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Trigonometric ratios for 45 degrees

Introduction to Trigonometry I - Concepts

Identifying Hypotenuses In A Right Triangle

Identifying Adjacent Side To An Angle

Identifying Opposite Side To An Angle

Trigonometric Ratios

Reciprocal Trigonometric Ratios

Trigonometric ratios for 30 degrees

Trigonometric ratios for 45 degrees

Trigonometric ratios for 60 degrees

Trigonometric ratios for 90 degrees

Simplifying Trigonometric Expressions

Class - 10th CBSE Subjects

Concept Explanation

Trigonometric ratios for 45 degrees

Trigonometric Ratios for 45 Degrees: Value for the Trigonometrical Ratios for certain angle such as $30^0$ , $60^0$ , $90^0$ and $45^0$ are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles. In this section we will derive the value of trigonometric ratios for $45^0$ .

Let $Delta ABC$ be a right - angled at B and let angle A is $45^0$

Now $angle B=90^o ; and ; angle A=45^o.$ Then, clearly, $angle C=45^o.$ [ Angle Sum Property]

$angle A=angle C ; Rightarrow ;AB=BC$ [sides opposite equal angles]

Let AB = BC = a units.

By the Pythagorean theoram

$AC=sqrt {AB^2+BC^2}; ; units ;; =sqrt {a^2+a^2}$

$AC=sqrt {2a^2} = sqrt 2; a$

Thus, Adjacent = AB = a units, Opposite = BC = a units and hypotenuse AC = $sqrt 2;$ a units.

Thus, we have

$sin ; 45^o =frac{Opposite}{Hypotenuse}=frac {BC}{AC}= frac {a}{sqrt {2}a}=frac {1}{sqrt {2}}$

$cos ; 45^o =frac{Adjacent}{Hypoteneuse}=frac {AB}{AC}=frac {a}{sqrt {2}a}=frac {1}{sqrt {2}}$

$tan ; 45^o =frac{Opposite}{Adjacent}=frac {BC}{AB}=frac {a}{a}=1$

$cosec ; 45^o=frac {1}{sin ; 45^o}=sqrt {2}$

$sec ; 45^o =frac {1}{cos; 45^o}=sqrt {2}$

$cot ; 45^o=frac {1}{tan ; 45^o}=1$

$large angle A$	sin A	cos A	tan A	cosec A	sec A	cot A
$large 45^{circ}$	$large frac{1}{sqrt{2}}$	$large frac{1}{sqrt{2}}$	1	$large {sqrt{2}}$	$large {sqrt{2}}$	$large 1$

Illustration: Simplify the given expression $sin ;45^0 + cos ;45^0;(1+ sin; 45^0;cos;45^0)$

Solution: To solve this expression we will substitute the value of the ratios at $45^0$

$sin ;45^0 + cos ;45^0;(1+ sin; 45^0;cos;45^0)$

$=frac{1}{sqrt 2}+frac{1}{sqrt 2}(1+frac{1}{sqrt 2}times frac{1}{sqrt2})$

$=frac{1}{sqrt 2}+frac{1}{sqrt 2}(1+frac{1}{2})$

$=frac{1}{sqrt 2}+frac{1}{sqrt 2}times frac{3}{2}$

$=frac{1}{sqrt 2}times (1+ frac{3}{2})$

$=frac{1}{sqrt 2}times frac{5}{2}$

$=frac{5}{2sqrt 2}$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

Solve the given expression : $sin;45^0+cos;45^0+(2+tan ;45^0+cot;45^0)$
A. $frac{5sqrt2+8}{2}$	B. $frac{2sqrt2+5}{2}$	C. $frac{2sqrt3+8}{3}$	D. $frac{2sqrt2+8}{2}$
Right Option : D
View Explanation

Question : 2

If $sin;theta -cos;theta =0,$ then the value of $(sin^4;theta +cos^4;theta )is$
A. 1	B. $frac {3}{4}$	C. $frac {1}{2}$	D. $frac {1}{4}$
Right Option : C
View Explanation

Question : 3

Solve the following expressions : $frac{5Sin^245^0+Cos^245^0-4Tan^245^0}{2Sin45^0 Cos45^0+Tan45^0}$
A. $large dpi{120}tan : : 90^o$	B. $large dpi{120}1$	C. $large dpi{120}sin: : 45^o$	D. $large dpi{120}sin: : 0^o$
Right Option : B
View Explanation

Chapters

Polynomial I

Polynomials II

Pair of Linear Equations I

Pair of Linear Equations II

Real Numbers I

Real Numbers II

Coordinate Geometry I

Coordinate Geometry II

Quadratic Equations I

Quadratic Equations II

Arithmetic Progression I

Arithmetic Progression II

Introduction to Trigonometry I

Introduction to Trigonometry II

Surface Area and Volume I

Surface Area and Volume II

Areas Related to Circles

Some Applications of Trigonometry

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Trigonometric ratios for 45 degrees

Trigonometric ratios for 45 degrees

Students / Parents Reviews [10]

Archit Segal

Ayan Ghosh

Cheshta

Aman Kumar Shrivastava

Om Umang

Prisha Gupta

Hiya Gupta

Manish Kumar

Yatharthi Sharma

Shreya Shrivastava