Convert to Sine and Cosine

Prove that $\tan x \cos x = \sin x$.

Choose the side to work on: The left side has two functions, the right side only one. Start from the left. = $\tan x \cos x$

Use the quotient identity: Replace $\tan x$ with $\frac{\sin x}{\cos x}$ = $\frac{\sin x}{\cos x} \cdot \cos x$

Cancel the common factor: $\cos x$ in the numerator and denominator cancel (where $\cos x \neq 0$) = $\sin x$

Compare with the right side: The left side now looks exactly like the right side = $\sin x = \sin x$ ✓

Answer: $\tan x \cos x = \sin x$ is an identity for every $x$ with $\cos x \neq 0$.

Combine Fractions, Then Use Pythagoras

Prove that $\sec x - \cos x = \sin x \tan x$.

Convert to sine and cosine: $\sec x = \frac{1}{\cos x}$ = $\frac{1}{\cos x} - \cos x$

Use a common denominator: $\cos x = \frac{\cos^2 x}{\cos x}$ = $\frac{1 - \cos^2 x}{\cos x}$

Apply the Pythagorean identity: $1 - \cos^2 x = \sin^2 x$ = $\frac{\sin^2 x}{\cos x}$

Split off one factor: $\frac{\sin^2 x}{\cos x} = \sin x \cdot \frac{\sin x}{\cos x}$ = $\sin x \tan x$ ✓

Answer: $\sec x - \cos x = \sin x \tan x$ is an identity (for $\cos x \neq 0$). Check at $x = 60°$: $2 - 0.5 = 1.5$ and $\frac{\sqrt{3}}{2} \cdot \sqrt{3} = 1.5$.

Multiply by the Conjugate

Prove that $\dfrac{\sin x}{1 - \cos x} = \dfrac{1 + \cos x}{\sin x}$.

Pick the trick: The denominator $1 - \cos x$ has the conjugate $1 + \cos x$. Multiply the left side by $\frac{1 + \cos x}{1 + \cos x}$, which equals 1. = $\frac{\sin x\,(1 + \cos x)}{(1 - \cos x)(1 + \cos x)}$

Multiply out the denominator: $(a - b)(a + b) = a^2 - b^2$ = $\frac{\sin x\,(1 + \cos x)}{1 - \cos^2 x}$

Apply the Pythagorean identity: $1 - \cos^2 x = \sin^2 x$ = $\frac{\sin x\,(1 + \cos x)}{\sin^2 x}$

Cancel one $\sin x$: $\frac{\sin x}{\sin^2 x} = \frac{1}{\sin x}$ = $\frac{1 + \cos x}{\sin x}$ ✓

Answer: The two fractions are equal for every $x$ where $\sin x \neq 0$. The conjugate turned $1 - \cos x$ into $\sin^2 x$.

Start From the Right Side

Prove that $\dfrac{\cos x}{1 - \sin x} = \sec x + \tan x$.

Choose the side: The right side has two different functions that can be combined, so start there. = $\sec x + \tan x$

Convert and combine: $\frac{1}{\cos x} + \frac{\sin x}{\cos x}$ = $\frac{1 + \sin x}{\cos x}$

Multiply by the conjugate of $1 + \sin x$: $\frac{1 + \sin x}{\cos x} \cdot \frac{1 - \sin x}{1 - \sin x} = \frac{1 - \sin^2 x}{\cos x\,(1 - \sin x)}$ = $\frac{\cos^2 x}{\cos x\,(1 - \sin x)}$

Cancel one $\cos x$: $\frac{\cos^2 x}{\cos x} = \cos x$ = $\frac{\cos x}{1 - \sin x}$ ✓

Answer: Starting from the right side, we reached the left side, so the identity is proved (where $\cos x \neq 0$ and $\sin x \neq 1$).

Disproving With One Angle

Is $\sin x + \cos x = 1$ an identity?

Test some easy angles: $x = 0°$: $0 + 1 = 1$. $x = 90°$: $1 + 0 = 1$. = True for these two angles

Test an angle in between: $x = 45°$: $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$ = $1.414 \neq 1$

Draw the conclusion: One counterexample is enough: the two sides differ at $45°$. = Not an identity

See it on a graph: $y = \sin x + \cos x$ is a wave and $y = 1$ is a line: they only meet at some points. = The graphs do not overlap

Answer: $\sin x + \cos x = 1$ is an equation (true at $0°$, $90°$, $360°$, ...), not an identity. The counterexample $x = 45°$ disproves it.

Mistake: Working on both sides like an equation, for example cross-multiplying $\frac{\sin x}{1 - \cos x} = \frac{1 + \cos x}{\sin x}$ to get $\sin^2 x = 1 - \cos^2 x$

Why: Cross-multiplying, moving terms or dividing both sides all start from the claim that the two sides are equal, which is exactly what you are trying to show.

Correct: Transform one side until it becomes the other side. The equals sign stays untouched until the very end.

Mistake: Checking one or two angles and calling the identity proved

Why: $\sin x + \cos x = 1$ works at $0°$ and $90°$ but fails at $45°$. A few true examples never prove that something holds for every angle.

Correct: Use numbers only to test or to disprove. A proof needs algebra with identities that hold for all $x$.

Mistake: Splitting a fraction at the denominator: $\frac{\sin x}{1 - \cos x} = \frac{\sin x}{1} - \frac{\sin x}{\cos x}$

Why: You may split a sum in the numerator, $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$, but never a sum in the denominator. At $x = 60°$ the left side is $1.732$ and the right side is $-0.866$.

Correct: To get rid of $1 - \cos x$ in a denominator, multiply numerator and denominator by the conjugate $1 + \cos x$.

Mistake: Writing $(1 - \cos x)(1 + \cos x) = 1 - \cos x$ or $1 + \cos^2 x$

Why: The product of conjugates is a difference of squares: both terms are squared and the sign is minus.

Correct: $(1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x$

How Computer Algebra Tests Identities

Math software checks whether two formulas are the same by plugging in many random angles before it tries a full symbolic proof. If even one angle gives different values, the formulas are different.

To compare $\sec x - \cos x$ and $\sin x \tan x$, a program evaluates both at angles like $17°$, $60°$ and $133°$ and finds the same values every time.

Physics: Simplifying a Ramp Formula

Physics formulas often come out of a calculation in a clumsy form. An identity turns them into a shorter form that is faster to compute and easier to understand.

A force on a ramp came out as $F = \dfrac{W\sin\theta}{\tan\theta}$. Since $\frac{\sin\theta}{\tan\theta} = \sin\theta \cdot \frac{\cos\theta}{\sin\theta} = \cos\theta$, the formula is simply $F = W\cos\theta$.

Computer Graphics: Rotations Keep Lengths

Games rotate points with $x' = x\cos\theta - y\sin\theta$ and $y' = x\sin\theta + y\cos\theta$. Expanding $x'^2 + y'^2$ and using $\sin^2\theta + \cos^2\theta = 1$ proves that $x'^2 + y'^2 = x^2 + y^2$ for every angle, so objects never stretch.

The point $(3, 4)$ rotated by $30°$, $90°$ or $217°$ always stays at distance 5 from the origin.

Q: Why can't I just do the same thing to both sides?

A: Doing the same operation to both sides starts from the statement that the sides are equal, which is what you want to prove. Some of those operations (squaring, multiplying by an expression that can be 0) can even turn a false claim into a true one. Working on one side avoids this.

Q: Which side should I start with?

A: Usually the more complicated one: the side with more terms, fractions, or functions like $\sec$, $\csc$, $\tan$, $\cot$. Simplifying is easier than inventing complications.

Q: What if I get stuck halfway?

A: Try another tool: convert to $\sin$ and $\cos$, combine fractions, factor, or multiply by a conjugate. You may also simplify both sides separately until they reach the same expression, as long as each side is transformed on its own.

Q: Is checking the graph a proof?

A: No. If the graphs of both sides overlap everywhere, the identity is very likely true, and if they do not overlap, it is false. But a picture only shows a limited window, so a proof still needs algebraic steps.