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प्रश्न
If sin A = `(-5)/13, pi < "A" < (3pi)/2` and cos B = `3/5, (3pi)/2 < "B" < 2pi` find cos (A – B)
उत्तर
Given, sin A = `(-5)/13`
We know that,
cos2A = 1 – sin2A = `1 - (-5/13)^2`
= `1 - 25/169`
= `144/169`
∴ cos A = `±12/13`
Since, `pi < "A" < (3pi)/2`
∴ ‘A’ lies in the 3rd quadrant
∴ cos A < 0
∴ cos A = `(-12)/13`
Also, cos B = `3/5`
∴ sin2B = 1 – cos2B = `1 - (3/5)^2`
= `1 - 9/25`
= `16/25`
∴ sin B = `±4/5`
Since, `(3pi)/2 < "B" < 2pi`
∴ ‘B’ lies in the 4th quadrant.
∴ sin B < 0
∴ sin B = `(-4)/5`
cos(A – B) = cosA cosB +sinA sinB
= `(-12/13)(3/5)+(-5/13)(-4/5)`
= `-36/65+20/65`
= `-16/65`
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