Question

Если sin2a+cos2a=1/3 Найдите sin 4 a

Answer 1

Возводим обе части равенства в квадрат

$1+2\sin2\alpha\cos2\alpha=\dfrac{1}{9}\\ \\ \sin4\alpha=\dfrac{1}{9}-1=-\dfrac{8}{9}$

Answer 2

To find the value of sin(4a), we can use trigonometric identities to express sin(4a) in terms of sin(2a) and cos(2a). Here's how to do it step by step:

Step 1: Double Angle Identity for Sine sin(2a) = 2 * sin(a) * cos(a)

Step 2: Square of Double Angle Identity for Sine sin^2(2a) = (2 * sin(a) * cos(a))^2 = 4 * sin^2(a) * cos^2(a)

Step 3: Double Angle Identity for Cosine cos(2a) = cos^2(a) - sin^2(a)

Step 4: Square of Double Angle Identity for Cosine cos^2(2a) = (cos^2(a) - sin^2(a))^2 = cos^4(a) - 2 * cos^2(a) * sin^2(a) + sin^4(a)

Step 5: Use the Given Equation Given sin^2(2a) + cos^2(2a) = 1/3, we can substitute the expressions from Step 2 and Step 4: 4 * sin^2(a) * cos^2(a) + cos^4(a) - 2 * cos^2(a) * sin^2(a) + sin^4(a) = 1/3

Step 6: Rearrange the equation to solve for sin^4(a) 4 * sin^2(a) * cos^2(a) - 2 * cos^2(a) * sin^2(a) + sin^4(a) + cos^4(a) = 1/3 sin^4(a) + 4 * sin^2(a) * cos^2(a) - 2 * cos^2(a) * sin^2(a) + cos^4(a) = 1/3

Step 7: Use the Pythagorean Identity sin^2(a) + cos^2(a) = 1 sin^4(a) + 4 * (1 - sin^2(a)) * sin^2(a) - 2 * (1 - sin^2(a)) * sin^2(a) + cos^4(a) = 1/3

Step 8: Simplify the equation sin^4(a) + 4sin^2(a) - 4sin^4(a) - 2sin^2(a) + 2sin^4(a) + cos^4(a) = 1/3

Step 9: Combine like terms -3sin^4(a) + 3sin^2(a) + cos^4(a) = 1/3

Step 10: Rearrange the equation to isolate sin^4(a) 3sin^4(a) - 3sin^2(a) + cos^4(a) = -1/3

Step 11: Use the Pythagorean Identity sin^2(a) + cos^2(a) = 1 to replace cos^2(a) 3sin^4(a) - 3sin^2(a) + (1 - sin^2(a)) = -1/3

Step 12: Simplify the equation 3sin^4(a) - 3sin^2(a) + 1 - sin^2(a) = -1/3

Step 13: Combine like terms 3sin^4(a) - 4sin^2(a) + 1 = -1/3

Step 14: Move the constant to the right side 3sin^4(a) - 4sin^2(a) + 1 + 1/3 = 0

Step 15: Common denominator 3sin^4(a) - 4sin^2(a) + 3/3 + 1/3 = 0

Step 16: Combine fractions 3sin^4(a) - 4sin^2(a) + 4/3 = 0

Now, we have a quadratic equation in terms of sin^2(a). Let's solve for sin^2(a) using the quadratic formula:

a = 3, b = -4, c = 4/3

sin^2(a) = (-b ± √(b^2 - 4ac)) / 2a sin^2(a) = (4 ± √(16 - 4 * 3 * 4/3)) / 6 sin^2(a) = (4 ± √(16 - 16)) / 6 sin^2(a) = (4 ± √0) / 6 sin^2(a) = (4 ± 0) / 6

Now, we have two possible values for sin^2(a):

1. sin^2(a) = (4 + 0) / 6 = 4 / 6 = 2 / 3
2. sin^2(a) = (4 - 0) / 6 = 4 / 6 = 2 / 3

Since sin^2(a) can only be between 0 and 1, the possible solution is sin^2(a) = 2 / 3.

Finally, we can find sin(4a) using the double angle identity for sine:

sin(4a) = 2 * sin(2a) * cos(2a) = 2 * √(2 / 3) * √(1 - 2 / 3) = 2 * √(2 / 3) * √(1 / 3) = 2 * √(2 / 9) = √(8 / 9) = √8 / 3

Therefore, the value of sin(4a) is √8 / 3.