Question

prove that cos π/32
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\large\underline\purple{\bold{Correct \: Statement :- }}$

$\sf \:  Prove \: that : cos(\dfrac{\pi}{32} ) = \dfrac{ \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2} } }} }{2}$

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$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\bf \: \blue{Evaluate} : cos(\dfrac{\pi}{32} )$

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$\huge{AηsωeR} ✍$

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$\large\underline\blue{\bold{Identities \: used :- :-  }}$

$\bf \: ❥︎ \: {cos}^{2} x = \dfrac{1 + cos2x}{2}$

$\bf \:❥︎ \: cos(\dfrac{\pi}{4} ) = \dfrac{1}{ \sqrt{2} } = \dfrac{ \sqrt{2} }{2}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \:❥︎ \: Step :- 1.$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{8} = \dfrac{1 + cos(2 \times \dfrac{\pi}{8} )}{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{8} = \dfrac{1 + cos\dfrac{\pi}{4} }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{8} =\dfrac{1 + \dfrac{ \sqrt{2} }{2} }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{8} =\dfrac{2 + \sqrt{2} }{4}$

$\bf\implies \:cos(\dfrac{\pi}{8} ) = \dfrac{ \sqrt{2 + \sqrt{2} } }{2}$

$\sf \:  (\because \: \dfrac{\pi}{8} \: lies \: in \: first \: quadrant.)\sf \:  ⟼cos(\dfrac{\pi}{8} ) > 0$

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$\bf \:❥︎ \: Step :- 2.$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{16} =\dfrac{1 + cos(2 \times \dfrac{\pi}{16}) }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{16} =\dfrac{1 + cos\dfrac{\pi}{8} }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{16} =\dfrac{1 + \dfrac{ \sqrt{2 + \sqrt{2} } }{2} }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{16} = \dfrac{2 + \sqrt{2 + \sqrt{2} } }{4}$

$\bf\implies \:cos(\dfrac{\pi}{16} ) = \sqrt{\dfrac{2 + \sqrt{2 + \sqrt{2} } }{4} }$

$\bf \:  ⟼ cos(\dfrac{\pi}{16} ) = \dfrac{ \sqrt{2 + \sqrt{2 + \sqrt{2} } } }{2}$

$\sf \: (\because \: \dfrac{\pi}{16} \: lies \: in \: first \: quadrant.)\sf \:  ⟼cos(\dfrac{\pi}{16} ) > 0$

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$\bf \:❥︎ \: Step :- 3.$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{32} =\dfrac{1 + cos(2 \times \dfrac{\pi}{32}) }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{32} = \dfrac{1 + cos \dfrac{\pi}{16} }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{32} = \dfrac{1 + \dfrac{ \sqrt{2 + \sqrt{2 + \sqrt{2} } } }{2} }{2}$

$\sf \:  ⟼ {cos}^{2} \dfrac{\pi}{32} = \dfrac{2 + \sqrt{2 + \sqrt{2 + \sqrt{2} } } }{4}$

$\sf \:  ⟼ (\because \: \dfrac{\pi}{32} \: lies \: in \: first \: quadrant.)\bf \:  ⟼ cos(\dfrac{\pi}{32} ) > 0$

$\bf \:cos(\dfrac{\pi}{32} ) = \dfrac{ \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2} } }} }{2}$

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${\boxed{\boxed{\bf{Hence, \sf \:cos(\dfrac{\pi}{32} ) = \dfrac{ \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2} } }} }{2} }}}}$

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