I saw this in a book (without explanation). Please show how. It is given that tan 2θ=(B/(A−C)) (A,B,C ∈R) . Find cos 2θ.


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Question Number 190151 by Matica last updated on 28/Mar/23

$I s a w t h i s i n a b o o k (w i t h o u t e x p l a n a t i o n) . P l e a s e s h o w h o w .$ $I t i s g i v e n t h a t \tan 2 θ = \frac{B}{A - C} (A, B, C \in R) . F i n d \cos 2 θ .$
Commented by mr W last updated on 28/Mar/23

$i s a w t h i s i n a n o t h e r b o o k (w i t h o u t$ $e x p l a n a t i o n, b e c a u s e n o b o d y n e e d s) .$ $\tan α i s g i v e n, h o w t o f i n d \cos α ?$
Commented by mr W last updated on 28/Mar/23

Commented by Matica last updated on 30/Mar/23

$i n t h a t b o o k, t h e a u t h o r c a l c u l a t e \cos 2 θ w i t h o u t f i n d i n g t h e v a l u e o f θ . T h a t m a k e s m e w o n d e r .$
Commented by Matica last updated on 30/Mar/23

$t h a n k y o u$
	Answered by ARUNG_Brandon_MBU last updated on 28/Mar/23

	$\tan^{2} x + 1 = \sec^{2} x = \frac{1}{\cos^{2} x}$ $\Rightarrow \cos x = \pm \frac{1}{\sqrt{\tan^{2} x + 1}}$ $\Rightarrow \cos 2 θ = \pm \frac{1}{\sqrt{\tan^{2} 2 θ + 1}}$ $= \pm \frac{∣ A - C ∣}{\sqrt{A^{2} + B^{2} + C^{2} - 2 A C}}$
	Commented by Matica last updated on 30/Mar/23

	$O h t h a n k y o u$
	Answered by manxsol last updated on 29/Mar/23
	$((sin^2 2θ)/(cos^2 2θ))=(B^2 /((A−C)^2 )) {+1} ((sin^2 2θ+cos^2 2θ)/(cos^2 2θ))=((B^2 +(A−C)^2 )/((A−C)^2 )) (1/(cos^2 2θ))=((B^2 +(A−C)^2 )/((A−C)^2 )) cos2θ=±((∣A−C∣)/( (√(B^2 +(A−C)^2 ))))$
	$\frac{{s i n}^{2} 2 θ}{{c o s}^{2} 2 θ} = \frac{B^{2}}{{(A - C)}^{2}} {+ 1}$ $\frac{{s i n}^{2} 2 θ + {c o s}^{2} 2 θ}{{c o s}^{2} 2 θ} = \frac{B^{2} + {(A - C)}^{2}}{{(A - C)}^{2}}$ $\frac{1}{{c o s}^{2} 2 θ} = \frac{B^{2} + {(A - C)}^{2}}{{(A - C)}^{2}}$ $c o s 2 θ = \pm \frac{∣ A - C ∣}{\sqrt{B^{2} + {(A - C)}^{2}}}$
	Commented by Matica last updated on 30/Mar/23

	$t h a n k y o u$
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