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Question Number 87424 by Rio Michael last updated on 04/Apr/20
prove that ((sin x − 2 sin x + sin 3x)/(sin x + 2sin x + sin 3x)) ≡ −tan^2 ((x/2))
$$\mathrm{prove}\:\mathrm{that}\:\:\:\frac{\mathrm{sin}\:{x}\:−\:\mathrm{2}\:\mathrm{sin}\:{x}\:+\:\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{sin}\:{x}\:+\:\mathrm{2sin}\:{x}\:+\:\mathrm{sin}\:\mathrm{3}{x}}\:\equiv\:−\mathrm{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$
Commented by john santu last updated on 04/Apr/20
((sin 3x−sin x)/(sin 3x+3sin x)) = ((2cos 2x sin x)/(3sin x−4sin^3 x+3sin x)) = ((2cos 2x)/(6−4sin^2 x)) = ((cos 2x)/(3−2sin^2 x)) = ((cos 2x)/(2+1−2sin^2 x)) = ((cos 2x)/(2+cos 2x))
$$\frac{\mathrm{sin}\:\mathrm{3x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{3x}+\mathrm{3sin}\:\mathrm{x}}\:=\:\frac{\mathrm{2cos}\:\mathrm{2x}\:\mathrm{sin}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{3sin}\:\mathrm{x}} \\ $$$$=\:\frac{\mathrm{2cos}\:\mathrm{2x}}{\mathrm{6}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{cos}\:\mathrm{2x}}{\mathrm{3}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\frac{\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}+\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}+\mathrm{cos}\:\mathrm{2x}} \\ $$
Commented by john santu last updated on 04/Apr/20
remember cos 2x = ((1−tan^2 x)/(1+tan^2 x)) ⇒ ((1−tan^2 x)/(1+tan^2 x)) ×((1+tan^2 x)/(3+tan^2 x)) = ((1−tan^2 x)/(3+tan^2 x))
$$\mathrm{remember}\:\mathrm{cos}\:\mathrm{2x}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:×\frac{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{3}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\: \\ $$$$=\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{3}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$
Commented by Rio Michael last updated on 04/Apr/20
okay sir thanks
$$\mathrm{okay}\:\mathrm{sir}\:\mathrm{thanks} \\ $$
Commented by Rio Michael last updated on 04/Apr/20
thanks sir, but please sir how come you substituted sin 3x − sin x in the denominator?
$$\mathrm{thanks}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{please}\:\mathrm{sir}\: \\ $$$$\:\mathrm{how}\:\mathrm{come}\:\mathrm{you}\:\mathrm{substituted}\:\mathrm{sin}\:\mathrm{3}{x}\:−\:\mathrm{sin}\:{x}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{denominator}? \\ $$
Commented by john santu last updated on 04/Apr/20
oo sorry. i misread
$$\mathrm{oo}\:\mathrm{sorry}.\:\mathrm{i}\:\mathrm{misread} \\ $$
Answered by Ar Brandon last updated on 04/Apr/20
Question should be ((sin x−2sin 2x+sin 3x)/(sin x+2sin 2x+sin 3x))≡−tan^2 ((x/2)) ((sin x−2sin 2x+sin 3x)/(sin x+2sin 2x+sin 3x))=((2sin(2x)cos(x)−2sin 2x)/(2sin(2x)cos(x)+2sin 2x)) =((cos x−1)/(cosx+1)) =(((cos^2 ((x/2))−sin^2 ((x/2)))−(cos^2 ((x/2))+sin^2 ((x/2))))/((cos^2 ((x/2))−sin^2 ((x/2)))+(cos^2 ((x/2))+sin^2 ((x/2))))) =((−2sin^2 ((x/2)))/(2cos^2 ((x/2))))=−tan^2 ((x/2))
$${Question}\:\:{should}\:\:{be} \\ $$$$\frac{{sin}\:{x}−\mathrm{2}{sin}\:\mathrm{2}{x}+{sin}\:\mathrm{3}{x}}{{sin}\:{x}+\mathrm{2}{sin}\:\mathrm{2}{x}+{sin}\:\mathrm{3}{x}}\equiv−{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\frac{{sin}\:{x}−\mathrm{2}{sin}\:\mathrm{2}{x}+{sin}\:\mathrm{3}{x}}{{sin}\:{x}+\mathrm{2}{sin}\:\mathrm{2}{x}+{sin}\:\mathrm{3}{x}}=\frac{\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left({x}\right)−\mathrm{2}{sin}\:\mathrm{2}{x}}{\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left({x}\right)+\mathrm{2}{sin}\:\mathrm{2}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{cos}\:{x}−\mathrm{1}}{{cosx}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)−\left({cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)}{\left({cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)+\left({cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}=−{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$
Commented by Rio Michael last updated on 04/Apr/20
thanks bro,for the correction
$$\mathrm{thanks}\:\mathrm{bro},\mathrm{for}\:\mathrm{the}\:\mathrm{correction} \\ $$
Commented by Ar Brandon last updated on 04/Apr/20
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