At this point, you might be familiar with the identities when the angle is moving anti-clockwise. At every quadrant, there are some specific identities for different trigonometric ratios. However, you might know that the angle moves anti-clockwise, the angle can vary but they all will remain positive, but this isn't the only case. The angles can also move in the clockwise direction as well! We call them negative angle angles. The angles will be written with a negative sign and the quadrants will have new angles for example, in the fourth quadrant, the angle ranges from { 270 }^{ \circ } to { 360 }^{ \circ } but when the angle is moving in the clockwise direction, the fourth quadrant's angle will range from { 0 }^{ \circ } to { -90 }^{ \circ }. This will ultimately change everything! from angle to the trigonometric values and many more.

If you know how to find angles from different quadrants, irrespective of angle movement, then you won't have any problems with angle moving anticlockwise direction as well. However, you will be needing some help with trigonometric identities which are called Even-Odd Identities. These identities are very easy to learn and not to mention that their concept is easy as well. Let's clear one thing, the value of \alpha will always be negative when the angle is moving clockwise.

Trigonometric Identities

Below are the even-odd trigonometric identities:

\sin { (- \alpha) } = - \sin { \alpha }

\cos { (- \alpha) } = \cos { \alpha }

\tan { (- \alpha) } = - \tan { \alpha }

For \alpha is an acute angle.

 

Let's interpret the above trigonometric identities. If the angle of the \sin is negative that means the value of the \sin for that angle(positive) will also change its sign. For example: \sin { ({ 45 }^{ \circ }) } is equal to \frac { 1 }{ \sqrt { 2 } } but if the angle's sign becomes negative then \sin { (- { 45 }^{ \circ }) } = - \frac { 1 }{ \sqrt { 2 } }. The reason is simple, the value of \sin { (\alpha) } is positive in { 1 }^{ st } and { 2 }^{ nd } quadrants.  For the \cos ratio, if the angle becomes negative angle, it will not have any impact on the value of \cos. In simple words, the value of \cos { (\alpha) } will remain the same. For example: \cos { ({ 45 }^{ \circ }) } is equal to \frac { 1 }{ \sqrt { 2 } }, however, \cos { (-{ 45 }^{ \circ }) } will also be equal to \frac { 1 }{ \sqrt { 2 } } because \cos is positive in { 1 }^{ st } and { 4 }^{ th } quadrants. Last but not least, if the angle moves clockwise direction that means the value of \tan with a negative angle will be equal to negative value. For example: \tan { ({ 45 }^{ \circ }) } is equal to 1 but if the angle is negative, hence \tan { (-{ 45 }^{ \circ }) } then the answer will be -1 because \tan { (\alpha) } is positive in { 1 }^{ st } and { 3 }^{ rd } quadrants.

 

 

 

 

 

 

 

\sin { (-{ 30 }^{ \circ }) } = - \sin { ({ 30 }^{ \circ }) } = - \frac { 1 }{ 2 }

\cos { (-{ 30 }^{ \circ }) } = \cos { ({ 30 }^{ \circ }) } = \frac { \sqrt { 3 } }{ 2 }

\tan { (-{ 30 }^{ \circ }) } = - \tan { ({ 30 }^{ \circ }) } = - \frac { \sqrt { 3 } }{ 3 }

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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.