Question

Sin²A-1/1-cos²A=? 2sin a/b cos a/b=? sin(П-a)+cos(3/2 П+a)=?

Answer 1

Ответ:

Объяснение:

(sin²A-1)/(1-cos²A)=-cos²A/sin²A=-(cosA/sinA)²=-ctg²A

2sin a/b cos a/b=sin(2·a/b)=sin(2a/b)

sin(π-a)+cos(3/2π+a)=sina+sina=2sina

Answer 2

Let's solve each of the trigonometric expressions step by step:

1. To simplify sin²A - 1 / (1 - cos²A):

First, we know the Pythagorean identity: sin²A + cos²A = 1

Rearrange it to find cos²A = 1 - sin²A

Now substitute this into the original expression:

sin²A - 1 / (1 - (1 - sin²A))

Now simplify further:

sin²A - 1 / sin²A

To simplify the expression, we can find a common denominator:

(sin²A * sin²A - 1) / sin²A

Now, using the difference of squares, we get:

(sin^4A - 1) / sin²A

So, sin²A - 1 / (1 - cos²A) simplifies to: (sin^4A - 1) / sin²A

2. To find 2sin(A/B)cos(A/B):

Using the double angle identity: sin(2θ) = 2sin(θ)cos(θ)

We can see that A/B is analogous to θ in this case. So the expression becomes:

2sin(A/B)cos(A/B) = sin(2 * A/B)

3. To simplify sin(π - A) + cos(3/2 * π + A):

4. sin(π - A) = sin(π)cos(A) - cos(π)sin(A) Since sin(π) = 0 and cos(π) = -1: sin(π - A) = -cos(A)

5. cos(3/2 * π + A) = cos(π + A/2) = -sin(A/2) (Using the cosine half-angle identity)

Now substitute the values back into the original expression:

-sin(A) - sin(A/2)

Please note that I simplified the expressions using basic trigonometric identities. If you need numerical results, make sure to provide specific values for A and B.