You need to group the sines together, such that:

`sin 25^o + sin 35^o = cos 5^o`

Converting the summation into a product, yields:

`2 sin((25^o + 35^o)/2)*cos((25^o - 35^o)/2) = cos 5^o`

`sin(30^o)*cos(-5^o) = (cos 5^o)/2`

`(1/2)*cos 5^o = (cos 5^o)/2`

**Hence, testing if the given expression holds yields **`sin 25^o = cos 5^o - sin 35^o .`

We'll re-write the identity, moving the term sin 35 to the right side, to create a sum of matching functions:

sin 25 + sin 35 = cos 5** **

Supposing that 25,35 and 5 are degrees, we'll transform the sum of matching trigonometric functions into a product.

We'll use the formula:

sin a + sin b = 2sin [(a+b)/2]*cos[(a-b)/2]

According to this formula, we'll obtain:

sin 25 + sin 35 = 2sin [(25+35)/2]*cos[(25-35)/2]

sin 25 + sin 35 = 2sin [(60)/2]*cos[(-10)/2]

sin 25 + sin 35 = 2sin 30*cos(-5)

Since the cosine function is even, we'll get:

sin 25 + sin 35 = 2sin 30*cos(5)

But sin 30 = 1/2

sin 25 + sin 35 = (2/2)*cos(5)

sin 25 + sin 35 = cos 5

** **