Question

1)Доказать тождество: (sin2A+sin5A-sin3A)/cosA+1-2*sin(квадрат)2A=2sinA 2)Упростить sin210+cos(-480)-√3*ctg480 Большое спасибо!

Answer 1

1)(2sinAcosA+2sinAcos4A)/(coaA+cos4A)=2sinA(cosA+cos4A)/(cosA+cos4A)=2sinA
2)sin(180+30)+cos(180-60)-√3ctg(180-60)=-sin30-cos60+√3ctg60=-1/2-1/2-√3*1/√3=-2

Answer 2

Proof of Identity: (sin2A+sin5A-sin3A)/cosA+1-2*sin^2(2A)=2sinA

To prove the given identity, let's simplify both sides of the equation and show that they are equal.

Starting with the left-hand side (LHS):

LHS = (sin2A + sin5A - sin3A) / (cosA + 1 - 2*sin^2(2A))

Using the sum-to-product trigonometric identities, we can rewrite the numerator as follows:

sin2A + sin5A - sin3A = 2sin(A + 3A)/2 * cos(A - 3A)/2 - sin(A + 3A)/2 * cos(A + 3A)/2

Simplifying further:

= 2sin(4A)/2 * cos(-2A)/2 - sin(4A)/2 * cos(4A)/2

= sin(4A) * cos(-A) - sin(4A) * cos(2A)

= sin(4A) * (cos(-A) - cos(2A))

Now, let's simplify the denominator:

cosA + 1 - 2*sin^2(2A) = cosA + 1 - 2*(1 - cos^2(2A))

= cosA + 1 - 2 + 2*cos^2(2A)

= 2*cos^2(2A) + cosA - 1

Now, let's rewrite the right-hand side (RHS):

RHS = 2sinA

Now, we can compare the LHS and RHS:

LHS = sin(4A) * (cos(-A) - cos(2A)) / (2*cos^2(2A) + cosA - 1)

RHS = 2sinA

To prove the identity, we need to show that LHS = RHS.

Unfortunately, I couldn't find a way to simplify the expression further to directly match the RHS. It's possible that the given identity is not correct.

If you have any other questions, feel free to ask!