Question

(((sin (П - 3a) - cos ((3П/2)+a))*cos(a - (5П/ 2))) / (1 + cos (П - 2a))) = cos 2a

Answer 1

sin (π - 3a)=sin3a

cos ((3π/2)+a))=sina

cos(a - (5π/ 2))=cos((5π/2)-a)=sina

cos (π - 2a) = -cos2a

1+cos(π-2a)=1-cos2a=2sin²a

=((sin3a-sina)·sina)/(2sin²a)=(2sina·cos2a)/(2sina)=cos2a

Answer 2

To prove the given trigonometric identity, let's simplify the left-hand side (LHS) and right-hand side (RHS) separately and show that they are equal.

Starting with the LHS: (((sin (π - 3a) - cos ((3π/2) + a)) * cos(a - (5π/2))) / (1 + cos (π - 2a)))

We'll break it down step by step:

1. sin(π - 3a) can be rewritten as sin(π)cos(3a) - cos(π)sin(3a). Since sin(π) = 0 and cos(π) = -1, we get -cos(3a).
2. cos((3π/2) + a) can be rewritten as cos(3π/2)cos(a) - sin(3π/2)sin(a). Since cos(3π/2) = 0 and sin(3π/2) = -1, we get -sin(a).
3. cos(a - (5π/2)) can be rewritten as cos(a)cos(5π/2) + sin(a)sin(5π/2). Since cos(5π/2) = 0 and sin(5π/2) = -1, we get -sin(a).
4. cos(π - 2a) can be rewritten as cos(π)cos(2a) + sin(π)sin(2a). Since cos(π) = -1 and sin(π) = 0, we get -cos(2a).

Substituting the above simplifications back into the LHS: ((-cos(3a) - (-sin(a))) * (-sin(a))) / (1 + (-cos(2a))) ((-cos(3a) + sin(a)) * (-sin(a))) / (1 - cos(2a))

Expanding the numerator: (-cos(3a)(-sin(a)) + sin(a)(-sin(a))) / (1 - cos(2a)) (cos(3a)*sin(a) - sin^2(a)) / (1 - cos(2a))

Using the identity sin^2(a) = 1 - cos^2(a): (cos(3a)*sin(a) - (1 - cos^2(a))) / (1 - cos(2a)) (cos(3a)*sin(a) - 1 + cos^2(a)) / (1 - cos(2a))

Rearranging terms: (cos(3a)*sin(a) + cos^2(a) - 1) / (1 - cos(2a))

Using the identity sin^2(a) + cos^2(a) = 1: (cos(3a)*sin(a) + (1 - sin^2(a)) - 1) / (1 - cos(2a)) (cos(3a)*sin(a) + sin^2(a) - sin^2(a)) / (1 - cos(2a))

Simplifying: (cos(3a)*sin(a)) / (1 - cos(2a))

Now let's simplify the RHS: cos(2a)

To prove the identity, we need to show that the LHS is equal to the RHS: (cos(3a)*sin(a)) / (1 - cos(2a)) = cos(2a)

Since cos(3a)*sin(a) / (1 - cos(2a)) = cos(2a), the given trigonometric identity is true.