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There are six trigonometric ratios which relate the sides of a right triangle to its angles. Trigonometric ratios are generally used to calculate the unknown lengths and angles in a right triangle. The six trigonometric ratios are tabulated below: -
1) Sinθ | \(\frac{perpendicuar}{hypotenuse}\) (\(\frac{p}{h}\)) |
2) Cosθ | \(\frac{base}{hypotenuse}\) (\(\frac{b}{h}\)) |
3) Tanθ | \(\frac{perpendicuar}{base}\) (\(\frac{p}{b}\)) |
4) Cosecθ | \(\frac{hypotenuse}{perpendicuar}\) (\(\frac{h}{p}\)) |
5) Secθ | \(\frac{hypotenuse}{base}\) (\(\frac{h}{b}\)) |
6) Cotθ | \(\frac{base}{perpendicuar}\) (\(\frac{b}{p}\)) |
NOTE: - Since, Sinθ ≠ Sin×θ, Cosθ ≠ Cos×θ, etc.
As we know that the trigonometric ratios are numbers like fraction, decimal or whole numbers. We can operate the operations of addition, subtraction, multiplication and division on them that we do in algebra.
Operations | In algebra | In Trigonometry |
Addition | 2a + 4a = 6a 4x^{2} + 7x^{2} = 11x^{2} |
2sinθ + 4sinθ = 6sinθ 4cos^{2}θ + 7cos^{2}θ = 11cos^{2}θ |
Subtraction | 11z − 6z = 5z 8x^{2}− 5x^{2} = 3x^{2} |
11tanθ − 6tanθ = 5tanθ 8sec^{2}θ − 5sec^{2}θ = 3sec^{2}θ |
Multiplication | 7a × 4a = 28a^{2}5y^{2}× 3y^{2}= 15y^{4} | 7sinθ × 4sinθ = 28sin^{2}θ 5tan^{2}θ × 4tan^{2}θ = 20tan^{4}θ |
Division | \(\frac{q^{5}}{q}\) = q^{4} 10b ÷ 5b = 2 |
\(\frac{cosθ^{5}}{cosθ}\) = cosθ^{4} 10secθ ÷ 5secθ = 2 |
Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions.
We can simplify or factorise the trigonometric expressions as in algebra. For examples,
Simply there are four fundamental relations of trigonometric ratios. They are,
Reciprocal Relation
Reciprocal relations of trigonometric ratios are explained to represent the relationship between the three pairs of trigonometric ratios as well as their reciprocals. The reciprocals relations are given below: -
Sinθ = \(\frac{p}{h}\) and Cosecθ = \(\frac{h}{p}\)
Then, sinθ × cosecθ = \(\frac{p}{h}\) × \(\frac{h}{p}\) = 1
∴ Sinθ × cosecθ = 1 |
sinθ = \(\frac{1}{cosecθ }\), cosecθ = \(\frac{1}{sinθ }\)
Also, cosθ = \(\frac{b}{h}\) and secθ = \(\frac{h}{b}\)
Then, cosθ × secθ = \(\frac{b}{h}\) × \(\frac{h}{b}\) = 1
∴ cosθ × secθ = 1 |
cosθ = \(\frac{1}{secθ}\), secθ \(\frac{1}{cosθ}\)
And, tanθ = \(\frac{p}{b}\), cotθ = \(\frac{b}{p}\)
Then tanθ × cotθ = 1
∴ tanθ × cotθ = 1 |
tanθ = \(\frac{1}{cotθ}\), cotθ = \(\frac{1}{tanθ}\)
Quotient Relation
We have, sinθ = \(\frac{p}{h}\), cosθ = \(\frac{b}{h}\)
Then, \(\frac{sinθ}{cosθ}\) = \(\frac{p}{h}\) × \(\frac{h}{b}\) = \(\frac{p}{b}\)
We have tanθ = \(\frac{p}{b}\)
∴ tanθ = \(\frac{sinθ}{cosθ}\)
Similarly,
cotθ = \(\frac{cosθ}{sinθ}\)
Pythagoras Relation
From the right angled triangle ABC,
CA^{2}= AB^{2} + BC^{2}
or, h^{2}= p^{2} + b^{2}
Dividing both sides
or, \(\frac{h^{2}}{h^{2}}\) = \(\frac{p^{2}+b^{2}}{h^{2}}\)
or, 1 = \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\)
or, \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\) = 1
or, (\(\frac{p^{2}}{h^{2}}\)) + (\(\frac{b^{2}}{h^{2}}\)) = 1
or, (sinθ)^{2} + (cosθ)^{2} = 1
∴ sin^{2}θ = 1 − cos^{2}θ
cos^{2}θ = 1 − sin^{2}θ
Also,
sinθ = \(\sqrt{1-cos^{2}}{θ}\)
or, cosθ = \(\sqrt{1−sin^{2}}{θ}\)
Again,
p^{2}+ b^{2} = h^{2}
or,h^{2}− p ^{2}= b^{2}
Dividing both sides by b^{2}
or, \(\frac{h^{2}}{b{2}}\) - \(\frac{p^{2}}{b^{2}}\) = \(\frac{b^{2}}{b^{2}}\)
or, (\(\frac{h}{b}\))^{2} + (\(\frac{p}{b}\))^{2}= 1
or, (secθ)^{2}− (tanθ)^{2} = 1
or, sec^{2}θ − tan^{2}θ = 1
or, sec^{2}θ = 1 + tan^{2}θ
or, tan^{2}θ = sec^{2}θ − 1
Also,
secθ = \(\sqrt{1+tan^{2}}{θ}\)
or, tanθ = \(\sqrt{sec^{2} θ − 1}\)
Again,
p^{2}+ b ^{2}= h^{2}
or, h^{2}− b ^{2}= p^{2}
Dividing both sides by p^{2}\(\frac{h^{2}}{p{2}}\) − \(\frac{b^{2}}{p^{2}}\) = \(\frac{p^{2}}{p^{2}}\)
or, (cosecθ)^{2}− (cotθ)^{2} = 1
or, cosec^{2}θ − cot^{2}θ = 1
or, cosec^{2}θ = 1 + cot^{2}θ
or, cot^{2}θ = cosec^{2}θ − 1
Also, cotθ = \(\sqrt{cosec^{2} θ − 1}\)
or, cosecθ = \(\sqrt{1+cot^{2}}{θ}\)
To prove trigonometric identities, we may follow any one of the following: -
With the use of two methods, we can express the trigonometric ratios in terms of other ratios of the same angle. The two methods are as followings: -
NOTE: - We can also find remaining ratios if the value of any trigonometric ratio of an angle is given. |
As trigonometry have different angles, for those different angles trigonometric ratios also have different values. In the trigonometry the angles 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° are taken as standard angles.
We can find the trigonometric ratios of 0° and 90°. The following table shows the values of trigonometric ratios of the angles 0°, 30°, 45°, 60° and 90°.
Angles^{→}↓ | 0° | 30° | 45° | 60° | 90° |
sinθ | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
cosθ | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{1}{2}\) | 0 |
tanθ | 0 | \(\frac{1}{\sqrt3}\) | 1 | \(\sqrt{3}\) | ∞ |
cosecθ | ∞ | 2 | \(\sqrt{2}\) | \(\frac{2}{\sqrt3}\) | 1 |
secθ | 1 | \(\frac{2}{\sqrt3}\) | \(\sqrt{2}\) | 2 | ∞ |
cotθ | ∞ | \(\sqrt{3}\) | 1 | \(\frac{1}{\sqrt3}\) | 0 |
The standard angles i.e 0°, 30°, 45°, 60°, 90° in prompt way is given below: -
Process 1: -
Put the number from 0 to 4 as follows: -
0° | 30° | 45° | 60° | 90° |
0 | 1 | 2 | 3 | 4 |
Process 2: -
Divide each number by 4
0° | 30° | 45° | 60° | 90° |
\(\frac{0}{4}\) | \(\frac{1}{4}\) | \(\frac{2}{4}\) | \(\frac{3}{4}\) | \(\frac{4}{4}\) |
Process 3: -
Take square root of all the numbers
0° | 30° | 45° | 60° | 90° |
\(\sqrt{\frac{0}{4}}\) | \(\sqrt{\frac{1}{4}}\) | \(\sqrt{\frac{2}{4}}\) | \(\sqrt{\frac{3}{4}}\) | \(\sqrt{\frac{4}{4}}\) |
Process 4: -
The values which are obtained at first are the values of sine of the standard angles.
Angles | 0° | 30° | 45° | 60° | 90° |
sin | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
Process 5: -
Then reversing the order, the value of cosine of theb standard angles are obtained.
Angles | 0° | 30° | 45° | 60° | 90° |
cos | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{1}{2}\) | 0 |
Process: -
After dividing the value of sine by the value of cosine, the values of tangent can be obtained: -
Angle | 0° | 30° | 45° | 60° | 90° |
tan | 0 | \(\frac{1}{\sqrt3}\) | 1 | \(\sqrt{3}\) | ∞ |
NOTE: - The reciprocal of sine, cosine and tangent are the values of cosecant, secant and cotangent respectively. |
Two angles are called complementary when those angles are added up to 90° or the sum of two angles is 90°.
For example,
60° and 30° are complementary angles.
5° and 85° are complementary angles.
30° and 60°, 50° and 40°, (90° - θ) and θ are complementary angles.
Here ABC is a right angle triangle right angled at B.
Let, ∠ABC = θ, then ∠CAB = 90° - θ
Now taking θ as an angle of reference,
AB = Perpendicular (p)
BC = Base (b)
CA = Hyptenuse (h)
Sinθ = \(\frac{p}{h}\) = \(\frac{AB}{CA}\) .............. .(i)
Cosθ = \(\frac{b}{h}\) = \(\frac{BC}{CA}\) ..............(ii)
Tanθ = \(\frac{p}{b}\) = \(\frac{AB}{BC}\) ..............(iii)
Cotθ = \(\frac{b}{p}\) = \(\frac{BC}{AB}\) ...............(iv)
Taking (90° - θ) as an angle of reference,
BC = Perpendicular (p)
AB = Base (b)
CA = Hypotenuse (h)
Now,
Sinθ(90° - θ) = \(\frac{p}{h}\) = \(\frac{BC}{CA}\) = Cosθ
Cosθ(90° - θ) = \(\frac{b}{h}\) = \(\frac{AB}{CA}\) = Sinθ
Tanθ(90° - θ) = \(\frac{p}{b}\) = \(\frac{BC}{AB}\) = Cotθ
On the other hand,
cosec(90° - θ) = secθ
sec(90° - θ) = cosecθ
cot(90° - θ) = tanθ
(sinθ)^{2} + (cosθ)^{2} = 1 | sec^{2}θ = 1+tan^{2}θ | cosec^{2}θ - cot^{2}θ = 1 |
sin^{2}θ = 1-cos^{2}θ | sec^{2}θ - tan^{2}θ =1 | cosec^{2}θ = 1 + cot^{2}θ |
, sinθ= \(\sqrt{1-cos^{2}}{θ}\) | tan^{2}θ= sec^{2}θ - 1 | cot^{2}θ = cosec^{2}θ - 1 |
cosθ = \(\sqrt{1-sin^{2}}{θ}\) |
secθ = \(\sqrt{1+tan^{2}}{θ}\) |
cotθ =\(\sqrt{cosec^{2} θ-1}\) |
sec^{2}θ - tan^{2}θ =1 | tanθ = \(\sqrt{sec^{2} θ-1}\) | cosecθ = \(\sqrt{1+cot^{2}}{θ}\) |
Solution:
From the given figure,
Hypotenuse (h) = 13cm
Perpendicular (p) = 5cm
Base (b) = 12cm
Now,
\(\therefore\) sinα = \(\frac{p}{h}\) = \(\frac{5}{13}\)
\(\therefore\) cosα = \(\frac{b}{h}\) = \(\frac{12}{13}\)
\(\therefore\) tanα = \(\frac{p}{b}\) = \(\frac{5}{12}\)
Solution:
From the given figure,
Hypotenuse (h) = 10cm
Perpendicular (p) = 6cm
Base (b) = 8cm
Now,
\(\therefore\) cosecθ = \(\frac{h}{p}\) = \(\frac{10}{6}\) = \(\frac{5}{3}\)
\(\therefore\) secθ = \(\frac{h}{b}\) = \(\frac{10}{8}\) = \(\frac{5}{4}\)
\(\therefore\) cotθ = \(\frac{b}{p}\) = \(\frac{8}{6}\) = \(\frac{4}{3}\)
Solution:
= 18sinθ − 7sinθ
= (18 − 7)sinθ
= 11sinθ
Solution:
L.H.S. = \(\frac{sinθ.cosecθ}{secθ}\)
= sinθ cosecθ÷ secθ
= sinθ× \(\frac{1}{sinθ}\)÷ \(\frac{1}{cosθ}\)
= sinθ× \(\frac{1}{sinθ}\)× cosθ
= cosθ
= R.H.S. proved.
Solution:
L.H.S. = secθ \(\sqrt{1− cos^2θ}\)
= \(\frac{1}{cosθ}\) . sinθ
= \(\frac{sinθ}{cosθ}\)
= tanθ
= R.H.S. proved.
Solution:
L.H.S. = sin^{2}θ + cos^{2}θ.tan^{2}θ
= sin^{2}θ + cos^{2}θ \(\frac{sin^2θ}{cos^2θ}\)
= sin^{2}θ + sin^{2}θ
= 2sin^{2}θ
= R.H.S. proved
Solution:
L.H.S. = (sinA + cosA)^{2} + (sinA − cosA)^{2}
= sin^{2}A + 2sinA.cosA + cos^{2}A + sin^{2}A − 2sinA.cosA + cos^{2}A
= (sin^{2}A + cos^{2}A) + (sin^{2}A + cos^{2}A)
= 1 + 1
= 2
= R.H.S. proved
Solution:
L.H.S. = \(\frac{1}{tanA + cotA}\)
= \(\frac {1}{\frac {sinA}{cosA} + \frac {cosA}{sinA}}\)
= \(\frac{1}{\frac{sin^2A + cos^2A}{sinA.cosA}}\)
= \(\frac{1}{\frac{1}{sinA.cosA}}\)
= sinA.cosA
= R.H.S. proved
Solution:
L.H.S. = \(\frac{sin^4θ− cos^4θ}{sinθ + cosθ}\)
= \(\frac{(sin^2θ)^2 − (cos^2θ)^2}{sinθ + cosθ}\)
= \(\frac{(sin^2θ + cos^2θ) (sin^2θ− cos^2θ)}{sinθ + cosθ}\)
= \(\frac{1× (sinθ + cosθ) (sinθ− cosθ)}{sinθ + cosθ}\)
= sinθ− cosθ
= R.H.S. proved
Solution:
L.H.S. = \(\frac{cosA}{1 + sinA} + \frac{cosA}{1− sinA}\)
= \(\frac{cosA (1− sinA) + cosA (1 + sinA)}{(1 + sinA) (1− sinA)}\)
= \(\frac{cosA− sinA.cosA + sinA.cosA}{1− sin^2A}\)
= \(\frac{2cosA}{cos^2A}\)
= \(\frac{2}{cosA}\)
= 2secA
= R.H.S. proved
Solution:
L.H.S.
= \(\frac{1− sin^4}{cos4A}\)
= \(\frac{1− sin^4A}{cos^4A}\) − \(\frac{sin^4A}{cos^4A}\)
= sec^{4}A − tan^{4}A
= (sec^{2}A)^{2}− (tan^{2}A)^{2}
= (sec^{2}A− tan^{2}A) (sec^{2}A + tan^{2}A)
= (1 + tan^{2}A + tan^{2}A)
= 1 + 2tan^{2}A
= R.H.S. proved
Solution:
L.H.S
= (xcosθ + ysinθ)^{2}+ (xsinθ− yccosθ)^{2}
= (xcosθ)^{2}+ 2xysinθ.cosθ + (ysinθ)^{2}+ (xsinθ)^{2}− 2xysinθ.cosθ + (ycosθ)^{2}
= x^{2}cos^{2}θ + y^{2}sin^{2}θ + x^{2}sin^{2}θ+ y^{2}cos^{2}θ
= x^{2}(cos^{2}θ + sin^{2}θ) + y^{2}(sin^{2}θ + cos^{2}θ)
= (x^{2}+ y^{2}) (cos^{2}θ + sin^{2}θ)
= (x^{2}+ y^{2}) \(\times\) 1 [\(\because\) (cos^{2}θ + sin^{2}θ) =1]
= (x^{2}+ y^{2}) R.H.S. proved
Solution:
L.H.S. =\(\sqrt\frac{1 + cosθ}{1 – cosθ}\)
= \(\sqrt{\frac{1 + cosθ}{1− cosθ}× \frac{1 + cosθ}{1− cosθ}}\)
= \(\sqrt\frac{(1 + cosθ)^2}{1− cos^2θ}\)
= \(\sqrt\frac{(1 + cosθ)^2}{sin^2θ}\)
= \(\frac{1 + cosθ}{sinθ}\)
= \(\frac{1}{sinθ} + \frac{cosθ}{sinθ}\)
= cosecθ + cotθ
= R.H.S. proved
sinθ is equal to ______.
cosθ is equal to ________.
tanθ is equal to _________.
cosecθ is equal to ______.
secθ is equal to _________.
cotθ is equal to ________.
Solve for x.
(frac{15}{18})=(frac{x}{6})
Solve for x.
(frac{12}{z})=(frac{8}{6})
Solve for y.
(frac{1}{3})=(frac{y}{24})
Find the value of trigonometrical ratios of sin ( heta), if tan ( heta) = (frac{3}{4})
Find the value of trigonometrical ratios of sin ( heta), if cosec ( heta) = 2
Find the value of trigonometrical ratios of cot ( heta), if sec ( heta) = 2
Solve for u.
(frac{u}{8})=(frac{27}{12})
While researching universities, Jack learns that a typical class at Clarksville Technical University has 15 boys and 9 girls, and a typical class at Lawrence County University has 12boys and 4 girls. Which university has classes with a lower ratio of boy to girl?
An auto dealership sells minivans and sedans. In January, they sold 18 minivans and 22 sedans. In February, they sold 9 minivans and 11 sedans. During which month did the auto dealership sell a lower ratio of minivans to sedans?
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