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Find the trigonometric function of: 45° - Mathematics and Statistics

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प्रश्न

Find the trigonometric function of:

45°

आकृति
योग

उत्तर

Trigonometric Functions of 45° :


Let measure of ∠XOA in standard position be 45°.

Its terminal arm (ray OA) intersects the standard unit circle in P (x, y), which lies in the first quadrant.

Draw segment PM perpendicular to the X-axis.

Then OM = | x | and MP = | y |.

Now, ΔOMP is a right-angled triangle in which m∠MOP=45°

∴ OM = MP

∴ | x | = | y |              ...(i)

By the distance formula,

x2 + y2 = 1

∴ x2 + x2 = 1               ...[∵ y = x] [From(i)]

∴ 2x2 = 1

∴ x2 = `1/2`

∴ x = `± 1/sqrt(2); y = ± 1/sqrt(2)`

But P lies in the first quadrant.

∴ x > 0 and y > 0

∴ x = `1/sqrt(2), y = 1/sqrt(2)`

∴ P is `(1/sqrt(2), 1/sqrt(2))`

∴ sin 45° = y = `1/sqrt(2)`

cos 45° = x = `(1/sqrt(2))`

tan 45° = `y/x = (1/sqrt(2))/(1/sqrt(2))` = 1

cosec 45° = `1/y = (1)/(1/sqrt(2)) = sqrt(2)`

sec 45° = `1/x = 1/(1/sqrt(2)) = sqrt(2)`

cot 45° = `x/y = (1/sqrt(2))/(1/sqrt(2))` = 1

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Trigonometric Functions of Specific Angles
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Trigonometry - 1 - EXERCISE 2.1 [पृष्ठ २१]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 2 Trigonometry - 1
EXERCISE 2.1 | Q 1) iii) | पृष्ठ २१
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