Find sin $$\dfrac{x}{2}$$, cos$$\dfrac{x}{2}$$ and tan$$\dfrac{x}{2}$$ if $$tanx=-\dfrac{4}{3}$$, x lies in 2nd quadrant

Asked by Pragya Singh | 1 year ago |  111

##### Solution :-

Here, x is in 2nd quadrant.

Therefore ,

$$\dfrac{\pi}{4}< \dfrac{x}{2}< \dfrac{\pi}{2}$$

hence

$$\dfrac{x}{2}$$ lies in 1st quadrant.

Therefore,

$$sin\dfrac{x}{2},cos\dfrac{x}{2}\;and\;tan\dfrac{x}{2}$$ are positive.

Given that tanx = $$- \dfrac{4}{3}$$

We know that,

$$sec^2x=1+tan^2x$$

$$=1+(-\dfrac{4}{3})^2$$

$$\dfrac{25}{9}$$

As x is in 2nd quadrant, secx is negative.

Therefore ,
$$secx=-\dfrac{5}{3}$$

Then

$$cosx=-\dfrac{3}{5}$$

Now we know that, $$2cos^2\dfrac{x}{2}=cosx+1$$

Computing we get,

$$2cos^2 \dfrac{x}{2}= \dfrac{2}{5}$$

$$cos\dfrac{x}{2}=\dfrac{1}{\sqrt{5}}$$

Now we know that, sin2x+cos2x=1

Therefore substituting $$cos\dfrac{x}{2}=\dfrac{1}{\sqrt{5}}$$ and computing,

$$sin\dfrac{x}{2}=\dfrac{2}{\sqrt{5}}$$

Hence ,

$$tan=\dfrac{x}{2}=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}$$ = 2

Thus, the respective values of $$sin\dfrac{x}{2},cos\dfrac{x}{2}\;and\;tan\dfrac{x}{2}$$ are

$$\dfrac{2\sqrt{5}}{5}, \dfrac{\sqrt{5}}{5},2$$

Answered by Abhisek | 1 year ago

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