Question

Cos (7pi-x) Sin (2/pi+x) Sin (3pi/2+x) Cos (2pi-x) Cos (pi+x)

Answer 1

$cos (7\pi-x)=-cosx; \\sin (2\pi+x)=sinx; \\sin (\frac{3\pi}{2}+x)= -cosx;\\cos (2\pi-x)=cosx; \\cos (\pi+x)=-cosx.$

Использовались формулы приведения.

Answer 2

To simplify this expression, we can use the following trigonometric identities:

• cos(a - b) = cos(a) cos(b) + sin(a) sin(b)
• sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

Applying these identities to the given expression, we get:

cos(7π - x) sin(2/π + x) sin(3π/2 + x) cos(2π - x) cos(π + x)

= cos(7π) cos(x) + sin(7π) sin(x) sin(2/π) cos(x) + cos(2/π) sin(x) sin(3π/2) cos(x) + cos(3π/2) sin(x) cos(2π) cos(x) + sin(2π) sin(x) cos(π) cos(x) - sin(π) sin(x)

= -cos(x) + cos(2/π) sin(x) - sin(x) + cos(x) - sin(x) - sin(x)

= -cos(x) + cos(2/π) sin(x) - 3sin(x)

Therefore, the simplified expression is:

-cos(x) + cos(2/π) sin(x) - 3sin(x)