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Prove that sin25°+sin35°=sin85°​

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Answer:

See below

Explanation:

To prove the identity
\sf \sin 25^\circ + \sin 35^\circ = \sin 85^\circ, we'll use the sum-to-product trigonometric identity for sine:


\boxed{\boxed{\sf \sin(A) + \sin(B) = 2 \sin\left((A+B)/(2)\right) \cos\left((A-B)/(2)\right) }}

Let
\sf A = 25^\circ and
\sf B = 35^\circ, then substitute into the identity:


\sf \sin 25^\circ + \sin 35^\circ = 2 \sin\left((25^\circ + 35^\circ)/(2)\right) \cos\left((25^\circ - 35^\circ)/(2)\right)

Simplify the expressions inside the sine and cosine functions:


\sf \sin 25^\circ + \sin 35^\circ = 2 \sin 30^\circ \cos(-5^\circ)

Now, simplify further:


\sf \sin 25^\circ + \sin 35^\circ = 2 \left((1)/(2)\right) \cos(-5^\circ)


\sf \sin 25^\circ + \sin 35^\circ = \cos(-5^\circ)

Now, recall that
\sf \cos(-\theta) = \cos(\theta)

So,


\sf \sin 25^\circ + \sin 35^\circ = \cos(5^\circ)

Now, use the fact that
\boxed{\sf \cos(90^\circ - \theta) = \sin(\theta) }:


\sf \sin 25^\circ + \sin 35^\circ = \sin(85^\circ)

Therefore,
\sf \sin 25^\circ + \sin 35^\circ = \sin 85^\circ, and the identity is proved using trigonometric identities.

User Georgy Grigoryev
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