Question

Даю 17 баллов 1)sin(a-b)+2cosa*sinb / sin(a+b)= 2)cos(a+b)+2sina*sinb / cos(a-b)= нужна помощь

Answer 1

1)$\frac{sinacosb-cosasinb+2cosasinb}{sin(a+b)}=\frac{sinacosb+cosasinb}{sin(a+b)}= \frac{sin(a+b)}{sin(a+b)}=1$
2)$\frac{cosacosb-sinasinb+2sinasinb}{cos(a-b)}= \frac{cosacosb+sinasinb}{cos(a-b)} = \frac{cos(a-b)}{cos(a-b)}=1$
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Answer 2

1) To solve the equation sin(a-b) + 2cos(a)sin(b) / sin(a+b) = 0, we can use trigonometric identities to simplify the expression. Let's break it down step by step:

First, let's rewrite the equation using the identity sin(x+y) = sin(x)cos(y) + cos(x)sin(y):

sin(a-b) + 2cos(a)sin(b) / sin(a+b) = 0

sin(a)cos(b) - cos(a)sin(b) + 2cos(a)sin(b) / sin(a)cos(b) + cos(a)sin(b) = 0

Next, let's combine like terms:

sin(a)cos(b) + cos(a)sin(b) + 2cos(a)sin(b) = sin(a)cos(b) + cos(a)sin(b)

Now, let's factor out sin(b) from the left side of the equation:

sin(b)(sin(a) + cos(a) + 2cos(a)) = sin(a)cos(b) + cos(a)sin(b)

sin(b)(3cos(a) + sin(a)) = sin(a)cos(b) + cos(a)sin(b)

Now, let's divide both sides of the equation by sin(b):

3cos(a) + sin(a) = cos(a)

Finally, let's rearrange the equation to solve for a:

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

2cos(a) = -sin(a)

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

Now, we can solve for a by taking the inverse cosine of both sides:

a = arccos(-1/2sin(a))

2) To solve the equation cos(a+b) + 2sin(a)sin(b) / cos(a-b) = 0, we can follow a similar process:

cos(a+b) + 2sin(a)sin(b) / cos(a-b) = 0

cos(a)cos(b) - sin(a)sin(b) + 2sin(a)sin(b) / cos(a)cos(b) + sin(a)sin(b) = 0

cos(a)cos(b) + sin(a)sin(b) + 2sin(a)sin(b) = cos(a)cos(b) + sin(a)sin(b)

cos(b)(cos(a) + sin(a) + 2sin(a)) = cos(a)cos(b) + sin(a)sin(b)

cos(b)(3sin(a) + cos(a)) = cos(a)cos(b) + sin(a)sin(b)

Dividing both sides by cos(b):

3sin(a) + cos(a) = sin(a)

2sin(a) = -cos(a)

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

Taking the inverse sine of both sides:

a = arcsin(-1/2cos(a))

Please note that these solutions are based on the given equations and the trigonometric identities used to simplify them. If you have any specific values for a and b, you can substitute them into the equations to find the exact solutions.