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If sin A = -513,π<A<3π2 and cos B = 35,3π2<B<2π find cos (A – B) - Mathematics and Statistics

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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)

योग
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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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Trigonometric Functions of Sum and Difference of Angles
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अध्याय 3: Trigonometry - 2 - Exercise 3.1 [पृष्ठ ४०]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 3 Trigonometry - 2
Exercise 3.1 | Q 3. (ii) | पृष्ठ ४०

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