Trigonometric identities

$\tan \theta =\frac{\sin \theta }{\cos \theta },\cos \theta \ne 0$

This follows from $\tan \theta =\frac{y}{x}$, $\sin \theta =\frac{y}{r}$ and $\cos \theta =\frac{x}{r}$.

Example. $\tan x\cos x=\frac{\sin x}{\cos x}\cdot \cos x=\sin x$

Rewrite $\tan$ as $\frac{\sin }{\cos }$ whenever an expression mixes $\tan$ with $\sin$ or $\cos$.

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

This follows from $x^2+y^2=r^2$, divided by $r^2$. Useful rearrangements:

• $1-{\sin }^2\theta ={\cos }^2\theta$
• $1-{\cos }^2\theta ={\sin }^2\theta$

Example. $\frac{1-{\sin }^{2}x}{\cos x}=\frac{{\cos }^{2}x}{\cos x}=\cos x$

Start with the more complicated side and simplify until you reach the other side.

• Change $\tan$ to $\frac{\sin }{\cos }$.
• Add fractions with a common denominator.
• Use the square identity, and look for $1-{\sin }^2$ or $1-{\cos }^2$.
• Factorise and cancel.

Example. Prove $\tan x+\frac{1}{\tan x}=\frac{1}{\sin x\cos x}$.

$\text{LHS}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}=\frac{1}{\sin x\cos x}=\text{RHS}$

Multiplying by a conjugate. To prove $\frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x}$, multiply the RHS by $\frac{1+\sin x}{1+\sin x}$. The denominator becomes $1-{\sin }^{2}x={\cos }^{2}x$.

An identity is undefined where any denominator is 0, or where $\tan$ is undefined.

Example. $\frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x}$ is undefined when:

• $\cos x=0$: $x=90°+k\cdot 180°$
• $1-\sin x=0$: $\sin x=1$, so $x=90°+k\cdot 360°$ (already included above)

So it is undefined for $x=90°+k\cdot 180°$, $k\in \mathbb{Z}$.

Check the original identity, not your simplified version.