Question

((sin2a-sin3a+sin4a)/(cos2a-cos3a+cos4a))*ctg3a

Answer 1

((sin2a-sin3a+sin4a)/(cos2a-cos3a+cos4a))*ctg3a=((2sin3a*cosa-sin3a)/(2cos3a*cosa-cos3a))*ctg3a=sin3a/cos3a*ctg3a=1

Answer 2

We can simplify the expression using trigonometric identities:

sin2a = 2sin(a)cos(a)

sin3a = sin(2a+a) = sin(2a)cos(a) + cos(2a)sin(a) = 2sin(a)cos^2(a) + (1-2sin^2(a))sin(a) = 3sin(a) - 4sin^3(a)

sin4a = sin(2(2a)) = 2sin(2a)cos(2a) = 4sin(a)cos(a)(1-2sin^2(a))

cos2a = cos^2(a) - sin^2(a) = 1-2sin^2(a)

cos3a = cos(2a+a) = cos(2a)cos(a) - sin(2a)sin(a) = (1-2sin^2(a))cos(a) - 2sin(a)cos^2(a) = 4cos^3(a) - 3cos(a)

cos4a = cos(2(2a)) = 2cos^2(2a) - 1 = 2(2cos^2(a)-1)^2 - 1 = 8cos^4(a) - 8cos^2(a) + 1

ctg3a = 1/tan(3a) = -1/tan(a)(1-3tan^2(a))

Substituting these expressions into the original expression, we get:

(sin2a - sin3a + sin4a)/(cos2a - cos3a + cos4a) * ctg3a

= [2sin(a)cos(a) - (3sin(a) - 4sin^3(a)) + 4sin(a)cos(a)(1-2sin^2(a))] / [(1-2sin^2(a)) - (4cos^3(a) - 3cos(a)) + (8cos^4(a) - 8cos^2(a) + 1)] * [-1/tan(a)(1-3tan^2(a))]

= [-4sin^3(a) + 10sin(a)cos^2(a) + 8sin(a)cos^4(a) - 8sin(a)cos^2(a)] / [-2sin^2(a) + 4cos^3(a) - 8cos^2(a) + 8cos^4(a) - 1] * [-1/tan(a)(1-3tan^2(a))]

= [-4sin^3(a) + 2sin(a)cos^2(a) + 8sin(a)cos^4(a)] / [-2sin^2(a) + 4cos^3(a) - 8cos^2(a) + 8cos^4(a) - 1] * [-1/tan(a)(1-3tan^2(a))]

= [2sin(a)(-2sin^2(a) + cos^2(a) + 4cos^4(a))] / [(2cos^2(a) - 1)(-2sin^2(a) + 4cos^3(a) - 8cos^2(a) + 8cos^4(a) - 1)] * [-1/tan(a)(1-3tan^2(a))]

= [2sin(a)(-2sin^2(a) + cos^2(a) + 4cos^4(a))] / [(2cos^2(a) - 1)(4cos^4(a) - 2cos^

Answer 3

To simplify this expression, we can use trigonometric identities to rewrite the numerator and denominator as a product of factors involving sines and cosines of angles in multiples of "a".

Starting with the numerator:

sin(2a) - sin(3a) + sin(4a)

We can use the formula for the difference of two sines:

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

Using this formula with x=3a and y=2a, we get:

sin(3a) - sin(2a) = 2 cos(5a/2) sin(a/2)

Therefore, we can rewrite the numerator as:

sin(2a) - sin(3a) + sin(4a) = sin(2a) + 2 cos(5a/2) sin(a/2) - 2 cos(2a) sin(a)

Now, let's work on the denominator:

cos(2a) - cos(3a) + cos(4a)

We can use the formula for the difference of two cosines:

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

Using this formula with x=3a and y=2a, we get:

cos(3a) - cos(2a) = -2 sin(5a/2) sin(a/2)

Using the same formula with x=4a and y=3a, we get:

cos(4a) - cos(3a) = -2 sin(7a/2) sin(a/2)

Therefore, we can rewrite the denominator as:

cos(2a) - cos(3a) + cos(4a) = cos(2a) + 2 sin(5a/2) sin(a/2) - 2 sin(7a/2) sin(a/2)

Now, let's substitute these expressions back into the original expression:

(sin(2a) + 2 cos(5a/2) sin(a/2) - 2 cos(2a) sin(a)) / (cos(2a) + 2 sin(5a/2) sin(a/2) - 2 sin(7a/2) sin(a/2)) * ctg(3a)

We can simplify this expression further by noting that ctg(3a) = 1/tan(3a) = cos(3a)/sin(3a). So we can cancel out the sin(3a) terms in the numerator and denominator:

(sin(2a) + 2 cos(5a/2) sin(a/2) - 2 cos(2a) sin(a)) / (cos(2a) + 2 sin(5a/2) sin(a/2) - 2 sin(7a/2) sin(a/2)) * cos(3a)/sin(3a)

Simplifying the denominator by factoring out sin(a/2), we get:

(cos(2a) + 2 sin(a/2) (cos(5a/2) - sin(7a/2))) / (sin(a/2) (2 cos(5a/2) - 2 sin(7a/2))) * cos(3a)/sin(3a)

Now, we can simplify further by canceling out the common factors of sin(a/2) and 2 in the numerator and denominator:

(cos(2a) + (cos(5a/2) - sin(7a/2))) / (cos(