Other Trigonometric Identities | Superprof

Learn Maths from the best

First Lesson Free!

Hamza

February 25, 2021

Chapters

Supplementary Angles
Angles That Differ by 180° or π Rad
Angles Greater Than 360º
Angles That Differ by 90° or π/2 Rad
Angles That Add Up to 270º or 3/2 π Rad
Angles That Differ by 270º or 3/2 π Rad

Trigonometric ratios behave in different quadrants. In some cases, they behave the same as well and that is why it becomes necessary for us to understand their behavior. To understand the quadrant effect on trigonometric ratios, you need to understand a few trigonometric identities. These trigonometric identities are split into different types:

Acute Angles Trigonometric Identities
Supplementary Angles Trigonometric Identities
Angles Greater Than ${ 180 }^{\circ}$ or $\pi$ Trigonometric Identities
Angles Greater Than ${ 360 }^{\circ}$ or $2 \pi$ Trigonometric Identities
Angles That Differ by ${ 90 }^{ \circ }$ or $\frac { \pi }{ 2 }$ Trigonometric Identities
Angles That Add Up to ${ 270 }^{ \circ }$ or $\frac { 3 }{ 2 } \pi$ Trigonometric Identities
Angles That Differ by ${ 270 }^{ \circ }$ or $\frac { 3 }{ 2 } \pi$ Trigonometric Identities

Supplementary Angles

Two angles are supplementary if the sum is ${ 180 }^{ \circ }$ or $\pi$ radians. So, if two supplementary Angles are added, a straight angle is obtained.

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

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

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

$\sin { { 150 }^{\circ} } = \sin { ({ 180 }^{\circ} - { 30 }^{\circ}) } = \sin { {30}^{\circ} } = \frac { 1 }{ 2 }$

$\cos { { 150 }^{\circ} } = \cos { ({ 180 }^{\circ} - { 30 }^{\circ}) } = -\cos { {30}^{\circ} } = -\frac { \sqrt { 3 } }{ 2 }$

$\tan { { 150 }^{\circ} } = \tan { ({ 180 }^{\circ} - { 30 }^{\circ}) } = -\tan { {30}^{\circ} } = -\frac { \sqrt { 3 } }{ 3 }$

The best Maths tutors available

1^st lesson free!

1^st lesson free!

4.9 (23 reviews)

1^st lesson free!

1^st lesson free!

4.9 (6 reviews)

1^st lesson free!

4.9 (11 reviews)

1^st lesson free!

4.9 (9 reviews)

1^st lesson free!

1^st lesson free!

1^st lesson free!

1^st lesson free!

4.9 (23 reviews)

1^st lesson free!

1^st lesson free!

4.9 (6 reviews)

1^st lesson free!

4.9 (11 reviews)

1^st lesson free!

4.9 (9 reviews)

1^st lesson free!

1^st lesson free!

First Lesson Free

Angles That Differ by 180° or π Rad

$\sin { (\pi + \alpha) } = - \sin { \alpha }$

$\cos { (\pi + \alpha) } = - \cos { \alpha }$

$\tan { (\pi + \alpha) } = \tan { \alpha }$

$\sin { { 210 }^{\circ} } = \sin { ({ 180 }^{\circ} + { 30 }^{\circ}) } = - \sin { {30}^{\circ} } = -\frac { 1 }{ 2 }$

$\cos { { 210 }^{\circ} } = \cos { ({ 180 }^{\circ} + { 30 }^{\circ}) } = -\cos { {30}^{\circ} } = -\frac { \sqrt { 3 } }{ 2 }$

$\tan { { 210 }^{\circ} } = \tan { ({ 180 }^{\circ} + { 30 }^{\circ}) } = \tan { {30}^{\circ} } = \frac { \sqrt { 3 } }{ 3 }$

Angles Greater Than 360º

$\sin { (\alpha + 2\pi k) } = \sin { \alpha }$

$\cos { (\alpha + 2\pi k) } = \cos { \alpha }$

$\tan { (\alpha + 2\pi k) } = \tan { \alpha }$

$\sin { { 750 }^{\circ} } = \sin { ({ 360 }^{\circ} \times 2 + { 30 }^{\circ}) } = \sin { { 30 }^{\circ} } = \frac { 1 }{ 2 }$

$\cos { { 750 }^{\circ} } = \cos { ({ 360 }^{\circ} \times 2 + { 30 }^{\circ}) } = \cos { { 30 }^{\circ} } = \frac { \sqrt { 3 } }{ 2 }$

$\tan { { 750 }^{\circ} } = \tan { ({ 360 }^{\circ} \times 2 + { 30 }^{\circ}) } = \tan { { 30 }^{\circ} } = \frac { \sqrt { 3 } }{ 3 }$

Angles That Differ by 90° or π/2 Rad

$\sin { (\alpha + \frac { \pi }{ 2 }) } = \cos { \alpha }$

$\cos { (\alpha + \frac { \pi }{ 2 }) } = -\sin { \alpha }$

$\tan { (\alpha + \frac { \pi }{ 2 }) } = -\cot { \alpha }$

$\sin { { 120 }^{\circ} } = \sin { ({ 90 }^{\circ} + { 30 }^{\circ}) } = \cos { { 30 }^{\circ} } = \frac { \sqrt { 3 } }{ 2 }$

$\cos { { 120 }^{\circ} } = \cos { ({ 90 }^{\circ} + { 30 }^{\circ}) } = - \sin { { 30 }^{\circ} } = - \frac { 1 }{ 2 }$

$\tan { { 120 }^{\circ} } = \tan { ({ 90 }^{\circ} + { 30 }^{\circ}) } = - \cot { { 30 }^{\circ} } = - \sqrt { 3 }$

Angles That Add Up to 270º or 3/2 π Rad

$\sin { (\frac { 3 \pi }{ 2 } - \alpha) } = -\cos { \alpha }$

$\cos { (\frac { 3 \pi }{ 2 } - \alpha) } = -\sin { \alpha }$

$\tan { (\frac { 3 \pi }{ 2 } - \alpha) } = \cot { \alpha }$

$\sin { { 240 }^{ \circ } } = \sin { ({ 270 }^{ \circ } - { 30 }^{ \circ }) } = - \cos { { 30 }^{ \circ } } = - \frac { \sqrt { 3 } }{ 2 }$

$\cos { { 240 }^{ \circ } } = \cos { ({ 270 }^{ \circ } - { 30 }^{ \circ }) } = - \cos { { 30 }^{ \circ } } = - \frac { 1 }{ 2 }$

$\tan { { 240 }^{ \circ } } = \tan { ({ 270 }^{ \circ } - { 30 }^{ \circ }) } = \cos { { 30 }^{ \circ } } = \sqrt { 3 }$

Angles That Differ by 270º or 3/2 π Rad

$\sin { (\frac { 3 \pi }{ 2 } + \alpha) } = -\cos { \alpha }$

$\cos { (\frac { 3 \pi }{ 2 } + \alpha) } = -\sin { \alpha }$

$\tan { (\frac { 3 \pi }{ 2 } + \alpha) } = \cot { \alpha }$

$\sin { { 300 }^{ \circ } } = \sin { ({ 270 }^{ \circ } + { 30 }^{ \circ }) } = - \cos { { 30 }^{ \circ } } = - \frac { \sqrt { 3 } }{ 2 }$

$\cos { { 300 }^{ \circ } } = \cos { ({ 270 }^{ \circ } + { 30 }^{ \circ }) } = \sin { { 30 }^{ \circ } } = \frac { 1 }{ 2 }$

$\tan { { 300 }^{ \circ } } = \tan { ({ 270 }^{ \circ } + { 30 }^{ \circ }) } = - \cot { { 30 }^{ \circ } } = - \sqrt { 3 }$

Need a Maths teacher?

Did you like the article?

2 Stars

3 Stars

4 Stars

5 Stars

3.00/5 - 2 vote(s)

Loading...

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.

Comments

Cancel reply