At this point, you might be familiar with the Pythagorean formula and how to find different parts of a triangle with the help of Pythagorean, if not then you might revise it because this lecture is purely based on the Pythagorean formula. The Pythagorean formula says that the sum of squared perpendicular and base equal to the square of the hypotenuse. In mathematical language, it is:

{ H }^{ 2 } = { P }^{ 2 } + { B }^{ 2 }

The hypotenuse is the largest part of a right-angle triangle. It is opposite to the { 90 }^{ \circ } angle. The rest is the adjacent and base of the triangle. The formula also gives birth to a new series of trigonometric identities which helps a lot in solving complex identities. Consider the image below.

Circle of radius one has a right-angle triangle and the lengths of axes are trigonometric ratios

The circle has a radius of 1 unit and with the help of the radius, we formed a triangle. The components of the triangle are \cos { \theta } and \sin { \theta }. Let's put this in the Pythagorean formula:

{ H }^{ 2 } = { P }^{ 2 } + { B }^{ 2 }

{ 1 }^{ 2 } = { \sin { \theta } }^{ 2 } + { \cos { \theta } }^{ 2 }

\sin^{ 2 }{ \theta } + \cos^{ 2 }{ \theta } = 1

Hence, we found that the sum of squred trigonometric ratios (\sin { \theta } and \cos { \theta }) are equal to 1. However, mathematicians didn't not stop on this, they found different identities as well from the above identities.

\sin^{ 2 }{ \theta } + \cos^{ 2 }{ \theta } = 1

Dividing the whole equation with \cos^{ 2 }{ \theta }:

\frac { \sin^{ 2 }{ \theta } }{ \cos^{ 2 }{ \theta } } + \frac { \cos^{ 2 }{ \theta } }{ \cos^{ 2 }{ \theta } } = \frac { 1 }{ \cos^{ 2 }{ \theta }}

As we know that \frac { \sin }{ \cos } = \tan and \frac { 1 }{ \cos } = \sec:

\tan^{ 2 }{ \theta } + 1 = \sec^{ 2 }{ \theta }

Hence, mathematicians found another identity but what if we divide the Pythagorean equation with \sin^{ 2 }{ \theta } instead of \cos^{ 2 }{ \theta }?

\frac { \sin^{ 2 }{ \theta } }{ \sin^{ 2 }{ \theta } } + \frac { \cos^{ 2 }{ \theta } }{ \sin^{ 2 }{ \theta } } = \frac { 1 }{ \sin^{ 2 }{ \theta } }

1 + \cot^{ 2 }{ \theta } = \csc^{ 2 }{ \theta }

 

Knowing that \sin { \alpha } = \frac { 3 }{ 5 }, and { 90 }^{ \circ } < \alpha < { 180 }^{ \circ }. Calculate the remaining trigonometric ratios of angle \alpha.

\sin { \alpha } = \frac { 3 }{ 5 } \qquad \csc { \alpha } = \frac { 5 }{ 3 }

{ 1 }^{ 2 } = { \sin { \alpha } }^{ 2 } + { \cos { \alpha } }^{ 2 }

{ 1 }^{ 2 } - { \sin { \alpha } }^{ 2 } = { \cos { \alpha } }^{ 2 }

\cos { \alpha } = \sqrt { 1 - { \frac { 3 }{ 5 } }^{ 2 } } = \frac { 4 }{ 5 }

Since, we are working in the second quadrant and \cos is negative in second quadrant, hence:

\cos { \alpha } = - \frac { 4 }{ 5 }

\sec { \alpha } = - \frac { 5 }{ 4 }

 

\frac { \sin }{ \cos } = \tan

\tan { \alpha } = - \frac { \frac { 3 }{ 5 } }{ \frac { 4 }{ 5 } } The minus sign is because of the 2nd quadrant

\tan { \alpha } = - \frac { 4 }{ 3 }

 

Knowing that \tan { \alpha } = 2, and { 180 }^{ \circ } < \alpha < { 270 }^{ \circ }. Calculate the remaining trigonometric ratios of angle \alpha.

\sec { \alpha } = - \sqrt { 1 + 4 } = - \sqrt { 5 } \qquad \cos { \alpha } = - \frac { 1 }{ \sqrt { 5 } } = - \frac { \sqrt { 5 } }{ 5 }

\sin { \alpha } = 2 . ( - \frac { \sqrt { 5 } }{ 5 } ) = - \frac { 2 \sqrt { 5 } }{ 5 } \qquad \csc { \alpha } = - \frac { \sqrt { 5 } }{ 2 }

\tan { \alpha } = 2 \qquad \cot { \alpha } = \frac { 1 }{ 2 }

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.