Find sin $$\dfrac{x}{2}$$, cos$$\dfrac{x}{2}$$ and tan$$\dfrac{x}{2}$$ if $$sinx=\dfrac{1}{4}$$, x lies in 2nd quadrant

Asked by Pragya Singh | 1 year ago |  153

##### Solution :-

Here, x lies 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 sinx = $$\dfrac{1}{4}$$

Now we know that, sin2x+cos2x=1

Therefore substituting $$si nx=\dfrac{1}{4}$$ and computing ,

$$cosx=-\dfrac{\sqrt{15}}{4}$$

since x lies in 2nd quadrant, cosx is negative.

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

Computing we get, $$2sin^2\dfrac{x}{2} =1+\dfrac{\sqrt{15}}{4}$$

Hence

$$sin\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}}$$

Now we know that, sin2x+cos2x=1

Therefore substituting

$$sin\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}}$$ and computing,

$$cos\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}}$$

Hence,

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

$$=\dfrac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}$$

$$4\pm\sqrt{15}$$

Answered by Abhisek | 1 year ago

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