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

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

$$\overset{} {\:}\mathrm{Given}\:\mathrm{that}\:{x}\:=\:\mathrm{tan}\:\mathrm{23}°,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\:\mathrm{of}\:\:\mathrm{cos}\:\mathrm{16}°\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{x}\underset{} {.} \\ $$

Answered by qaz last updated on 27/Jun/21

x=tan 23°=((tan 30°−tan 7°)/(1+tan 30°∙tan 7°)) ⇒tan 7°=((tan 30°−x)/(1+xtan 30°)) tan 16°=((tan 23°−tan 7°)/(1+tan 23°∙tan 7°))=((x−tan 7°)/(1+xtan 7°)) ⇒cos 16°=(1/( (√(1+tan^2 16°)))) =(1/( (√(1+(((x−tan 7°)/(1+xtan 7°)))^2 )))) =(1/( (√(1+(((x−((tan 30°−x)/(1+xtan 30°)))/(1+x∙((tan 30°−x)/(1+xtan 30°)))))^2 )))) continue...

$$\mathrm{x}=\mathrm{tan}\:\mathrm{23}°=\frac{\mathrm{tan}\:\mathrm{30}°−\mathrm{tan}\:\mathrm{7}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{30}°\centerdot\mathrm{tan}\:\mathrm{7}°} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{7}°=\frac{\mathrm{tan}\:\mathrm{30}°−\mathrm{x}}{\mathrm{1}+\mathrm{xtan}\:\mathrm{30}°} \\ $$$$\mathrm{tan}\:\mathrm{16}°=\frac{\mathrm{tan}\:\mathrm{23}°−\mathrm{tan}\:\mathrm{7}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°\centerdot\mathrm{tan}\:\mathrm{7}°}=\frac{\mathrm{x}−\mathrm{tan}\:\mathrm{7}°}{\mathrm{1}+\mathrm{xtan}\:\mathrm{7}°} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{16}°=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\mathrm{16}°}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}−\mathrm{tan}\:\mathrm{7}°}{\mathrm{1}+\mathrm{xtan}\:\mathrm{7}°}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{x}−\frac{\mathrm{tan}\:\mathrm{30}°−\mathrm{x}}{\mathrm{1}+\mathrm{xtan}\:\mathrm{30}°}}{\mathrm{1}+\mathrm{x}\centerdot\frac{\mathrm{tan}\:\mathrm{30}°−\mathrm{x}}{\mathrm{1}+\mathrm{xtan}\:\mathrm{30}°}}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{continue}... \\ $$

Commented by nadovic last updated on 27/Jun/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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 ))

$$\mathrm{cos}\:\mathrm{16}°=\mathrm{cos}\:\left(\mathrm{23}°−\mathrm{7}°\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{cos}\:\mathrm{23}°\:\mathrm{cos}\:\mathrm{7}°+\mathrm{sin}\:\mathrm{23}°\:\mathrm{sin}\:\mathrm{7}° \\ $$$$\mathrm{tan}\:\mathrm{7}°=\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{x}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{x}}=\frac{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}+\mathrm{x}}\: \\ $$$$\begin{cases}{\mathrm{sin}\:\mathrm{23}°=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}}\\{\mathrm{cos}\:\mathrm{23}°=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}}\end{cases} \\ $$$$\begin{cases}{\mathrm{sin}\:\mathrm{7}°=\frac{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}}\\{\mathrm{cos}\:\mathrm{7}°=\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\:\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}}\end{cases} \\ $$$$\mathrm{cos}\:\mathrm{16}°=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}.\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\:\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}.\frac{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{3}}}{\:\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$=\:\frac{\mathrm{x}+\sqrt{\mathrm{3}}+\mathrm{x}−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}=\frac{\mathrm{2x}+\sqrt{\mathrm{3}}−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}+\mathrm{2x}^{\mathrm{2}} } \\ $$$$ \\ $$

Commented by liberty last updated on 27/Jun/21

for x = tan 23° ≈ .424 cos 16°=((2(.424)+(√3)−(√3)(.424)^2 )/(2(1+(.424)^2 ))) ≈ .9615

$$\mathrm{for}\:\mathrm{x}\:=\:\mathrm{tan}\:\mathrm{23}°\:\approx\:.\mathrm{424} \\ $$$$\mathrm{cos}\:\mathrm{16}°=\frac{\mathrm{2}\left(.\mathrm{424}\right)+\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\left(.\mathrm{424}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}+\left(.\mathrm{424}\right)^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\:.\mathrm{9615}\: \\ $$$$ \\ $$

Commented by nadovic last updated on 27/Jun/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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