Question

The value of expression (1+cosπ/10)(1+cos3π/10)(1+cos7π/10)(1+cos9π/10) is

Answer 1

Here π= 180°
So,
1+cos180°/10. = 1+(-1)/10. = 0
As the whole expression is multiplied, so 0×something=0
Ans. 0

Answer 2

Answer:

Value of the given Expression is 1/16.

Step-by-step explanation:

Given Expression:

$(1+cos\,\frac{\pi}{10})(1+cos\,\frac{3\pi}{10})(1+cos\,\frac{7\pi}{10})(1+cos\,\frac{9\pi}{10})$

To find: Value of the given Expression

We know that

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

Also, $cos\,(\frac{9\pi}{10})=cos\,(\pi-\frac{\pi}{10})=-cos\,(\frac{\pi}{10})$

Consider,

$(1+cos\,({\pi}{10}))(1+cos\,(\frac{3\pi}{10}))(1-cos\,(\frac{3\pi}{10}))(1+cos\,(\frac{9\pi}{10}))$

$=(1+\cos\,(\frac{\pi}{10}))(1-cos\,(\frac{\pi}{10}))(1+cos\,(\frac{3\pi}{10}))(1-cos\,(\frac{3\pi}{10}))$

using, a² - b² = ( a - b )( a + b )

$=(1-cos^2\,(\frac{\pi}{10}))(1-cos^2\,(\frac{3\pi}{10}))$

using, sin² x + cos² x = 1  ⇒ sin² x = 1 - cos² x

$=sin^2\,(\frac{\pi}{10})\:sin^2\,(\frac{3\pi}{10})$

Also we know that,

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

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

Hence,

$sin\,(\frac{\pi}{10})\:sin\,(\frac{3\pi}{10})=\frac{1}{16}\times((\sqrt{5}-1)(\sqrt{5}+1))$

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

                              $=\frac{4}{16}$

                              $\frac{1}{4}$

Hence,

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

Therefore, Value of the given Expression is 1/16.