solve-the-equation-tan-3-cot-1-0-for-0-180-b-show-that-if-cos-2-is-not-zero-then-cos-2-sec-2-2-cos-4-sin-4-cos-4-sin-4-c-find-the-limit-of-tan-3-3-as-0-

Question Number 50892 by peter frank last updated on 21/Dec/18

s o l v e t h e e q u a t i o n

\tan 3 θ c o t θ + 1 = 0 f o r

0 ⩽ θ ⩽ 180

b) s h o w t h a t i f \cos 2 θ i s n o t z e r o

t h e n

\cos 2 θ + \sec 2 θ = 2 [\frac{\cos^{4} θ + \sin^{4} θ}{\cos^{4} θ - \sin^{4} θ}]

c) f i n d t h e l i m i t o f

\frac{\tan \frac{θ}{3}}{3 θ} a s θ \to 0

Answered by kaivan.ahmadi last updated on 21/Dec/18

\frac{\cot θ}{\cot 3 θ} + 1 = 0 \Rightarrow \frac{\cot θ}{\cot 3 θ} = - 1 \Rightarrow \cot θ = - \cot 3 θ

\Rightarrow \cot θ = \cot (- 3 θ)

θ = k π - 3 θ \Rightarrow θ = \frac{k π}{4}

θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}

2 (\frac{\cos^{4} θ + \sin^{4} θ}{\cos^{4} θ - \sin^{4} θ}) = 2 (\frac{{(\cos^{2} θ + \sin^{2} θ)}^{2} - {2 \sin}^{2} θ \cos^{2} θ}{(\cos^{2} θ - \sin^{2} θ) (\cos^{2} θ + \sin^{2} θ)}) =

\frac{2 - \sin^{2} 2 θ}{\cos 2 θ} = \frac{(1 - \sin^{2} 2 θ) + 1}{\cos 2 θ} = \frac{\cos^{2} 2 θ + 1}{\cos 2 θ} = \cos 2 θ + \sec 2 θ

Answered by peter frank last updated on 21/Dec/18

\frac{\sin 3 θ \cos θ}{\cos 3 θ \sin θ} + 1 = 0

\frac{\sin 3 θ \cos θ + \cos 3 θ \sin θ}{\cos 3 θ \sin θ} = 0

\frac{\sin 4 θ}{\cos 3 θ \sin θ} = 0

θ = 45 a n d 135

Answered by peter frank last updated on 21/Dec/18

\lim_{x \to 0} \frac{\sin \frac{θ}{3}}{\frac{\cos \frac{θ}{3}}{3 θ} .}

\lim_{x \to 0} \frac{\sin \frac{θ}{3}}{\frac{\cos \frac{θ}{3}}{3 θ (\frac{3}{3})} .} q

\lim_{x \to 0} \frac{\sin \frac{θ}{3}}{\frac{θ}{3} . 9} . \frac{1}{\cos \frac{θ}{3}} = \lim_{x \to 0} 1 . \frac{1}{9} = \frac{1}{9}

\frac{1}{9}