Answer:
Explanation:
We can use the identity for the cosine of the sum of two angles:
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
In this case, we can write:
cos(60°) = cos(55° + 5°) = cos(55°)cos(5°) - sin(55°)sin(5°)
We know that cos(60°) = 1/2, and we also know that cos(55°) = p. We can solve for cos(5°):
1/2 = p cos(5°) - sin(55°)sin(5°)
sin(55°)sin(5°) = p cos(5°) - 1/2
We also know that sin²(55°) + cos²(55°) = 1, so:
sin²(55°) = 1 - cos²(55°) = 1 - p²
Substituting this into the previous equation:
(1 - p²)sin(5°)² = p cos(5°) - 1/2
Rearranging and factoring:
p cos(5°) = 1/2 - (1 - p²)sin(5°)²
Using the fact that sin²(5°) + cos²(5°) = 1, we can substitute cos²(5°) = 1 - sin²(5°):
p cos(5°) = 1/2 - (1 - p²)(1 - cos²(5°))
Simplifying:
p cos(5°) = 1/2 - (1 - p²) + (1 - p²)cos²(5°)
p cos(5°) = 1 - 2p² + (2p² - 1)cos²(5°)
Solving for cos²(5°):
cos²(5°) = (1 - p cos(5°) + 2p²) / (2p² - 1)
Substituting the value we found for p, cos²(55°) = p²:
cos²(5°) = (1 - p cos(5°) + 2cos²(55°)) / (2cos²(55°) - 1)
Finally, substituting p = cos(55°):
cos²(5°) = (1 - cos²(55°) cos(5°) + 2cos²(55°)) / (2cos²(55°) - 1)
cos²(5°) = (2cos²(55°) - cos²(55°) cos(5°) + 1) / (2cos²(55°) - 1)
cos²(5°) = (cos²(55°) + 1) / (2cos²(55°) - 1)
Therefore, the value of cos 5° in terms of p = cos 55° is:
cos 5° = ±sqrt[(cos²(55°) + 1) / (2cos²(55°) - 1)]
Note that the sign of the square root is determined by the quadrant in which 5° lies. Since 5° is in the first quadrant, cos 5° is positive.