Question

cos pi by 11 cos 2 pi by 11cos 3 pie by 11 cos 4 pie by11 cos 5 pie by 11 =lambda then 32 lambda equal to​

Answer 1

Step-by-step explanation:

[tex]cos(π−113π)=−cos118π

cos(π+115π)=−cos1116π

The given expression would become

cos11πcos112πcos114πcos118πcos1116π

Let A=11π. Then the equation becomes

=cosAcos2Acos22Acos23Acos24A

Using the formula

cosAcos2Acos(22A)cos(23A) \\ cos(2n−1A)=2nsinAsin2nA

Thus,

=25sinAsin25A

=32sinAsin32A=32sinAsin(33A−A)∵11A=π

=32sinAsin(3π−A)

=(32sinAsinA)

Hence, option 'D' is correct.

=321



Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{cos\bigg(\dfrac{\pi}{11}\bigg)cos\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{3\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=cos\bigg(\dfrac{\pi}{11}\bigg)cos\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{3\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=cos\bigg(\dfrac{\pi}{11}\bigg)cos\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\pi-\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-cos\bigg(\dfrac{\pi}{11}\bigg)cos\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{2sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\dfrac{\pi}{11}\bigg)cos\bigg(\dfrac{\pi}{11}\bigg)\cdot\,cos\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{2sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{4sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\dfrac{2\pi}{11}\bigg)cos\bigg(\dfrac{2\pi}{11}\bigg)\cdot\,cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{4sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{8sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\dfrac{4\pi}{11}\bigg)cos\bigg(\dfrac{4\pi}{11}\bigg)\cdot\,cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{8sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{16sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\dfrac{8\pi}{11}\bigg)cos\bigg(\dfrac{8\pi}{11}\bigg)\cdot\,cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{16sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\dfrac{16\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{32sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\dfrac{16\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=-\dfrac{1}{32sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\pi+\dfrac{5\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=\dfrac{1}{32sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,2sin\bigg(\dfrac{5\pi}{11}\bigg)cos\bigg(\dfrac{5\pi}{11}\bigg)}$

$\sf{=\dfrac{1}{32sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\dfrac{10\pi}{11}\bigg)}$

$\sf{=\dfrac{1}{32sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\pi-\dfrac{\pi}{11}\bigg)}$

$\sf{=\dfrac{1}{32sin\bigg(\dfrac{\pi}{11}\bigg)}\cdot\,sin\bigg(\dfrac{\pi}{11}\bigg)}$

$\sf{=\dfrac{1}{32}}$

So,

$\rm{\bold{\blue{32\lambda=1}}}$