Question

Cos(п/3+b)+cos(п/3-b),если cosb=2

Answer 1

cosb не может быть равен 2, т.к. принадлежит промежутку [-1;1]

Answer 2

To solve this expression, we can use the given information that cos(b) = 2.

Substituting cos(b) = 2 into the expression, we have:

cos(π/3 + b) + cos(π/3 - b)

Using the cosine addition formula, which states that cos(A + B) = cos(A)cos(B) - sin(A)sin(B), we can rewrite the expression as:

[cos(π/3)cos(b) - sin(π/3)sin(b)] + [cos(π/3)cos(-b) - sin(π/3)sin(-b)]

Since cos(-b) = cos(b) and sin(-b) = -sin(b), we can simplify further:

[cos(π/3)cos(b) - sin(π/3)sin(b)] + [cos(π/3)cos(b) + sin(π/3)sin(b)]

Since cos(π/3) = 1/2 and sin(π/3) = √3/2, we substitute these values:

[(1/2)(2) - (√3/2)(sin(b))] + [(1/2)(2) + (√3/2)(sin(b))]

Simplifying, we have:

[1 - (√3/2)(sin(b))] + [1 + (√3/2)(sin(b))]

The terms with sin(b) cancel out:

1 + 1

Finally, the expression simplifies to:

2