Trigonometry/Identities

Let us take a right angled triangle with hypotenuse length 1. If we mark one of the acute angles as $\theta$, then using the definition of the sine ratio, we have

$\sin \theta =\frac{opposite}{hypotenuse}$

As the hypotenuse is 1,

$\sin \theta =\frac{opposite}{1}=opposite$

Repeating the same process using the definition of the cosine ratio, we have

$\cos \theta =\frac{adjacent}{hypotenuse}=\frac{adjacent}{1}=adjacent$

Pythagorean identities

Since this is a right triangle, we can use the Pythagorean Theorem:

$x^2+y^2=r^2$

$\frac{x^2}{r^2}+\frac{y^2}{r^2}=\frac{r^2}{r^2}$

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

This is the most fundamental identity in trigonometry.

$\frac{x^2}{y^2}+\frac{y^2}{y^2}=\frac{r^2}{y^2}$

${\cot }^{2}x+1={\csc }^2$

$\frac{x^2}{x^2}+\frac{y^2}{x^2}=\frac{r^2}{x^2}$

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

From this identity, if we divide through by squared cosine, we are left with:

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

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

${\sec }^2\theta -{\tan }^2\theta =1$

If instead we divide the original identity by squared sine, we are left with:

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

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

${\csc }^2\theta -{\cot }^2\theta =1$

There are basically 3 main trigonometric identities. The proofs come directly from the definitions of these functions and the application of the Pythagorean theorem:

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

${\sec }^2\theta -{\tan }^2\theta =1$

${\csc }^2\theta -{\cot }^2\theta =1$

Angle sum-difference identities

$\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$

$\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

Cofunction identities

$\cos (90-\theta )=\sin \theta$

$\sec (90-\theta )=\csc \theta$

$\tan (90-\theta )=\cot \theta$

$\sin (90-\theta )=\cos \theta$

$\csc (90-\theta )=\sec \theta$

$\cot (90-\theta )=\tan \theta$

Multiple angle identities

$\cos 2A={\cos }^{2}A-{\sin }^{2}A$

$\sin 2A=2\sin A\cos A$

$\sin 2\theta =\frac{tan2\theta *tan\theta }{tan2\theta -tan\theta }$

$\cos 2\theta =\frac{tan\theta }{tan2\theta -tan\theta }$

$\tan 2\theta =tan\theta (\frac{1}{cos2\theta }+1)$

$\tan 2\theta =\frac{2si{n}^2\theta }{sin2\theta -tan\theta }$