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Question Number 84019 by mathocean1 last updated on 08/Mar/20
Show that: 1• tan3x=((3−tan^2 x)/(1−3tan^2 x)) using cos3x=4cos^4 x−3cosx sin3x=−4sin^3 x+3sinx Thanks...
$${Show}\:{that}: \\ $$$$\mathrm{1}\bullet\:\:\:{tan}\mathrm{3}{x}=\frac{\mathrm{3}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}} \\ $$$${using}\:{cos}\mathrm{3}{x}=\mathrm{4}{cos}^{\mathrm{4}} {x}−\mathrm{3}{cosx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}\mathrm{3}{x}=−\mathrm{4}{sin}^{\mathrm{3}} {x}+\mathrm{3}{sinx} \\ $$$${Thanks}… \\ $$
Commented by MJS last updated on 08/Mar/20
it′s wrong: tan 3x =−((sin x)/(cos x))×((1−4cos^2 x)/(1−4sin^2 x)) ((3−tan^2 x)/(1−3tan^2 x))=((1−4cos^2 x)/(1−4sin^2 x))
$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}: \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:=−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}×\frac{\mathrm{1}−\mathrm{4cos}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}} \\ $$$$\frac{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \:{x}}=\frac{\mathrm{1}−\mathrm{4cos}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}} \\ $$
Commented by $@ty@m123 last updated on 08/Mar/20
There is a typo in question It is tan 3x=(((3−tan^2 x)tan x)/(1−3tan^2 x))
$${There}\:{is}\:{a}\:{typo}\:{in}\:{question} \\ $$$${It}\:{is}\:\:\:\:\:\mathrm{tan}\:\mathrm{3}{x}=\frac{\left(\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}\right)\mathrm{tan}\:{x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \:{x}} \\ $$
Answered by Rio Michael last updated on 08/Mar/20
b) cos 3x = cos (2x + x) = cos 2xcos x − sin 2x sin x = (2cos^2 x −1)cos x − 2sin^2 x cos x = 2cos^3 x−cos x −2 (1−cos^2 x)cos x = 2cos^3 x−cos x −2cos x +2cos^3 x = 4cos^3 x−3cos x
$$\left.{b}\right)\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{cos}\:\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:{x}\:−\:\mathrm{sin}\:\mathrm{2}{x}\:\mathrm{sin}\:{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{2cos}\:^{\mathrm{2}} {x}\:−\mathrm{1}\right)\mathrm{cos}\:{x}\:−\:\mathrm{2sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:{x}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2cos}\:^{\mathrm{3}} {x}−\mathrm{cos}\:{x}\:−\mathrm{2}\:\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2cos}\:^{\mathrm{3}} {x}−\mathrm{cos}\:{x}\:−\mathrm{2cos}\:{x}\:+\mathrm{2cos}\:^{\mathrm{3}} {x}\:=\:\mathrm{4cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:{x} \\ $$
Commented by mathocean1 last updated on 08/Mar/20
please sir we want to show 1•
$${please}\:{sir}\:{we}\:{want}\:{to}\:{show}\:\mathrm{1}\bullet\: \\ $$
Answered by $@ty@m123 last updated on 08/Mar/20
See Q. No. 76497
$${See}\:{Q}.\:{No}.\:\mathrm{76497} \\ $$

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