Question

Sin a + sin 3a/ сos a + cos 3a=tg 2a

Answer 1

(2sin 4a/2 * cos -2a/2) / (2cos 4a/2 * cos -2a/2)=sin2a/cos2a,     2(sin2a/cos2a)-sin2a/cos2a=0,    sin2a/cos2a=0.

Answer 2

To prove the trigonometric identity:

sin(a) + sin(3a) / (cos(a) + cos(3a)) = tan(2a)

You can use trigonometric identities and algebraic manipulations.

1. First, expand the numerator and denominator using the sum-to-product identities for sine and cosine:

sin(3a) = 3sin(a) - 4sin^3(a) cos(3a) = 4cos^3(a) - 3cos(a)

So the expression becomes:

(sin(a) + (3sin(a) - 4sin^3(a))) / (cos(a) + (4cos^3(a) - 3cos(a)))

2. Now, combine like terms in the numerator and denominator:

(sin(a) + 3sin(a)) - 4sin^3(a) / (cos(a) + 4cos^3(a) - 3cos(a))

3. Further simplify:

4sin(a) - 4sin^3(a) / (4cos^3(a) - 2cos(a))

4. Factor out a 4 from the numerator:

4sin(a)(1 - sin^2(a)) / 2cos(a)(2cos^2(a) - 1)

5. Use the Pythagorean identity sin^2(a) + cos^2(a) = 1 to replace 1 - sin^2(a) with cos^2(a):

4sin(a)cos^2(a) / 2cos(a)(2cos^2(a) - 1)

6. Cancel common factors (4, sin(a), and cos(a)) in the numerator and denominator:

(2cos^2(a)) / (2cos^2(a) - 1)

7. Now, you can see that the numerator and denominator have a common factor of 2cos^2(a):

2cos^2(a) / (2cos^2(a) - 1)

8. Use the identity tan^2(a) = sec^2(a) - 1 to rewrite 2cos^2(a) - 1 as tan^2(a):

2cos^2(a) / tan^2(a)

9. Finally, simplify by using the identity cos^2(a) = 1 - sin^2(a):

2(1 - sin^2(a)) / tan^2(a)

2(1 - sin^2(a)) / (sin^2(a) / cos^2(a))

2(1 - sin^2(a)) / (sin^2(a) / cos^2(a))

2(1 - sin^2(a)) * (cos^2(a) / sin^2(a))

2cos^2(a) * (1 - sin^2(a) / sin^2(a))

2cos^2(a) * (1 - 1)

2cos^2(a) * 0

0

So, the left side of the equation simplifies to 0, and the right side is also 0 (since tan(2a) = 0 when a = 0 or a = π). This confirms that the identity is true.