Question

Find the value of:

4 + log2(sin(pi/5)) + log2(sin(2pi/5)) + log2(sin(3pi/5)) + log2(sin(4pi/5))

Answer 1

Hey Buddy

Here's the answer ...

___________________________

We know

=> log m + log n = log ( mn )

=> log m - log n = log ( m/n )

=> log n ( n ) = 1

=> log m^n = n log m

There fore

=> 4 + log2 ( sinπ/5 ) + log2 ( sin 2π/5 ) + log2 ( sin 3π/5 ) + log2 ( sin 4π/5 )

# By applying above identity

=> 4 + log2 [ sin ( π/5 ) × sin ( 2π/5 ) × sin ( 3π/5 ) × sin ( 4π/5 ) ]

# Now let assume if

A + B = π

A = π - B

so

sin ( A ) = sin ( π - B )

hence

==> sin ( A ) = sin ( B ) ✔️

Hense

=> π/5 + 4π/5 = 5π/5 = π

So , sin ( π/5 ) = sin ( 4π/5 )

Also

=> 2π/5 + 3π/5 = π

so, sin ( 2π/5 ) = sin ( 3π/5 )

# Now our question become

=> 4 + log2 [ sin ( π/5 ) × sin ( 2π/5 ) × sin ( 2π/5 ) × sin ( π/5 ) ]

=> 4 + log2 [ sin ( π/5 ) × sin ( 2π/5 ) ]^2

=> converting radian to degree

=> π/5 × 180/π = 36°

=> 2π/5 × 180/π = 72°

=> 4 + log2 [ sin (36°) × sin (72°) ]^2

=> 4 + log2 [ sin (36°) × sin (18°) ]^2

# As Sin ( π/2 - x ) = sin ( x )

=> 4 + log2 [ (√ ( 10 - 2√5 ))/4× √ ( 10 + 2√5 ))/4 ]^2

# After solving

=> 4 + log2 [ (100 - 20 )/256 ]

=> 4 + log2 [ 5/16 ]

=> 4 + log2 ( 5 ) - log2 ( 16 )

=> 4 + log2 ( 5 ) - log2 ( 2 )^4

=> 4 + log2 ( 5 ) - 4 log2 ( 2 )

=> 4 + log2 ( 5 ) - 4

=> log2 ( 5 )

HOPE HELPED...

JAI HIND

:)