^ Given that x = tan 23°, find the value of cos 16° in terms of x.


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Question Number 144614 by nadovic last updated on 27/Jun/21

$\overset{}{} Given that x = \tan 23 °, find the value$ $of \cos 16 ° in terms of x \underset{}{.}$
	Answered by qaz last updated on 27/Jun/21

	$x = \tan 23 ° = \frac{\tan 30 ° - \tan 7 °}{1 + \tan 30 ° \cdot \tan 7 °}$ $\Rightarrow \tan 7 ° = \frac{\tan 30 ° - x}{1 + xtan 30 °}$ $\tan 16 ° = \frac{\tan 23 ° - \tan 7 °}{1 + \tan 23 ° \cdot \tan 7 °} = \frac{x - \tan 7 °}{1 + xtan 7 °}$ $\Rightarrow \cos 16 ° = \frac{1}{\sqrt{1 + \tan^{2} 16 °}}$ $= \frac{1}{\sqrt{1 + {(\frac{x - \tan 7 °}{1 + xtan 7 °})}^{2}}}$ $= \frac{1}{\sqrt{1 + {(\frac{x - \frac{\tan 30 ° - x}{1 + xtan 30 °}}{1 + x \cdot \frac{\tan 30 ° - x}{1 + xtan 30 °}})}^{2}}}$ $continue . . .$
	Commented by nadovic last updated on 27/Jun/21

	$T h a n k y o u S i r$
	Answered by imjagoll last updated on 27/Jun/21
	$cos 16°=cos (23°−7°) = cos 23° cos 7°+sin 23° sin 7° tan 7°=(((1/( (√3)))−x)/(1+(1/( (√3)))x))=((1−x(√3))/( (√3)+x)) { ((sin 23°=(x/( (√(1+x^2 )))))),((cos 23°=(1/( (√(1+x^2 )))))) :} { ((sin 7°=((1−x(√3))/(2(√(x^2 +1)))))),((cos 7°=((x+(√3))/( 2(√(x^2 +1)))))) :} cos 16°=(1/( (√(1+x^2 )))).((x+(√3))/( 2(√(1+x^2 ))))+(x/( (√(1+x^2 )))).((1−x(√3))/( 2(√(1+x^2 )))) = ((x+(√3)+x−x^2 (√3))/(2(1+x^2 )))=((2x+(√3)−x^2 (√3))/(2+2x^2 ))$
	$\cos 16 ° = \cos (23 ° - 7 °)$ $= \cos 23 ° \cos 7 ° + \sin 23 ° \sin 7 °$ $\tan 7 ° = \frac{\frac{1}{\sqrt{3}} - x}{1 + \frac{1}{\sqrt{3}} x} = \frac{1 - x \sqrt{3}}{\sqrt{3} + x}$ ${\begin{cases} \sin 23 ° = \frac{x}{\sqrt{1 + x^{2}}} \\ \cos 23 ° = \frac{1}{\sqrt{1 + x^{2}}} \end{cases}$ ${\begin{cases} \sin 7 ° = \frac{1 - x \sqrt{3}}{2 \sqrt{x^{2} + 1}} \\ \cos 7 ° = \frac{x + \sqrt{3}}{2 \sqrt{x^{2} + 1}} \end{cases}$ $\cos 16 ° = \frac{1}{\sqrt{1 + x^{2}}} . \frac{x + \sqrt{3}}{2 \sqrt{1 + x^{2}}} + \frac{x}{\sqrt{1 + x^{2}}} . \frac{1 - x \sqrt{3}}{2 \sqrt{1 + x^{2}}}$ $= \frac{x + \sqrt{3} + x - x^{2} \sqrt{3}}{2 (1 + x^{2})} = \frac{2 x + \sqrt{3} - x^{2} \sqrt{3}}{2 + {2 x}^{2}}$
	Commented by liberty last updated on 27/Jun/21

	$for x = \tan 23 ° \approx . 424$ $\cos 16 ° = \frac{2 (. 424) + \sqrt{3} - \sqrt{3} {(. 424)}^{2}}{2 (1 + {(. 424)}^{2})}$ $\approx . 9615$
	Commented by nadovic last updated on 27/Jun/21

	$T h a n k y o u S i r$
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