Question

Prove that sin25°+sin35°=sin85°​

Answer 1

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.