cos(a-b)cos(b)-sin(a-b)sin(b)

Question

i dont have the greek symbols but what would this be simplified using only one expression???

Answer 1

We can use the Angle Sum and Difference Formulas to help solve this problem. The key Angle Sum and Difference Formulas are:

1. cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)
2. cos(a - b) = cos(a)*cos(b) + sin(a)*sin(b)
3. sin(a + b) = sin(a)*cos(b) + cos(a)*sin(b)
4. sin(a - b) = sin(a)*cos(b) - cos(a)*sin(b)

We can replace cos(a-b) and sin(a - b) in your given problem such that

cos(a - b)*cos(b) - sin(a - b)*sin(b)

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

Distributing the cos(b) to the first set of parenthesis and the sin(b) to the second set of parenthesis, we have:

cos(a)*cos²(b) + sin(a)*sin(b)*cos(b) - [sin(a)*sin(b)*cos(b) - cos(a)*sin²(b)]

We still need to distribute the negative, doing so gives us:

cos(a)*cos²(b) + sin(a)*sin(b)*cos(b) - sin(a)*sin(b)*cos(b) + cos(a)*sin²(b)

The sin(a)*sin(b)*cos(b) - sin(a)*sin(b)*cos(b) cancels out to 0. This leaves us with:

cos(a)*cos²(b) + cos(a)*sin²(b)

Both terms have cos(a) in common, so we can factor the cos(a) to the outside:

cos(a)*[cos²(b) + sin²(b)]

The Pythagorean Identity tells us that cos²(b) + sin²(b) = 1, by substituting 1 in for cos²(b) + sin²(b), we have:

cos(a) * 1

Which we can write as cos(a).

Therefore,

cos(a - b)*cos(b) - sin(a - b)*sin(b) = cos(a)

I hope this helps! Please let me know if you have any other questions.

Answer 2

Substitute C=A-B into cos(A-B)cos(B)-sin(A-B)sin(B) results in cos(C)cos(B)-sin(C)sin(B) which is equivalent to cos(C+B) by the trig identity for the cosine of a sum of two angles. Now substitute A-B for C into the last expression producing cos(A-B+B) which equals cos(A).