 ## Trigonometrical Ratios

Subject: Optional Maths #### Overview

The ratios which are generally used to calculate the unknown lengths and angles in a right triangle is called trigonometric ratios. Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions. This note has information about trigonometric ratios, its operations, trigonometric expressions and reciprocals relations.

##### Trigonometrical Ratios 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.

#### Operation of Trigonometric Ratios

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 = 6a4x2 + 7x2 = 11x2 2sinθ + 4sinθ = 6sinθ4cos2θ + 7cos2θ = 11cos2θ Subtraction 11z − 6z = 5z8x2− 5x2 = 3x2 11tanθ − 6tanθ = 5tanθ8sec2θ − 5sec2θ = 3sec2θ Multiplication 7a × 4a = 28a25y2× 3y2= 15y4 7sinθ × 4sinθ = 28sin2θ5tan2θ × 4tan2θ = 20tan4θ Division $\frac{q^{5}}{q}$ = q410b ÷ 5b = 2 $\frac{cosθ^{5}}{cosθ}$ = cosθ410secθ ÷ 5secθ = 2

#### Trigonometric Expressions

Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions.

#### Simplification and factorisation

We can simplify or factorise the trigonometric expressions as in algebra. For examples,

1. (a+b) (a−b) = a2− b2
(SinA + CosA) (SinA − CosA) = Sin2A − Cos2A
2. (a+b)2= a2+ 2ab + b2
(SinA + CosA)2= Sin2A + 2SinA.CosA + Cos2A
3. a3+b3= (a+b) (a2−ab+b2)
Sin3A + Cos3A = (SinA + CosA)(Sin2A − 2SinA.CosA + Cos2A)
4. a3−b3= (a+b) (a2+ab+b2)
Sin3A − Cos3A = (SinA − CosA) (Sin2A + SinA.CosA + Cos2A)

#### Relation between the Trigonometric Ratios of an Angle

Simply there are four fundamental relations of trigonometric ratios. They are,

1. Reciprocal Relation
2. Quotient Relation
3. Pythagoras Relation
4. Derived Relation

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,
CA2= AB2 + BC2
or, h2= p2 + b2
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
∴ sin2θ = 1 − cos2θ
cos2θ = 1 − sin2θ
Also,
sinθ = $\sqrt{1-cos^{2}}{θ}$
or, cosθ = $\sqrt{1−sin^{2}}{θ}$

Again,
p2+ b2 = h2
or,h2− p 2= b2
Dividing both sides by b2
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, sec2θ − tan2θ = 1
or, sec2θ = 1 + tan2θ
or, tan2θ = sec2θ − 1
Also,
secθ = $\sqrt{1+tan^{2}}{θ}$
or, tanθ = $\sqrt{sec^{2} θ − 1}$

Again,
p2+ b 2= h2
or, h2− b 2= p2
Dividing both sides by p2
$\frac{h^{2}}{p{2}}$ − $\frac{b^{2}}{p^{2}}$ = $\frac{p^{2}}{p^{2}}$
or, (cosecθ)2− (cotθ)2 = 1
or, cosec2θ − cot2θ = 1
or, cosec2θ = 1 + cot2θ
or, cot2θ = cosec2θ − 1
Also, cotθ = $\sqrt{cosec^{2} θ − 1}$
or, cosecθ = $\sqrt{1+cot^{2}}{θ}$

#### Methods of proving Trigonometric Identities

To prove trigonometric identities, we may follow any one of the following: -

• Begin from the left-hand side (L.H.S.) and deduct it to the right-hand side (R.H.S.), if L.H.S. is more complex.
• Begin from R.H.S. and deduct it to L.H.S., if L.H.S. is complex.
• Reduce both the L.H.S. and the R.H.S. to the same expression if both the expression are complex.
• By transposition or cross multiplication, change the identity into the appropriate form. Then show that the new L.H.S. = the new R.H.S.

#### Conversion of a Trigonometric Ratios

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: -

• Using trigonometric relations i.e Trigonometric formulae.
• Using Pythagoras theorem.
 NOTE: - We can also find remaining ratios if the value of any trigonometric ratio of an angle is given.

#### Trigonometric Ratios of some angles

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

#### Procedures to find the trigonometric ratios

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.

#### Complementary Angles

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.

#### Trigonometric Ratios of 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

##### Things to remember
 (sinθ)2 + (cosθ)2 = 1 sec2θ = 1+tan2θ cosec2θ - cot2θ = 1 sin2θ = 1-cos2θ sec2θ - tan2θ =1 cosec2θ = 1 + cot2θ , sinθ= $\sqrt{1-cos^{2}}{θ}$ tan2θ= sec2θ - 1 cot2θ = cosec2θ - 1 cosθ = $\sqrt{1-sin^{2}}{θ}$ secθ = $\sqrt{1+tan^{2}}{θ}$ cotθ =$\sqrt{cosec^{2} θ-1}$ sec2θ - tan2θ =1 tanθ = $\sqrt{sec^{2} θ-1}$ cosecθ = $\sqrt{1+cot^{2}}{θ}$
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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:

= (3+7+10)sinA

= 20sinA

Solution:

= 18sinθ − 7sinθ

= (18 − 7)sinθ

= 11sinθ

Solution:

= 3tanA × 7tan4A

= 21tan1+4A

= 21tan5A

= (3sinθ + 2cosθ) (3sinθ− 2cosθ)

= (3sinθ)2− (2cosθ)2

= 9sin2θ − 4cos2θ

Solution:

= 12tan7A÷ 3tan3A

= $\frac{12tan^7A}{3tan^3A}$

= 4tan7−3A

= 4tan4A

= sinθ(1− cosθ) + cosθ(1 + sinθ)

= sinθ − sinθ.cosθ + cosθ + sinθ.cosθ

= sinθ + cosθ

= $\frac{(sinA− cosA)^2}{sin^2− cos^2}$

= $\frac{(sinA− cosA)(sinA− cosA)}{(sinA + cosA)(sinA− cosA)}$

= $\frac{sinA− cosA}{sinA + cosA}$

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. = sin2θ + cos2θ.tan2θ

= sin2θ + cos2θ $\frac{sin^2θ}{cos^2θ}$

= sin2θ + sin2θ

= 2sin2θ

= R.H.S. proved

Solution:

L.H.S. = (sinA + cosA)2 + (sinA − cosA)2

= sin2A + 2sinA.cosA + cos2A + sin2A − 2sinA.cosA + cos2A

= (sin2A + cos2A) + (sin2A + cos2A)

= 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.
= $\frac{1− sin^4}{cos4A}$

= $\frac{1− sin^4A}{cos^4A}$ − $\frac{sin^4A}{cos^4A}$

= sec4A − tan4A

= (sec2A)2− (tan2A)2

= (sec2A− tan2A) (sec2A + tan2A)

= (1 + tan2A + tan2A)

= 1 + 2tan2A

= 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

= x2cos2θ + y2sin2θ + x2sin2θ+ y2cos2θ

= x2(cos2θ + sin2θ) + y2(sin2θ + cos2θ)

= (x2+ y2) (cos2θ + sin2θ)

= (x2+ y2) $\times$ 1 [$\because$ (cos2θ + sin2θ) =1]

= (x2+ y2) 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

Solution:

Let, ABC be a right angled triangle where ∠B = 90° and ∠C = θ = k

Let, sinθ = k

Then,

or, sinθ = $\frac{k}{1}$ = $\frac{p}{h}$

∴ p = k, h = 1

Now, from right angled ΔABC,

or, b = $\sqrt{h^2 − p^2}$

or, b = $\sqrt{1 − k^2}$

or, b = $\sqrt{1− sin^2θ}$

Then,

cosecθ = $\frac{h}{p}$ = $\frac{1}{k}$ = $\frac{1}{sinθ}$

cosθ = $\frac{b}{h}$ = $\sqrt\frac{1− sin^2θ}{1}$ = $\sqrt{1−sin^2θ}$

secθ = $\frac{h}{b}$ = $\frac{1}{\sqrt{1− k^2}}$

tanθ = $\frac{p}{b}$ = $\frac{k}{\sqrt{1− k^2}}$ = $\frac{sinθ}{\sqrt{1− sin^2θ}}$

cotθ = $\frac{b}{p}$ = $\frac{\sqrt{1− sin^2θ}}{sinθ}$

Solutions:

Here,

cosα = $\frac{5}{13}$ = $\frac{b}{h}$

Then,

b = 5, h = 13, p= ?

By using Pythagoras theorem, we have,

or, p2= h2− b2

or, p = $\sqrt{h^2 − b^2}$

or, p = $\sqrt{13^2 − 5^2}$

or, p = $\sqrt{169 − 25}$

or, p = $\sqrt{144}$

∴ p = 12

Now,

secα = $\frac{h}{b}$ = $\frac{13}{5}$

sinα = $\frac{p}{h}$ = $\frac{12}{13}$

cosecα = $\frac{h}{p}$ = $\frac{13}{12}$

tanα = $\frac{p}{b}$ = $\frac{12}{5}$

cotα = $\frac{b}{p}$ = $\frac{5}{12}$

Solution:

Here,

or, sinθ − cosθ = 0

or, sinθ = cosθ

or, tanθ = 1

We have,

tanθ = $\frac{p}{b}$ = $\frac{1}{1}$

Then,

p = 1, b = 1

By Pythagoras theorem,

or, h2 = p2 + b2

or, h = $\sqrt{p^2 + b^2}$

or, h = $\sqrt{1^2 +1^2}$

or, h = $\sqrt{1 + 1}$

or, h = $\sqrt{2}$

∴ h = $\sqrt{2}$

Now,

cosecθ = $\frac{h}{p}$ = $\frac{\sqrt{2}}{1}$ = $\sqrt{2}$

sinθ = $\frac{p}{h}$ = $\frac{1}{\sqrt{2}}$

Solution:

Here,

or, sinθ = $\frac{m}{n}$ = $\frac{p}{h}$

Then,

p = m, h = n

We have,

or, b2 = h2− p2

or, b2 = n2− m2

or, b = $\sqrt{n^2 − m^2}$

Now,

or, tanθ = $\frac{p}{b}$

or, tanθ = $\frac{m}{\sqrt{n^2− m^2}}$

or, $\sqrt{n^2− m^2}$ tanθ = m

$\therefore$ L.H.S = R.H.S proved

Solution:

Here,

or, 3cotθ = 4

We have,

or, cotθ = $\frac{b}{p}$ = $\frac{4}{3}$

Then,

b = 4, p = 3, h = ?

By Pythagoras theorem,

or, h2 = p2 + b2

or, h2 = 32 + 42

or, h2 = 9 + 16

or, h2 = $\sqrt{25}$

∴ h = 5

Now,

= $\frac{3cosθ + 2sinθ}{3cosθ− 2sinθ}$

= {3 $(\frac{b}{h}) + 2 (\frac{p}{h})$}÷ {3 $(\frac{b}{h})− 2 (\frac{p}{h})$}

= (3× $\frac{4}{5}$ + 2× $\frac{3}{5}$)÷ (3× $\frac{4}{5}$− 2× $\frac{3}{5}$)

= ($\frac{12}{5} + \frac{6}{5}$)÷ ($\frac{12}{5}− \frac{6}{5}$)

= $\frac{18}{5}$÷ $\frac{6}{5}$

= $\frac{18}{5}$× $\frac{5}{6}$

= 3