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If sin x=5/8, x in quadrant I....

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I dont understand

felrose13  Feb 17, 2017
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Set a right-angled triangle with one of the angle = x.

sin x = 5/8 means opposite of x = 5 and hypotenuse = 8.

By means of Pythagoras theorem we get the adjacent of x = $$\sqrt{8^2 - 5^2}$$ = $$\sqrt{39}$$

Therefore cos x = adjacent/hypotenuse = $$\dfrac{\sqrt{39}}{8}$$

tan x = sin x / cos x = $$\dfrac{5}{\sqrt{39}}$$ = $$\dfrac{5\sqrt{39}}{39}$$

By formula we get:

$$\sin(2x)=2\sin x \cos x = 2\left(\dfrac{5}{8}\right)\left(\dfrac{\sqrt{39}}{8}\right)=\dfrac{5}{32}\sqrt{39}$$

$$\cos(2x)=1-2\sin^2 x = 1- 2\cdot\left(\dfrac{5}{8}\right)^2=\dfrac{7}{32}$$

$$\tan(2x)=\dfrac{\sin (2x)}{\cos (2x)}=\dfrac{\frac{5}{32}\sqrt{39}}{\frac{7}{32}}=\dfrac{5}{7}\sqrt{39}$$

MaxWong  Feb 18, 2017

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