Show that: 1• tan3x=((3−tan^2 x)/(1−3tan^2 x)) using cos3x=4cos^4 x−3cosx sin3x=−4sin^3 x+3sinx Thanks...


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Question Number 84019 by mathocean1 last updated on 08/Mar/20

$S h o w t h a t :$ $1 ∙ t a n 3 x = \frac{3 - {t a n}^{2} x}{1 - 3 {t a n}^{2} x}$ $u s i n g c o s 3 x = 4 {c o s}^{4} x - 3 c o s x$ $s i n 3 x = - 4 {s i n}^{3} x + 3 s i n x$ $T h a n k s . . .$
Commented by MJS last updated on 08/Mar/20

${it}^{'} s wrong :$ $\tan 3 x = - \frac{\sin x}{\cos x} \times \frac{1 - {4 \cos}^{2} x}{1 - {4 \sin}^{2} x}$ $\frac{3 - \tan^{2} x}{1 - {3 \tan}^{2} x} = \frac{1 - {4 \cos}^{2} x}{1 - {4 \sin}^{2} x}$
Commented by $@ty@m123 last updated on 08/Mar/20

$T h e r e i s a t y p o i n q u e s t i o n$ $I t i s \tan 3 x = \frac{(3 - \tan^{2} x) \tan x}{1 - {3 \tan}^{2} x}$
	Answered by Rio Michael last updated on 08/Mar/20

	$b) \cos 3 x = \cos (2 x + x) = \cos 2 x \cos x - \sin 2 x \sin x$ $= (2 \cos^{2} x - 1) \cos x - 2 \sin^{2} x \cos x$ $= 2 \cos^{3} x - \cos x - 2 (1 - \cos^{2} x) \cos x$ $= 2 \cos^{3} x - \cos x - 2 \cos x + 2 \cos^{3} x = 4 \cos^{3} x - 3 \cos x$
	Commented by mathocean1 last updated on 08/Mar/20

	$p l e a s e s i r w e w a n t t o s h o w 1 ∙$
	Answered by $@ty@m123 last updated on 08/Mar/20

	$S e e Q . N o . 76497$
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