Subject: Optional Mathematics

Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs. Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.

Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.

**Trigonometrical Ratio of 45° **

Let ABC be a right-angledisosceles trianglewhere \(\angle\)B = 90° and \(\angle\)A = \(\angle\)C = 45°

Also let BC = AC = \(\alpha\).

We know, in right angled \(\triangle\)ABC

h^{2} = p^{2} + b^{2}

or, (AC)^{2} = (AB)^{2} + (BC)^{2}

or, (AC)^{2} = \(\alpha\)^{2} + \(\alpha\)^{2}

or, (AC)^{2} = 2 \(\alpha\)^{2}

∴ AC = \(\alpha\)\(\sqrt{2}\)

Taking \(\angle\)C as the reference angle, we get,

sin 45 = \(\frac{AB}{AC}\)= \(\frac{α}{α\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)

cos 45 = \(\frac{BC}{AC}\)= \(\frac{α}{α\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)

tan 45 = \(\frac{AB}{BC}\)= \(\frac{α}{α}\) = 1

cosec 45 = \(\frac{AC}{AB}\)= \(\frac{α\sqrt{2}}{α}\) = \(\sqrt{2}\)

sec 45 = \(\frac{AC}{AB}\)= \(\frac{α\sqrt{2}}{α}\) =\(\sqrt{2}\)

cot 45 = \(\frac{BC}{AB}\) = \(\frac{α}{α}\) = 1

Let ABC be an equilateral triangle where\(\angle\)A = \(\angle\)B = \(\angle\)C = 60°

and AB = BC = CA = 2\(\alpha\)

Now, let's draw AD perpendicular to BC so that,

BD = DC = a and \(\angle\)BAD = \(\angle\)DAC = 30°

Now, in right angled \(\triangle\)ADC,

(AC)^{2} = (AD)^{2} + (DC)^{2}

or, (AD)^{2} = (AC)^{2} + (DC)^{2}

= (2α)^{2} - (α)^{2}

= 4\(\alpha\)^{2} - \(\alpha\)^{2} = 3\(\alpha\)^{2}

\(\therefore\) AD = \(\alpha\)\(\sqrt{3}\)

Now, to find the trigonometric ratio for 60°, lets take \(\angle\)C = 60° as the reference angle, we get,

sin 60° = \(\frac{AD}{AC}\) = \(\frac{\alpha\sqrt{3}}{2α}\) = \(\frac{\sqrt{3}}{2}\)

cos 60° = \(\frac{DC}{AC}\)= \(\frac{α}{2α}\) =\(\frac{1}{2}\)

tan 60° = \(\frac{AD}{DC}\)=\(\frac{\alpha\sqrt{3}}{2α}\) = \(\sqrt{3}\)

cosec 60° = \(\frac{AC}{AD}\)= \(\frac{2α}{\alpha\sqrt{3}}\) = \(\frac{2}{\sqrt{3}}\)

sec 60° = \(\frac{AC}{DC}\)= \(\frac{2α}{α}\) = 2

cot 60° = \(\frac{DC}{AD}\)= \(\frac{α}{\alpha\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)

Now, to find out the trigonometric ratio for 30° , let's take \(\angle\)DAC = 30° as the reference angle, we get,

sin 30° =\(\frac{DC}{AC}\)= \(\frac{α}{2α}\) =\(\frac{1}{2}\)

cos 30° = \(\frac{AD}{AC}\) = \(\frac{\alpha\sqrt{3}}{2α}\) = \(\frac{\sqrt{3}}{2}\)

tan 30° =\(\frac{DC}{AD}\)= \(\frac{α}{\alpha\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)

cosec 30° =\(\frac{AC}{DC}\)= \(\frac{2α}{α}\) = 2

sec 30° = \(\frac{AC}{AD}\)= \(\frac{2α}{\alpha\sqrt{3}}\) = \(\frac{2}{\sqrt{3}}\)

cot 30° =\(\frac{AD}{DC}\)=\(\frac{\alpha\sqrt{3}}{2α}\) = \(\sqrt{3}\)

**Trigonometrical Ratio of 0°**

Let ABC be a right angled triangle where \(\angle\)B = 90° and \(\angle\)C = \(\theta\).

If \(\theta\) tends to 0°

i.e \(\theta\)→0°, AC coincides with BC.

or, AC \(\approx\)BC

so, AC= BC = a (say)

Now, using pythagoras theorem

h^{2}= p^{2}+ b^{2}

or, (AC)^{2}=(AB)^{2}+ (BC)^{2}

or, a^{2} = (AB)^{2} + a^{2}

or, (AB)^{2} = a^{2} - a^{2} = 0

\(\therefore\) AB = 0

Now, taking right angled \(\triangle\)ABC

sin 0° = \(\frac{p}{h}\) =\(\frac{AB}{AC}\) =\(\frac{0}{a}\) = 0

cos 0° =\(\frac{b}{h}\) = \(\frac{BC}{AC}\) = \(\frac{a}{a}\) = 1

tan 0° =\(\frac{p}{b}\) = \(\frac{AB}{BC}\) = \(\frac{0}{a}\) = 0

cosec 0° =\(\frac{h}{p}\) = \(\frac{AC}{AB}\) = \(\frac{a}{0}\) = \(\infty\)

sec 0° =\(\frac{h}{b}\) = \(\frac{AC}{BC}\) = \(\frac{a}{a}\) = 1

cot 0° =\(\frac{b}{p}\) = \(\frac{BC}{AB}\) = \(\frac{a}{0}\) = \(\infty\)

**Trigonometric Ratio of 90°**

Let ABC be a right angled triangle where \(\angle\)B = 90° and \(\angle\)C = \(\theta\) be the reference angle.

If \(\theta\) tends to 90°

i.e \(\theta\)→90°, AC coincides with BC.

or, AC \(\approx\)AB

so, AC= AB = a (say)

Now, using pythagoras theorem

h^{2}= p^{2}+ b^{2}

or, (AC)^{2}=(AB)^{2}+ (BC)^{2}

or, a^{2} = (BC)^{2} + a^{2}

or, (BC)^{2} = a^{2} - a^{2} = 0

\(\therefore\) BC = 0

Now, taking right angled \(\triangle\)ABC

sin 90° = \(\frac{p}{h}\) =\(\frac{AB}{AC}\) =\(\frac{a}{a}\) =1

cos 90° =\(\frac{b}{h}\) = \(\frac{BC}{AC}\) = \(\frac{0}{a}\) = 0

tan 90° =\(\frac{p}{b}\) = \(\frac{AB}{BC}\) = \(\frac{a}{0}\) =\(\infty\)

cosec 90° =\(\frac{h}{p}\) = \(\frac{AC}{AB}\) = \(\frac{a}{a}\) =1

sec 90° =\(\frac{h}{b}\) = \(\frac{AC}{BC}\) = \(\frac{a}{0}\) = \(\infty\)

cot 90° =\(\frac{b}{p}\) = \(\frac{BC}{AB}\) = \(\frac{0}{a}\) =0

Two angles are Complementary when they add up to 90 degrees.

Let'a see the following example,

What is the angle added to 60° and 90°?

Let the angle be x.

Then, x + 60° = 90°

or, x = 90° - 60° = 30°

Hence, 30° and 60° when added together gives us 90°, So, 30° and 60° are called complements of each other.

The angles are said to be complementary if the sum of the angles is 90°.

**Complementary Angles in Trigonometry**

Let ABC be a right angledtriangle where \(\angle\)ABC = 90° and \(\angle\)ACB = \(\theta\)

Now, we know,

\(\angle\)ACB +\(\angle\)ABC +\(\angle\)BAC = 180° (sum of angles of a \(\triangle\))

or, \(\theta\) + 90° + \(\angle\)BAC = 180°

or, \(\angle\)BAC = 180° - 90° - \(\theta\) = 90° - \(\theta\)

So, \(\angle\)ACB and\(\angle\)CAB are complementary angles.

Now, taking\(\angle\)A = 90° - \(\theta\) as the reference angle, we get,

BC = perpendicular (p)

AC = hypotenuse (h)

AB = base (b)

So,

sin(90° - \(\theta\)) = \(\frac{p}{h}\) =\(\frac{BC}{AC}\) = cos\(\theta\) (For reference angle \(\theta\))

cos (90° - \(\theta\)) = \(\frac{b}{h}\) =\(\frac{AB}{AC}\) = sin\(\theta\)

tan(90° - \(\theta\)) =\(\frac{p}{h}\) =\(\frac{BC}{AB}\) = cot\(\theta\)

cot(90° - \(\theta\)) =\(\frac{b}{p}\) =\(\frac{AB}{BC}\) = tan\(\theta\)

sec(90° - \(\theta\)) =\(\frac{h}{b}\) =\(\frac{AC}{AB}\) = cosec\(\theta\)

cosec(90° - \(\theta\)) =\(\frac{h}{p}\) =\(\frac{AB}{BC}\) = sec\(\theta\)

Hence,

sin(90° - \(\theta\)) = cos\(\theta\)

cos(90° - \(\theta\)) = sin\(\theta\)

tan(90° - \(\theta\)) =cot\(\theta\)

cot(90° - \(\theta\)) =tan\(\theta\)

sec(90° - \(\theta\)) = cosec\(\theta\)

cosec(90° - \(\theta\)) =sec\(\theta\)

- h
^{2}= p^{2}+ b^{2} - Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

sin420^{0} cos390^{0} + cos390^{0} + cos (-300^{0}) sin (-330^{0}) = 1

soln: L.H.S = sin420^{0} cos 390^{0} + cos (-300^{0}) sin (-330^{0})

= sin420^{0} cos390^{0} + cos300^{0} (-sin330^{0})

= sin420^{0} cos390^{0} - cos300^{0} sin330^{0}

= sin (360^{0} + 60^{0}) cos (360^{0} + 30^{0}) - cos (360^{0}-60^{o}) sin(360^{0}-30^{0})

= sin60^{0} cos30^{0} cos60^{0} (-sin 30^{0}) = sin60^{0} cos30^{0} + cos60^{0}sin30^{0}

= \(\frac{(\sqrt{3})}{2}\)×\(\frac{(\sqrt{3})}{2}\)+\(\frac{1}{2}\)×\(\frac{1}{2}\)=\(\frac{3}{4}\)+\(\frac{1}{4}\)=\(\frac{3+1}{4}\)=\(\frac{4}{4}\)=1= R.H.S. proved.

If aA =30, then verify that sin3A = 3sinA - 4sin^{3}A

Given that A = 30°

Then, LHS = sin3A

= sin3 × 30°

= sin90°

= 1

RHS = 3sinA - 4sin^{3}A

= 3sin30° - 4(sin30°)^{3}

= 3\(\frac{1}{2}\) - 4(\(\frac{1}{2}\))^{3}

= \(\frac{3}{2}\) - 4 \(\frac{1}{8}\)

= \(\frac{3}{2}\) - \(\frac{1}{2}\)

= \(\frac{2}{2}\)

= 1

\(\therefore\) LHS = RHS verified.

Cos 60° = cos^{2}30° - sin^{2}30°

Solution

LHS = cos 60

= \(\frac{1}{2}\)

RHS = cos^{2}30° - sin^{2}30°

(\(\frac{\sqrt(3)}{2}\))^{2} - (\(\frac{1}{2}\))^{2}

\(\frac{3}{4}\) - \(\frac{1}{4}\)

\(\frac{2}{4}\)

\(\frac{1}{2}\)

\(\therefore\) LHS = RHS proved.

x.sin 30 × cos^{2}45° = \(\frac{cot^230° sec60° tan45°}{cosec^245°×cosec30°}\)

Solution

Here, x×sin30° cos^{2}45° = \(\frac{cot^230° sec60° tan45°}{cosec^245°×cosec30°}\)

or, x×\(\frac{1}{2}\)×(\(\frac{1}{\sqrt(2)}\))^{2} = \(\frac{(\sqrt(3)^2×2×1)}{(\sqrt(2)^ 2×2}\)

or, x = \(\frac{3×2×2×2}{2×2}\) = 6

cot 3\(\theta\) = tan 7\(\theta\)

Solution

Here, cot 3\(\theta\) = tan 7\(\theta\)

or, cot 3\(\theta\) = cot (90° -7\(\theta\)) [cot(90° - \(\theta) = tan\theta\)]

or, 3\(\theta\) = 90° - 7\(\theta\)

or,10\(\theta\) = 90°

\(\therefore\) \(\theta\) = 9°

Solve the triangle ABC if B = 90°, a = \(\sqrt(3)\) and c = 1.

Solution

Here, In ABC

B = 90° (Since Ac is the hypotenuse.)

So, using Pythagoras theorem

(AC)2 = (AB)2 + (BC)2

= (1)2 +(\(\sqrt(3)\))2

= 1 + 3 = 4

or, AC = 2 = b

Now, sinA =\(\frac{BC}{AC}\)

or, sinA = \(\frac{\sqrt(3)}{2}\)

or, sinA = sin 60°

\(\therefore\) A = 60°

Again,

\(\angle\)A + \(\angle\)B + \(\angle\)C = 180°

or, 60° + 90° + \(\angle\)C = 180°

C = 180° - 150° = 30°

Hence, b = 2, A = 60° and C = 30°

Evaluate if (x^{c} = 180°)sec^{2}\(\frac{π}{4}\) sec^{2}\(\frac{π}{3}\)(cosec\(\frac{π}{6}\) - cosec\(\frac{π}{2}\))

Solution

sec^{2}\(\frac{π}{4}\) sec^{2}\(\frac{π}{3}\)(cosec\(\frac{π}{6}\) - cosec\(\frac{π}{2}\))

= sec^{2} \(\frac{180°}{4}\) × sec^{2}\(\frac{180°}{3}\) (cosec\(\frac{180°}{6}\) - cosec\(\frac{180°}{2}\))

= sec^{2}45° × sec^{2}60° (cosec30° - cosec90°)

= (\(\sqrt(2)\))^{2} × (2)2 (2-1)

= 2×4×1

= 8

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