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Find sin(x – y), given that sin x = 817 with 0 < x < π2, and cos y = -2425, x < y < 3π2

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प्रश्न

Find sin(x – y), given that sin x = `8/17` with 0 < x < `pi/2`, and cos y = `- 24/25`, x < y < `(3pi)/2`

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उत्तर


sin x = `8/17`, 0 < x < `pi/2`

⇒ Where x is in I quadrant

∴ sin x, cos x are +ve

From ΔABX,

AX = `sqrt(17^2 - 8^2)`

= `sqrt((17 + 8)(17 - 8))`

= `sqrt((25)(9))`

= 5 × 3

= 15

∴ sin x = `8/17` and cos x = `15/17`

cos y = `- 24/25, pi < y < (3pi)/2`

⇒ Where y is in III quadrant

So, sin y and cos y are –ve

From ΔALY,

AL = `sqrt(25^2 - 24^2)`

= `sqrt(49)`

= 7

∴ cos y = `- 24/25` and sin y = `- 7/25`

sin(x – y) = sin x cos y – cos x sin y

= `(8/17)(- 24/25) - (15/17)(- 7/25)`

= `- 192/425 + 105/425`

= `- 87/425`

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.4 [पृष्ठ १०९]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.4 | Q 4 | पृष्ठ १०९

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