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Question Number 113781 by Ar Brandon last updated on 15/Sep/20

Answered by Dwaipayan Shikari last updated on 15/Sep/20

sin(π/(12))sin((7π)/(12))=(1/2)(cos(π/2)−cos((8π)/(12)))=(1/2).(1/2)=(1/4)

$${sin}\frac{\pi}{\mathrm{12}}{sin}\frac{\mathrm{7}\pi}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\frac{\pi}{\mathrm{2}}−{cos}\frac{\mathrm{8}\pi}{\mathrm{12}}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

tan^(−1) (1/n)−tan^(−1) (1/(n+p)) =tan^(−1) (((1/n)−(1/(n+p)))/(1+(1/(n(n+p)))))=tan^(−1) (p/(n^2 +pn+1)) (Which is true)

$${tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{n}}−{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{n}+{p}} \\ $$$$={tan}^{−\mathrm{1}} \frac{\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+{p}}}{\mathrm{1}+\frac{\mathrm{1}}{{n}\left({n}+{p}\right)}}={tan}^{−\mathrm{1}} \frac{{p}}{{n}^{\mathrm{2}} +{pn}+\mathrm{1}}\:\:\left({Which}\:{is}\:{true}\right) \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

2(sin^4 (π/8)+sin^4 ((3π)/8)) sin((3π)/8)=sin((5π)/8), sin(π/8)=sin((7π)/8) 2(sin^2 (π/8)+sin^2 ((3π)/8))^2 −4sin^2 (π/8)sin^2 ((3π)/8) 2(((1−cos(π/4))/2)+((1+cos(π/4))/2))^2 −(cos(π/4)+cos(π/2))^2 sin((3π)/8)=cos(π/8)) 2−((1/( (√2)))+0)^2 =(3/2)

$$\mathrm{2}\left({sin}^{\mathrm{4}} \frac{\pi}{\mathrm{8}}+{sin}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:\:\:\:\:\:\:\:\:\:\:{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}={sin}\frac{\mathrm{5}\pi}{\mathrm{8}},\:\:{sin}\frac{\pi}{\mathrm{8}}={sin}\frac{\mathrm{7}\pi}{\mathrm{8}} \\ $$$$\mathrm{2}\left({sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}+{sin}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{8}}\right)^{\mathrm{2}} −\mathrm{4}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}{sin}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\left.\mathrm{2}\left(\frac{\mathrm{1}−{cos}\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\frac{\mathrm{1}+{cos}\frac{\pi}{\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({cos}\frac{\pi}{\mathrm{4}}+{cos}\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}={cos}\frac{\pi}{\mathrm{8}}\right) \\ $$$$\mathrm{2}−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{0}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

((sin(A+(B/2)))/(sin(B/2)))=k ((sin(A+(B/2))+sin(B/2))/(sin(A+(B/2))−sin(B/2)))=((k+1)/(k−1)) ((sin(((A+B)/2))cos(A/2))/(cos(((A+B)/2))sin(A/2)))=((k+1)/(k−1)) ((sin(C/2))/(cos(C/2))).((sin(A/2))/(cos(A/2)))=((k−1)/(k+1)) tan(A/2).tan(C/2)=((k−1)/(k+1)) cos((A+B)/2)=sin(C/2), sin((A+B)/2)=cos(C/2)

$$\frac{{sin}\left({A}+\frac{{B}}{\mathrm{2}}\right)}{{sin}\frac{{B}}{\mathrm{2}}}={k} \\ $$$$\frac{{sin}\left({A}+\frac{{B}}{\mathrm{2}}\right)+{sin}\frac{{B}}{\mathrm{2}}}{{sin}\left({A}+\frac{{B}}{\mathrm{2}}\right)−{sin}\frac{{B}}{\mathrm{2}}}=\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\ $$$$\frac{{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\right){cos}\frac{{A}}{\mathrm{2}}}{{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right){sin}\frac{{A}}{\mathrm{2}}}=\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}} \\ $$$$\frac{{sin}\frac{{C}}{\mathrm{2}}}{{cos}\frac{{C}}{\mathrm{2}}}.\frac{{sin}\frac{{A}}{\mathrm{2}}}{{cos}\frac{{A}}{\mathrm{2}}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${tan}\frac{{A}}{\mathrm{2}}.{tan}\frac{{C}}{\mathrm{2}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\frac{{A}+{B}}{\mathrm{2}}={sin}\frac{{C}}{\mathrm{2}},\:{sin}\frac{{A}+{B}}{\mathrm{2}}={cos}\frac{{C}}{\mathrm{2}} \\ $$

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