{2}} \\ $$ Commented by Prithwish sen last updated on 22/Oct/19 $$\mathrm{Using}\:\mathrm{componendo}\:\mathrm{dividendo} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}\:=\:\frac{\left(\mathrm{1}−\mathrm{cos}\beta\right)\left(\mathrm{x}+\mathrm{y}\right)}{\left(\mathrm{1}+\mathrm{cos}\beta\right)\left(\mathrm{x}−\mathrm{y}\right)} \\ $$$$\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\theta}}{\mathrm{2}}\:=\:\sqrt{\frac{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}}\:\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\beta}}{\mathrm{2}}…

How-many-idempotent-matrices-can-be-formed-from-a-diagonal-matrix-A-with-the-elements-a-i-i-for-i-1-2-3-n-

Question Number 6316 by sanusihammed last updated on 23/Jun/16 $${How}\:{many}\:{idempotent}\:{matrices}\:{can}\:{be}\:{formed}\:{from}\:{a} \\ $$$${diagonal}\:{matrix}\:{A}\:{with}\:{the}\:{elements}\: \\ $$$${a}\left({i},{i}\right)\:{for}\:{i}\:=\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},…….{n}\right\} \\ $$ Answered by nburiburu last updated on 23/Jun/16 $$\:{in}\:\mathbb{R}^{{n}×{n}} :\:…