Cos π/7+cos 6π/7+ sin π/5 - sin 4π/5

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вопрос

To simplify this expression, we can use the sum-to-product identities for cosine and sine functions.

We know that:
cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)
sin(A) - sin(B) = 2cos((A+B)/2)sin((A-B)/2)

Therefore,
cos(π/7) + cos(6π/7) = 2cos((π/7 + 6π/7)/2)cos((π/7 - 6π/7)/2)
= 2cos(7π/14)cos(-5π/14)
= 2cos(π/2)cos(-5π/14)
= 20cos(-5π/14)
= 0

Also,
sin(π/5) - sin(4π/5) = 2cos((π/5 + 4π/5)/2)sin((π/5 - 4π/5)/2)
= 2cos(5π/10)sin(-3π/10)
= 2cos(π/2)sin(-3π/10)
= 20sin(-3π/10)
= 0

Therefore, the expression simplifies to 0 + 0 = 0.