cosA (-7/25) pi < A < 3pi/2 sin B= -3/5 3pi/2 <B<2pi
To find the value of cos(A) when -7/25π < A < 3π/2, we need to consider the signs of cos(A) in different quadrants of the unit circle.
First, let's determine the quadrants in which the angle A lies. The given range -7/25π < A < 3π/2 indicates that A is in the third quadrant and extends beyond 3π/2 but does not reach 2π.
In the third quadrant (180° < θ < 270°), cos(θ) is negative. Therefore, cos(A) will also be negative.
Now, we can find the value of cos(A) by using the given value of A in radians (-7/25π) and taking the cosine of that angle. Since cos(A) is negative, we'll add a negative sign in front of the result.
cos(-7/25π) ≈ -cos(7/25π)
To solve for cos(7/25π), we can use a calculator or a trigonometric identity: cos(θ) = cos(θ + 2πn), where "n" is an integer.
cos(7/25π) ≈ cos(2π + 7/25π)
≈ cos(49/25π)
Since the value of cos(2π) is 1, cos(49/25π) will have the same value as cos(π/25).
cos(π/25) ≈ 0.9912.
Therefore, cos(A) ≈ -0.9912.
Moving on to the second part of the question, sin(B) = -3/5 when 3π/2 < B < 2π. Since B lies in the fourth quadrant of the unit circle, where sine is negative, sin(B) will also be negative.
Thus, sin(B) = -3/5.
To summarize:
- cos(A) ≈ -0.9912, where -7/25π < A < 3π/2.
- sin(B) = -3/5, where 3π/2 < B < 2π.