Question

Prove that
sin 70°-cos 40°|
cos 50°-sin 20°|
1lrout+3​

Answer 1

Answer:

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

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

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

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

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