Question

sin²(2π/7)+sin² (3π/14)+sin²(11π/14)+sin²(5π/7)=2​

Answer 1

Answer:

Step-by-step explanation:sin^2(2pi/7)+sin^2(5pi/7)

=sin^2(2pi/7)+sin^2(2pi/7)   (sin(A)=sin(pi-A))

Then,

sin^2(3pi/14)+sin^2(11pi/14)

=sin^2(3pi/14)+sin^2(3pi/14)  sin(A)=sin(pi-A)

Then,

2[sin^2(2pi/7)+sin^2(3pi/14)]

=2[sin^2(2pi/7)+cos^2(2pi/7)]       sin(B)=cos(pi/2-B)

=2

Answer 2

${sin}^{2} ( \frac{2\pi}{7} ) + {sin}^{2} ( \frac{3\pi}{14} ) + {sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{5\pi}{7} ) = 2$

Solution :-

Let's take LHS →

$\longrightarrow {sin}^{2} ( \frac{2\pi}{7} ) + {sin}^{2} ( \frac{3\pi}{14} ) + {sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{5\pi}{7} )$

So, we can write -

• $\frac{2\pi}{7} \: = \: (\pi - \frac{5\pi}{7} )$
• $\frac{3\pi}{14} = (\pi - \frac{11\pi}{14} )$

So,

$\longrightarrow {sin}^{2} ( \frac{2\pi}{7} ) + {sin}^{2} ( \frac{3\pi}{14} ) + {sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{5\pi}{7} )$

$\longrightarrow {sin}^{2} ( \pi - \frac{5\pi}{7} ) + {sin}^{2} ( \pi - \frac{11\pi}{14} ) + {sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{5\pi}{7} )$

We know that :-

• $sin(\pi - a) = sin \: a$
• ${sin}^{2} ( \frac{5\pi}{7} ) + {sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{5\pi}{7} )$

$\longrightarrow 2{sin}^{2} (\frac{11\pi}{14} ) + 2{sin}^{2} ( \frac{5\pi}{7} )$

$\longrightarrow 2({sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{5\pi}{7} ))$

Now :-

• $\frac{5\pi}{7} = \frac{\pi}{2} - \frac{11\pi}{14}$

We know that :-

• $sin \: ( \frac{\pi}{2} - a) = cos \: a$

$\longrightarrow 2({sin}^{2} (\frac{11\pi}{14} ) + {sin}^{2} ( \frac{\pi}{2} - \frac{11\pi}{14} ))$

$\longrightarrow 2({sin}^{2} (\frac{11\pi}{14} ) + {cos}^{2} ( \frac{11\pi}{14} ))$

Using the identity :-

• ${sin}^{2} a + {cos}^{2} a = 1$

$\longrightarrow 2 \times 1$

$\longrightarrow 2$

R.H.S

Hence proved !!