Trigonometric Ratios of Some Standard Angles

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

h² = p² + b²

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

h²= p²+ b²

or, (AC)²=(AB)²+ (BC)²

or, a² = (AB)² + a²

or, (AB)² = a² - a² = 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

h²= p²+ b²

or, (AC)²=(AB)²+ (BC)²

or, a² = (BC)² + a²

or, (BC)² = a² - a² = 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

Complementary Angles

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\(\theta\)