If tanx=-3/4 and 3Π/2<x<2Π,then value of sin2x?

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If tanx=-3/4 and 3Π/2<x<2Π,then value of sin2x?

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Answer 1 · 2018-04-02T16:57:06

$sin (2 x) = - \frac{24}{25}$ $sin(2x)=-24/25$

Explanation:

We seek $sin (2 x)$ $sin(2x)$ when $tan (x) = - \frac{3}{4}$ $tan(x)=-3/4$

By the double angle identity

$sin (2 x) = 2 sin (x) cos (x)$ $color(blue)(sin(2x)=2sin(x)cos(x)$

Let's express this in term of tangens

$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x)=2sin(x)cos(x)$

$sin (2 x) = \frac{2 sin (x) cos (x)}{1}$ $color(white)(sin(2x))=(2sin(x)cos(x))/1$

$sin (2 x) = \frac{2 sin (x) cos (x) {sec}^{2} (x)}{{sec}^{2} (x)}$ $color(white)(sin(2x))=(2sin(x)cos(x)sec^2(x))/sec^2(x)$

$sin (2 x) = \frac{2 tan (x)}{{sec}^{2} (x)}$ $color(white)(sin(2x))=(2tan(x))/sec^2(x)$

$sin (2 x) = \frac{2 tan (x)}{1 + {tan}^{2} (x)}$ $color(white)(sin(2x))=(2tan(x))/(1+tan^2(x))$

Now let $tan (x) = - \frac{3}{4}$ $tan(x)=-3/4$

$sin (2 x) = \frac{2 (- \frac{3}{4})}{1 + {(- \frac{3}{4})}^{2}}$ $sin(2x)=(2(-3/4))/(1+(-3/4)^2)$

$sin (2 x) = - \frac{\frac{3}{2}}{1 + \frac{9}{16}}$ $color(white)(sin(2x))=-(3/2)/(1+9/16)$

$sin (2 x) = - \frac{\frac{3}{2} \cdot 16}{16 + 9}$ $color(white)(sin(2x))=-(3/2*16)/(16+9)$

$sin (2 x) = - \frac{3 \cdot 8}{25}$ $color(white)(sin(2x))=-(3*8)/(25)$

$sin (2 x) = - \frac{24}{25}$ $color(white)(sin(2x))=-(24)/(25)$

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If tanx=-3/4 and 3Π/2<x<2Π,then value of sin2x?

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