Oblicz: sinα, cosα, tgα, jeśli : a) cos2α=3/4 i α∈(0;π/2) b) cos2α=3/4 i α∈(3π/2 ; 2π) c) cos2α= -1/.... Question from @Akuka54

December 2018 1 9 Report

Oblicz: sinα, cosα, tgα, jeśli :
a) cos2α=3/4 i α∈(0;π/2)
b) cos2α=3/4 i α∈(3π/2 ; 2π)
c) cos2α= -1/3 i α∈(π/2 ; π)
d) cos2α= -1/3 i α∈(3π/2 ; 2π)

ghe Wzory:
$cos2\alpha=1-2sin^2\alpha$

$2sin^2\alpha=1-cos2\alpha \ /:2$

$sin^2\alpha= \frac{1-cos2\alpha}{2}$
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$sin^2\alpha+cos^2\alpha=1$

$cos^2\alpha=1-sin^2\alpha$
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a)
$cos2 \alpha =\frac{3}{4}$
$\alpha \in\left(0; \frac{\pi}{2}\right)$ - I ćwiartka

Obliczam $sin\alpha$
$sin^2\alpha= \frac{1-cos2\alpha}{2}$

$sin^2\alpha= \frac{1-\frac{3}{4}}{2}=\frac{\frac{1}{4}}{2}=\frac{1}{8}$

$sin\alpha= \sqrt{\frac{1}{8}}=\frac{1}{ \sqrt{8} }=\frac{1}{2 \sqrt{2} }=\frac{ \sqrt{2} }{4}$

Obliczam $cos\alpha$
$cos^2\alpha=1-sin^2\alpha$

$cos^2\alpha=1-\frac{1}{8}=\frac{7}{8}$

$cos\alpha= \sqrt{\frac{7}{8}}=\frac{ \sqrt{7} }{ \sqrt{8} }=\frac{ \sqrt{7} }{2 \sqrt{2} }=\frac{ \sqrt{14} }{4}$

obliczam $tg\alpha$
$tg\alpha=\frac{sin\alpha}{cos\alpha}$

$tg\alpha=\frac{\frac{ \sqrt{2} }{4}}{\frac{ \sqrt{14} }{4}}=\frac{ \sqrt{2}} { \sqrt{14} }=\frac{ \sqrt{28}} {14}=\frac{2 \sqrt{7}} {14}=\frac{ \sqrt{7}} {7}$

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b)
$cos2 \alpha = \frac{3}{4}$
$\alpha \in\left( \frac{3\pi}{2};2\pi\right)$ - III ćwiartka

Obliczam $sin\alpha$
$sin^2\alpha= \frac{1-cos2\alpha}{2}$

$sin^2\alpha= \frac{1-\frac{3}{4}}{2}= \frac{\frac{1}{4}}{2}=\frac{1}{8}$

$sin\alpha=- \sqrt{\frac{1}{8}}=-\frac{1}{ \sqrt{8} }=-\frac{1}{2 \sqrt{2} }=-\frac{ \sqrt{2} }{4}$

Obliczam $cos\alpha$
$cos^2\alpha=1-sin^2\alpha$

$cos^2\alpha=1-\frac{1}{8}=\frac{7}{8}$

$cos\alpha=- \sqrt{\frac{7}{8}}=-\frac{ \sqrt{7} }{ \sqrt{8} }-\frac{ \sqrt{7} }{2 \sqrt{2} }=-\frac{ \sqrt{14} }{4}$

obliczam $tg\alpha$
$tg\alpha=\frac{sin\alpha}{cos\alpha}$

$tg\alpha=\frac{-\frac{ \sqrt{2} }{4}}{-\frac{ \sqrt{14} }{4}}=\frac{ \sqrt{2}} { \sqrt{14} }=\frac{ \sqrt{28}} {14}=\frac{2 \sqrt{7}} {14}=\frac{ \sqrt{7}} {7}$
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c)
$cos2 \alpha =- \frac{1}{3}$
$\alpha \in\left( \frac{\pi}{2};\pi\right)$ - II ćwiartka

Obliczam $sin\alpha$
$sin^2\alpha= \frac{1-cos2\alpha}{2}$

$sin^2\alpha= \frac{1-\left(-\frac{1}{3}\right) }{2}= \frac{1+\frac{1}{3}}{2}= \frac{\frac{4}{3}}{2}=\frac{2}{3}$

$sin\alpha= \sqrt{\frac{2}{3}}=\frac{ \sqrt{2} }{ \sqrt{3} }=\frac{ \sqrt{6} }{3}$

Obliczam $cos\alpha$
$cos^2\alpha=1-sin^2\alpha$

$cos^2\alpha=1-\frac{2}{3}=\frac{1}{3}$

$cos\alpha=- \sqrt{\frac{1}{3}}=-\frac{1}{ \sqrt{3} }=-\frac{ \sqrt{3} }{3 }$

Obliczam $tg\alpha$
$tg\alpha=\frac{sin\alpha}{cos\alpha}$

$tg\alpha=\frac{\frac{ \sqrt{6} }{3}}{-\frac{ \sqrt{3} }{3 }}=-\frac{ \sqrt{6}} { \sqrt{3} }=-\frac{ \sqrt{18}} {3}=-\frac{3\sqrt{2}} {3}=-\sqrt{2}$
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d)
$cos2 \alpha =- \frac{1}{3}$

$\alpha \in\left( \frac{3\pi}{2};2\pi\right)$ - III ćwiartka

Obliczam $sin\alpha$
$sin^2\alpha= \frac{1-cos2\alpha}{2}$

$sin^2\alpha= \frac{1-\left(-\frac{1}{3}\right) }{2}= \frac{1+\frac{1}{3}}{2}= \frac{\frac{4}{3}}{2}=\frac{2}{3}$

$sin\alpha=- \sqrt{\frac{2}{3}}=-\frac{ \sqrt{2} }{ \sqrt{3} }=-\frac{ \sqrt{6} }{3}$

Obliczam $cos\alpha$
$cos^2\alpha=1-sin^2\alpha$

$cos^2\alpha=1-\frac{2}{3}=\frac{1}{3}$

$cos\alpha=- \sqrt{\frac{1}{3}}=-\frac{1}{ \sqrt{3} }=-\frac{ \sqrt{3} }{3 }$

Obliczam $tg\alpha$
$tg\alpha=\frac{sin\alpha}{cos\alpha}$

$tg\alpha=\frac{-\frac{ \sqrt{6} }{3}}{-\frac{ \sqrt{3} }{3 }}=\frac{ \sqrt{6}} { \sqrt{3} }=\frac{ \sqrt{18}} {3}=\frac{3\sqrt{2}} {3}=\sqrt{2}$

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