How do I prove that #2 sin ((C+D)/2) cos ((C-D)/2) = sin C+sin D#?
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See the proof below
We need
Therefore,
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To prove that (2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) = \sin C + \sin D), we can use the sum-to-product identities for sine and cosine.
The sum-to-product identity for sine states that (2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B)), and for cosine, it's (2\cos(A)\sin(B) = \sin(A+B) - \sin(A-B)).
Let's use the sum-to-product identity for sine on (C+D) and (C-D):
For (C+D): [2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) = \sin\left(\frac{C+D}{2} + \frac{C-D}{2}\right) + \sin\left(\frac{C+D}{2} - \frac{C-D}{2}\right)]
[= \sin\left(\frac{2C}{2}\right) + \sin\left(\frac{2D}{2}\right)] [= \sin C + \sin D]
Therefore, (2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) = \sin C + \sin D), as required.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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