# What is another form of cos(30degrees - A) + sin (60degrees + A)?Show all the steps

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You should use the formulas `cos(alpha-beta)= alpha*cos beta + sin alpha*sin beta` and `sin(alpha+beta) = sin alpha*cos beta+sin beta*cos alpha` such that:

`cos(30^o - A) = cos 30^o*cos A + sin 30^o*sin A`

`sin(60^o + A) = sin 60^o*cos A + sin A*cos 60^o`

You need to substitute `1/2` for `sin` `30^o` and `cos 60^o` and `sqrt3/2` for cos `30^o` and sin `60^o` such that:

`cos(30^o - A) = (cos A*sqrt3)/2 + (sin A)/2`

`sin(60^o + A) = (cos A*sqrt3)/2 + (sin A)/2`

You need to add `cos(30^o - A)` and `sin(60^o + A)` such that:

`(cos A*sqrt3)/2 + (sin A)/2 + (cos A*sqrt3)/2 + (sin A)/2 = 2(cos A*sqrt3)/2 + 2(sin A)/2`

`cos(30^o - A) + sin(60^o + A) = cos A*sqrt3 + sin A`

**Hence, evaluating the sum `cos(30^o - A) + sin(60^o + A)` yields `cos(30^o - A) + sin(60^o + A) = cos A*sqrt3 + sin A.` **