Question

Sin2(180°-x)+sin2(270°-x)​

Answer 1

это правильно

Объяснение:

♥...........

Answer 2

To simplify the expression sin(2(180° - x)) + sin(2(270° - x)), you can use trigonometric identities. First, let's apply the double angle identity for sine:

sin(2θ) = 2sin(θ)cos(θ)

Now, let's apply this identity to both terms:

1. For sin(2(180° - x)), θ = 180° - x: sin(2(180° - x)) = 2sin(180° - x)cos(180° - x)

Now, let's evaluate sin(180° - x) and cos(180° - x):

sin(180° - x) = sin(180°)cos(x) - cos(180°)sin(x) sin(180°) = 0 (since sine of 180° is 0) cos(180°) = -1 (since cosine of 180° is -1)

So, sin(180° - x) = 0 - (-1)sin(x) = sin(x)

cos(180° - x) = cos(x) (cosine function is even)

Now, we have:

sin(2(180° - x)) = 2sin(x)cos(x)

2. For sin(2(270° - x)), θ = 270° - x: sin(2(270° - x)) = 2sin(270° - x)cos(270° - x)

Now, let's evaluate sin(270° - x) and cos(270° - x):

sin(270° - x) = sin(270°)cos(x) - cos(270°)sin(x) sin(270°) = -1 (since sine of 270° is -1) cos(270°) = 0 (since cosine of 270° is 0)

So, sin(270° - x) = (-1)cos(x) - 0sin(x) = -cos(x)

cos(270° - x) = sin(x) (cosine function is odd)

Now, we have:

sin(2(270° - x)) = 2(-cos(x))sin(x) = -2sin(x)cos(x)

Now, you can rewrite the original expression with these simplified terms:

sin(2(180° - x)) + sin(2(270° - x)) = 2sin(x)cos(x) - 2sin(x)cos(x) = 0

So, the simplified expression is 0.