Question

Find sinx/2,cosx/2,tanx/2 if cosec x = 4 , x is in quadrant II

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{cosec(x)=4\,\,\,\,\,\&\,\,\,\,\,\dfrac{\pi}{2}<x<\pi}$

Since sine is positive in the 2nd quadrant, so,

$\rm{\implies\,sin(x)=\dfrac{1}{4}}$

Also, cosine is negative in the 2nd quadrant, so,

$\rm{\implies\,cos(x)=-\sqrt{1-\dfrac{1}{16}}}$

$\rm{\implies\,cos(x)=-\sqrt{\dfrac{16-1}{16}}}$

$\rm{\implies\,cos(x)=-\sqrt{\dfrac{15}{16}}}$

$\rm{\implies\,cos(x)=-\dfrac{\sqrt{15}}{4}}$

Since x lies in the 2nd quadrant, so,

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

$\sf{\implies\,\dfrac{\pi}{4}<\dfrac{x}{2}<\dfrac{\pi}{2}}$

So, x/2 lies in the 1st quadrant, and in the 1st quadrant all T-ratios are positive,

Now,

$\rm{sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{1-cos(x)}{2}}}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{1+\dfrac{\sqrt{15}}{4}}{2}}}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{\dfrac{4+\sqrt{15}}{4}}{2}}}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{4+\sqrt{15}}{8}}}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{8+2\sqrt{15}}{16}}\,\,\,\,\,\,\,\,\,\,\,\,[\because\,multiplying\,\,2\,\,in\,\,num.\,\,and\,\,den.]}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{5+3+2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{\left(\sqrt{5}\right)^{2}+\left(\sqrt{3}\right)^{2}+2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\rm{\implies\,sin\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{16}}}$

$\implies\boxed{\green{\rm{sin\left(\dfrac{x}{2}\right)=\dfrac{\sqrt{5}+\sqrt{3}}{4}}}}$

And,

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{1-sin^{2}(x)}=\sqrt{1-\left(\dfrac{\sqrt{5}+\sqrt{3}}{4}\right)^{2}}}$

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{1-\dfrac{5+3+2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{1-\dfrac{8+2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{16-8-2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{8-2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{5+3-2\cdot\sqrt{5}\cdot\sqrt{3}}{16}}}$

$\implies\rm{cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{16}}}$

$\implies\boxed{\pink{\rm{cos\left(\dfrac{x}{2}\right)=\dfrac{\sqrt{5}-\sqrt{3}}{4}}}}$

Also,

$\rm{tan\left(\dfrac{x}{2}\right)=\dfrac{sin\left(\dfrac{x}{2}\right)}{cos\left(\dfrac{x}{2}\right)}}$

$\rm{\implies\,tan\left(\dfrac{x}{2}\right)=\dfrac{\dfrac{\sqrt{5}+\sqrt{3}}{4}}{\dfrac{\sqrt{5}-\sqrt{3}}{4}}}$

$\rm{\implies\,tan\left(\dfrac{x}{2}\right)=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\rm{\implies\,tan\left(\dfrac{x}{2}\right)=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}}$

$\rm{\implies\,tan\left(\dfrac{x}{2}\right)=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}}$

$\rm{\implies\,tan\left(\dfrac{x}{2}\right)=\dfrac{5+3+2\cdot\sqrt{5}\cdot\sqrt{3}}{5-3}}$

$\rm{\implies\,tan\left(\dfrac{x}{2}\right)=\dfrac{8+2\sqrt{15}}{2}}$

$\implies\boxed{\rm{\purple{tan\left(\dfrac{x}{2}\right)=4+\sqrt{15}}}}$