Answer:
(a) 16/65
Explanation:
You want cos(θ-φ) where cos(θ) = -4/5, sin(φ) = -12/13, θ is a 3rd-quadrant angle, and φ is a 4th-quadrant angle.
Sin(θ)
In the 3rd quadrant, sin(θ) = -√(1 -cos(θ)²), so we have ...
sin(θ) = -√(1 -(-4/5)²) = -√((25 -16)/25) = -3/5
Cos(φ)
In the 4th quadrant, cos(φ) = √(1 -sin(φ)²), so we have ...
cos(φ) = √(1 -(-12/13)²) = √((169 -144)/169) = 5/13
Cos(θ-φ)
The cosine of the difference of the angles is given by the identity ...
cos(θ-φ) = cos(θ)cos(φ) +sin(θ)sin(φ)
cos(θ-φ) = (-4/5)(5/13) +(-3/5)(-12/13) = (-20 +36)/65
cos(θ-φ) = 16/65
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Additional comment
In general, these problems make use of the various trig identities relating different functions of the angles. You need to know where the trig functions are positive and where they are negative when computing some functions from others (sine from cosine, for example). In the attachment, functions not shown as positive are negative.
Sometimes a calculator can be of help.
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