Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

Voor een hoek α die tussen 0° en 90° ligt geldt dat sin α = . Hoe groot is dan tan 2α ?	A. \(\boldsymbol{\frac{24}{7}}\)
B. \(\boldsymbol{\frac{40}{9}}\)
C. \(\boldsymbol{\frac{4}{3}}\)
D. \(\boldsymbol{\frac{8}{3}}\)
E. \(\boldsymbol{-\,\frac{24}{7}}\)

Angle α measures between 0° and 90°. If sin α = then the value of tan 2α is	A. \(\boldsymbol{\frac{24}{7}}\)
B. \(\boldsymbol{\frac{40}{9}}\)
C. \(\boldsymbol{\frac{4}{3}}\)
D. \(\boldsymbol{\frac{8}{3}}\)
E. \(\boldsymbol{-\,\frac{24}{7}}\)

1^ste manier :
$\\\cos\alpha=\sqrt{1-\frac{16}{24}}=\sqrt{\frac{9}{25}}=\frac35\\\frac{\sin\alpha}{\cos\alpha}=\frac{\frac45}{\frac35}=\frac45.\frac53=\frac43$
$\text{Vermits}\;\;\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}\;\;\text{is}\;\;\tan2\alpha=\frac{2.\frac43}{1-\frac{16}{9}}=\frac{\frac83}{-\frac79}=-\frac83.\frac97=-\frac{24}{7}$
2^de manier :
$\\\cos\alpha=\sqrt{1-\frac{16}{24}}=\sqrt{\frac{9}{25}}=\frac35\\\tan2\alpha=\frac{\sin2\alpha}{\cos2\alpha}=\frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}=\frac{2.\frac45.\frac35}{\frac{9}{25}-\frac{16}{25}}=\frac{\frac{24}{25}}{-\frac{7}{25}}=-\frac{24}{7}$
3^de manier :
$\\\cos^2\alpha=1-\sin^2\alpha=1-\frac{16}{25}=\frac{9}{25}\\\frac{1}{\cos^2\alpha}=\frac{25}{9}=1+\tan^2\alpha\;\Rightarrow\;\tan^2\alpha=\frac{25}{9}-1=\frac{16}{9}$
$\Rightarrow\;\tan^2\alpha=\frac{25}{9}-1=\frac{16}{9}\;\Rightarrow\tan\alpha=\frac43\\\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{\frac83}{1-\frac{16}{9}}=\frac{\frac83}{-\frac79}=\frac83(-\frac97)=-\frac{24}{7}$
4^de manier :
Door de gegevens kunnen we al uitmaken dat 45° < α < 90° (sin 45° = v2/2

≈ 0,7)
Dus 90°< 2α < 180° ⇒ tan 2α < 0
Vermits er maar één van de vijf alternatieven negatief is, is dit dus het juiste antwoord.