tan-76-4-sin-2-14-

Question Number 143680 by mathlove last updated on 17/Jun/21

$$\mathrm{tan}\:\mathrm{76}=\mathrm{4} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{14}=? \\ $$

Commented by mr W last updated on 17/Jun/21

tan 14=(1/(tan 76))=(1/a) sin 14=(1/( (√(1^2 +a^2 ))))=(1/( (√(1+a^2 )))) sin^2 14=(1/( 1+a^2 )) a≈4 sin^2 14≈(1/( 17))

$$\mathrm{tan}\:\mathrm{14}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{76}}=\frac{\mathrm{1}}{{a}} \\ $$$$\mathrm{sin}\:\mathrm{14}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +{a}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{14}=\frac{\mathrm{1}}{\:\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$${a}\approx\mathrm{4} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{14}\approx\frac{\mathrm{1}}{\:\mathrm{17}} \\ $$

Answered by TheHoneyCat last updated on 17/Jun/21

tan=((sin)/(cos)) ⇔cos^2 tan^2 =sin^2 ⇔cos^2 tan^2 =1−cos^2 ⇔cos^2 (tan^2 +1)=1 ⇔cos^2 =(1/(tan^2 +1)) sin(x)=cos(x−90)=cos(90−x) thus : sin^2 (x)=cos^2 (90−x)=(1/(tan^2 (90−x)+1)) evaluating for x=14 sin^2 14=(1/(tan^2 (76) +1)) sin^2 14=(1/5) _(according to your own result) Howerver it is not a exact value, tan76≈4,010780934... But it is indeed a good approximation. and make a fun exercise ;•)

$${tan}=\frac{{sin}}{{cos}} \\ $$$$\Leftrightarrow{cos}^{\mathrm{2}} {tan}^{\mathrm{2}} ={sin}^{\mathrm{2}} \\ $$$$\Leftrightarrow{cos}^{\mathrm{2}} {tan}^{\mathrm{2}} =\mathrm{1}−{cos}^{\mathrm{2}} \\ $$$$\Leftrightarrow{cos}^{\mathrm{2}} \left({tan}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{1} \\ $$$$\Leftrightarrow{cos}^{\mathrm{2}} =\frac{\mathrm{1}}{{tan}^{\mathrm{2}} +\mathrm{1}} \\ $$$${sin}\left({x}\right)={cos}\left({x}−\mathrm{90}\right)={cos}\left(\mathrm{90}−{x}\right) \\ $$$$\mathrm{thus}\::\:{sin}^{\mathrm{2}} \left({x}\right)={cos}^{\mathrm{2}} \left(\mathrm{90}−{x}\right)=\frac{\mathrm{1}}{{tan}^{\mathrm{2}} \left(\mathrm{90}−{x}\right)+\mathrm{1}} \\ $$$$\mathrm{evaluating}\:\mathrm{for}\:{x}=\mathrm{14} \\ $$$${sin}^{\mathrm{2}} \mathrm{14}=\frac{\mathrm{1}}{{tan}^{\mathrm{2}} \left(\mathrm{76}\right)\:+\mathrm{1}} \\ $$$${sin}^{\mathrm{2}} \mathrm{14}=\frac{\mathrm{1}}{\mathrm{5}}\:_{\mathrm{according}\:\mathrm{to}\:\mathrm{your}\:\mathrm{own}\:\mathrm{result}} \\ $$$$\mathrm{Howerver}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{exact}\:\mathrm{value}, \\ $$$${tan}\mathrm{76}\approx\mathrm{4},\mathrm{010780934}… \\ $$$$\mathrm{But}\:\mathrm{it}\:\mathrm{is}\:\mathrm{indeed}\:\:\mathrm{a}\:\mathrm{good}\:\mathrm{approximation}. \\ $$$$\left.\mathrm{and}\:\mathrm{make}\:\mathrm{a}\:\mathrm{fun}\:\mathrm{exercise}\:;\bullet\right) \\ $$

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