Question

What is the value of sin3pi/10

How to find it?

Answer 1

Answer:

Step-by-step explanation:

We know,

$\tt{sin\bigg(\dfrac{\pi}{2}-\theta\bigg)=cos(\theta)}$

So,

$\sf{sin\bigg(\dfrac{3\pi}{10}\bigg)=sin\bigg(\dfrac{\pi}{2}-\dfrac{\pi}{5}\bigg)=cos\bigg(\dfrac{\pi}{5}\bigg)}$

$\rm{\green{Let\,\,x=\dfrac{\pi}{5}}}$

$\sf{\implies\,5x=\pi}$

$\sf{\implies\,2x+3x=\pi}$

$\sf{\implies\,2x=\pi-3x}$

$\sf{\implies\,cos(2x)=cos(\pi-3x)}$

$\sf{\implies\,2cos^2(x)-1=-cos(3x)}$

$\sf{\implies\,2cos^2(x)-1=-\{4cos^3(x)-3cos(x)\}}$

$\sf{\implies\,2cos^2(x)-1=3cos(x)-4cos^3(x)}$

$\sf{\implies\,4cos^3(x)+2cos^2(x)-3cos(x)-1=0}$

$\sf{\implies\,4cos^3(x)+4cos^2(x)-2cos^2(x)-2cos(x)-cos(x)-1=0}$

$\sf{\implies\,4cos^2(x)\{cos(x)+1\}-2cos(x)\{cos(x)+1\}-1\{cos(x)+1\}=0}$

$\sf{\implies\,\{cos(x)+1\}\{4cos^2(x)-2cos(x)-1\}=0}$

$\rm{\to\purple{If\,\,\blue{cos(x)+1=0},\,then,\,\,\blue{x=\dfrac{3\pi}{2}},which\,\,not\,\,possible.}}$

So,

$\sf{4cos^2(x)-2cos(x)-1=0}$

Applying the quadratic formula,

$\sf{cos(x)=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot4\cdot(-1)}}{2\cdot4}}$

$\sf{\implies\,cos(x)=\dfrac{2\pm\sqrt{4+16}}{2\cdot4}}$

$\sf{\implies\,cos(x)=\dfrac{2\pm\sqrt{20}}{2\cdot4}}$

$\sf{\implies\,cos(x)=\dfrac{2\pm2\sqrt{5}}{2\cdot4}}$

$\sf{\implies\,cos(x)=\dfrac{1\pm\sqrt{5}}{4}}$

$\rm{Since\,\,x=\dfrac{\pi}{5},\,\,which\,\,lies\,\,in\,\,1st\,\,quadrant\,\,}$

$\rm{So,\,\,cos(x)\,\,will\,\,be\,\,positive}$

$\sf{\implies\,cos(x)=\dfrac{1+\sqrt{5}}{4}}$

$\sf{\red{hence,\,\,cos\bigg(\dfrac{\pi}{5}\bigg)=sin\bigg(\dfrac{3\pi}{10}\bigg)=\dfrac{1+\sqrt{5}}{4}}}$