Trigonometry
Class-9-Mathematics-2-Chapter-8-Maharashtra Board
Notes
Topics to be learn :
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Introduction to Trigonometry :
- Trigonometry is a branch of mathematics that measures distances and angles in triangles.
- It is useful in fields like Engineering, Astronomy, and Navigation.
- The term 'Trigonometry' comes from the Greek words 'Tri' meaning three, 'gona' meaning sides, and 'metron' meaning measurement.
- It involves determining the remaining angles and sides of a triangle by comparing the angles and sides given.
- The study of trigonometry is crucial in various fields like engineering, astronomy, and navigation.
Terms related to right angled triangle :
In right angled Δ ABC, ∠ B = 90°, ∠ A and ∠ C are acute angles.
Trigonometric ratios :
In Δ ABC, (below fig.)
(i) \(\frac{AC}{AB}=\frac{\text{Opposite side of ∠ B}}{Hypotenuse}\)
This ratio is called the 'sine' ratio of ∠ B and is written in brief as sin B.
(ii) \(\frac{BC}{AB}=\frac{\text{Adjacent side of ∠ B}}{Hypotenuse}\)
This ratio is called the 'cosine' ratio of ∠ B and is written in brief as cos B.
(iii) \(\frac{AC}{BC}=\frac{\text{Opposite side of ∠ B}}{\text{Adjacent side of ∠ B}}\)
This ratio is called the 'tangent' ratio of ∠ B and is written in brief as tan B.
Sometimes we write measures of acute angles of a right angled triangle by using Greek letters θ (Theta), α (Alpha), β (Beta) etc.
In the below figure of Δ ABC, measure of acute angle C is denoted by the letter θ.
So we can write the ratios sin C, cos C, tan C as sin θ, cos θ, tan θ respectively.
sin C = sin θ = \(\frac{AB}{AC}\),
cos C = cos θ = \(\frac{BC}{AC}\),
tan C = tan θ = \(\frac{AB}{BC}\),
Remember :
sin ratio = \(\frac{\text{Opposite side}}{Hypotenuse}\) sin θ = \(\frac{\text{Opposite side of ∠ θ}}{Hypotenuse}\)
cos ratio = \(\frac{\text{Adjacent side}}{Hypotenuse}\) cos θ = \(\frac{\text{Adjacent side of ∠ θ}}{Hypotenuse}\)
tan ratio = \(\frac{\text{Opposite side}}{\text{Adjacent side}}\) tan θ = \(\frac{\text{Opposite side of ∠ θ}}{\text{Adjacent side of ∠ θ}}\)
Relation among trigonometric ratios :
cos (90 - θ) = sin θ,
sin (90 - θ) = cos θ
\(\frac{ sin\,θ}{cos\,θ}\) = tan θ,
tan θ × tan (90 - θ) = 1
cosec θ, sec θ and cot θ are inverse ratios of sin θ, cos θ and tan θ respectively.
cosec θ = \(\frac{1}{sin\,θ}\), sec θ = \(\frac{1}{cos\,θ}\), cot θ = \(\frac{ cos\,θ}{sin\,θ}\)
sec θ = cosec (90 - θ), cosec θ = sec (90 - θ),
tan θ = cot (90 - θ), cot θ = tan (90 - θ)
Trigonometric ratios of 30° and 60° angles :
Theorem of 30°- 60°-90° triangle :
We know that if the measures of angles of a triangle are 30°,60°, 90° then side opposite to 30° angle is half of the hypotenuse and side opposite to 60° angle is \(\frac{\sqrt{3}}{2}\) of hypotenuse.
In the Fig. Δ ABC is a right angled triangle.
∠ C = 30°, ∠ A = 60°, ∠ B = 90°.
∴ AB = \(\frac{1}{2}\) AC and BC = \(\frac{\sqrt{3}}{2}\) AC
Measures of angles →
……………. Ratios ↓ |
0° |
30° |
45° |
60° |
90° |
sin | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
cos | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{2}\) | 0 |
tan | 0 | \(\frac{1}{\sqrt{3}}\) | 1 | \(\sqrt{3}\) | Undefined |
Prove : (sin θ)2 + (cos θ)2 = 1.
Proof :
Δ PQR is a right angled triangle.
∠ PQR = 90°, ∠ R = θ
sin θ = \(\frac{PQ}{PR}\) ..........(I)
and cos θ = \(\frac{QR}{PR}\) ..........(II)
Using Pythagoras’ theorem,
PQ2 + QR2 = PR2
∴ \(\frac{(PQ)^2}{(PR)^2}+\frac{(QR)^2}{(PR)^2}=\frac{(PR)^2}{(PR)^2}\) .... dividing each term by PR2
∴ \(\frac{(PQ)^2}{(PR)^2}+\frac{(QR)^2}{(PR)^2}\) = 1
∴ (sin θ)2 + (cos θ)2 = 1 …..from (I) & (II)
Important Equation in Trigonometry :
(sin θ)2 + (cos θ)2 = 1.
(sin θ)2 means square of sin θ. It is written as sin2 θ.
∴ sin2 θ + cos2 θ = 1.
This equation is true even when θ = 0° or θ = 90°.
(i) 0≤ sin θ ≤ 1,0 ≤ sin2 θ ≤ 1
(ii) 0≤ cos θ ≤ 1,0 ≤ cos2 θ ≤ 1.
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