Question

Sin7a+sin5a\cos5a-cos7a=

Answer 1

$\frac{Sin7 \alpha +Sin5 \alpha }{Cos5 \alpha -Cos7 \alpha } = \frac{2Sin \frac{7 \alpha +5 \alpha }{2}Cos \frac{7 \alpha -5 \alpha }{2} }{-2Sin \frac{5 \alpha +7 \alpha }{2}Sin \frac{5 \alpha -7 \alpha }{2} }= \frac{Sin6 \alpha Cos \alpha }{Sin6 \alpha Sin \alpha } =Ctg \alpha$

Answer 2

Sin7a+sin5a /cos5a-cos7a= 2 sin6acosa / -2 sin(-a)sin6a=cosa / sina)= ctga

Answer 3

Let's simplify the given expression step by step:

1. Recall the trigonometric identities:

   • Sum-to-Product: $\sin(A) + \sin(B) = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
   • Cosine Double-Angle: $\cos(2C) = 1 - 2 \sin^2(C)$

2. Apply the Sum-to-Product identity to the first two terms: $\sin(7a) + \sin(5a) = 2 \sin \left( \frac{7a+5a}{2} \right) \cos \left( \frac{7a-5a}{2} \right) = 2 \sin(6a) \cos(a)$

3. Substitute the result into the expression: $2 \sin(6a) \cos(a) \cos(5a) - \cos(7a)$

4. Apply the Cosine Double-Angle identity to the term $\cos(2C)$: $\cos(2 \cdot 3.5a) = 1 - 2 \sin^2(3.5a)$

5. Substitute this back into the expression: $2 \sin(6a) \cos(a) \cos(5a) - (1 - 2 \sin^2(3.5a))$

6. Distribute the negative sign: $2 \sin(6a) \cos(a) \cos(5a) - 1 + 2 \sin^2(3.5a)$

So, the simplified expression is: $2 \sin(6a) \cos(a) \cos(5a) - 1 + 2 \sin^2(3.5a)$