Question

(1+cosπ/10)(1+cos3π/10)(1+cos7π/10)(1+cos9π/10)

Answer 1

if u have studied maths of class 11th.....then u will definitely understand that what I have done !!

Answer 2

Answer:

Hence, the value of the expression is:

$\dfrac{1}{16}$

Step-by-step explanation:

We have to find the value of:

(1+cosπ/10)(1+cos3π/10)(1+cos7π/10)(1+cos9π/10)

We know that:

$\cos (\dfrac{7\pi}{10})=\cos (\pi-\dfrac{3\pi}{10})=-\cos (\dfrac{3\pi}{10})$

Also,

$\cos (\dfrac{9\pi}{10})=\cos (\pi-\dfrac{\pi}{10})=-\cos (\dfrac{\pi}{10})$

Hence, the expression is converted to:

$=(1+\cos ({\pi}{10}))(1+\cos(\dfrac{3\pi}{10}))(1-\cos (\dfrac{3\pi}{10}))(1+cos (\dfrac{9\pi}{10}))\\\\\\= (1+\cos (\dfrac{\pi}{10}))(1-\cos (\dfrac{\pi}{10}))(1+\cos (\dfrac{3\pi}{10}))(1-\cos(\dfrac{3\pi}{10}))\\\\\\=(1-\cos^2 (\dfrac{\pi}{10}))(1-\cos^2 (\dfrac{3\pi}{10}))$  

( since (a-b)(a+b)=a^2-b^2  )

$=\sin^2 (\dfrac{\pi}{10})\sin^2 (\dfrac{3\pi}{10})$

since:

$\sin^2 (\theta)+\cos^2 (\theta)=1$

⇒ $1-\cos^2 (\theta)=\sin^2 (\theta)$

Also we know that:

$\sin(\dfrac{\pi}{10})=\dfrac{1}{4}\times (\sqrt{5}-1)$  

$\sin (\dfrac{3\pi}{10}) =\dfrac{1}{4}\times (\sqrt{5}+1)$

Hence,

$\sin(\dfrac{\pi}{10})\sin(\dfrac{3\pi}{10})=\dfrac{1}{16}\times [(\sqrt{5}-1)(\sqrt{5}+1)]$

                             $=\dfrac{1}{16}\times (5-1)$

                            $=\dfrac{4}{16}$

                             $\dfrac{1}{14}$

Hence,

$(\sin (\dfrac{\pi}{10})\sin (\dfrac{3\pi}{10}))^2=\sin^2 (\dfrac{\pi}{10})\sin^2 (\dfrac{3\pi}{10})\\\\\\=(\dfrac{1}{4})^2=\dfrac{1}{16}$

Hence, the value of the expression is:

1/16