Trigonometric identities - Maths & Stats - Trigonometry - The Library at Leeds Beckett University

Learning Outcomes

If you work through this section you should be able to:

According to Pythagoras’ theorem,

Diagram showing trigonometric identities.

AB²+ AC²⁼BC²

Dividing both sides by BC² gives:

${(\frac{A B}{B C})}^{2} + {(\frac{A C}{B C})}^{2} = 1$

Because $\frac{A B}{B C} = \sin θ$ and $\frac{A C}{B C} = \cos θ$ this equation can be converted to:

$\sin^{2} θ + \cos^{2} θ = 1$

Two more identities can be derived from the above identity:

$1 + \tan^{2} θ = \sec^{2} θ$

$1 + \cot^{2} θ = \csc^{2} θ$

The trigonometric formulas of the sum or difference of two angles can be given in terms of the ratios of the individual angles as below:

\sin (a \pm b) = \sin a \cos b \pm \cos a \sin b

$\cos (a \pm b) = \cos a \cos b \mp \sin a \sin b$

$\tan (a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$

\sin 75 ° = \sin (45 ° + 30 °)

$= \sin 45 ° \cos 30 ° + \cos 45 ° \sin 30 °$

$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}$

$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Double-angle identities are derived from the sum identities when $a = b$ :

$\sin 2 a = \sin a \cos a = \frac{2 \tan a}{1 + \tan^{2} a}$

$\cos 2 a = \cos^{2} a - \sin^{2} a = 2 \cos^{2} a - 1 = 1 - 2 \sin^{2} a = \frac{1 - \tan^{2} a}{1 + \tan^{2} a}$

$\tan 2 a = \frac{2 \tan a}{1 - \tan^{2} a}$

Half-angle identities can be written as below:

$\sin \frac{a}{2} = \pm \sqrt{\frac{1 - \cos a}{2}}$

$\cos \frac{a}{2} = \pm \sqrt{\frac{1 + \cos a}{2}}$

$\tan \frac{a}{2} = \pm \sqrt{\frac{1 - \cos a}{1 + \cos a}} = \frac{\sin a}{1 + \cos a} = \frac{1 - \cos a}{\sin a}$

$\sin 15 ° = \sin (\frac{30 °}{2})$

$= \sqrt{\frac{1 - \cos 30 °}{2}}$

$= \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$

$= \sqrt{\frac{2 - \sqrt{3}}{4}}$

$= \sqrt{\frac{{(\sqrt{3} - 1)}^{2}}{8}}$

$= \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

You can download a version of this Trigonometric identities activity in Word format:

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