How do you use the double angle or half angle formulas to find the exact value of cos2x, when $sin x = \frac{\sqrt{5}}{3}$ $sinx=(sqrt 5)/3$ and $\frac{π}{2} \leq x \leq π$ $pi/2 <= x <= pi$ ?

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How do you use the double angle or half angle formulas to find the exact value of cos2x, when $sin x = \frac{\sqrt{5}}{3}$ $sinx=(sqrt 5)/3$ and $\frac{π}{2} \leq x \leq π$ $pi/2 <= x <= pi$ ?

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Answer 1 · 2015-10-18T05:32:41

Find cos 2x knowing $sin x = \frac{\sqrt{5}}{3}$ $sin x = sqrt5/3$

Ans: cos 2x = -0.11

Explanation:

Apply the trig identity: $cos 2 x = 1 - 2 {sin}^{2} x$ $cos 2x = 1 - 2sin^2 x$ . We get:

$cos 2 x = 1 - 2 (\frac{5}{9}) = \frac{9 - 10}{9} = - \frac{1}{9} = - 0.11$ $cos 2x = 1 - 2(5/9) = (9 - 10)/9 = -1/9 = - 0.11$

Check by calculator.
$sin x = \frac{\sqrt{5}}{3} = \frac{2.24}{3} = 0.745$ $sin x = sqrt5/3 = 2.24/3 = 0.745$ --> arc $x = 48^{\circ} 16$ $x = 48^@16$ -->
--> $2 x = 96^{\circ} 32$ $2x = 96^@32$ --> $cos 2 x = - 0.11 = - \frac{1}{9}$ $cos 2x = - 0.11 = -1/9$ . OK

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How do you use the double angle or half angle formulas to find the exact value of cos2x, when $sin x = \frac{\sqrt{5}}{3}$ $sinx=(sqrt 5)/3$ and $\frac{π}{2} \leq x \leq π$ $pi/2 <= x <= pi$ ?

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How do you use the double angle or half angle formulas to find the exact value of cos2x, when sinx=(sqrt 5)/3sinx=√53sinx=(sqrt 5)/3 and pi/2 <= x <= piπ2≤x≤πpi/2 <= x <= pi?

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Explanation:

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How do you use the double angle or half angle formulas to find the exact value of cos2x, when $sin x = \frac{\sqrt{5}}{3}$ $sinx=(sqrt 5)/3$ and $\frac{π}{2} \leq x \leq π$ $pi/2 <= x <= pi$ ?