Proving Trigonometric Identities

Answer: $\sin^2 x + \cos^2 x = 1$

$\sin^2 x + \cos^2 x = 1$ is the fundamental Pythagorean identity and holds for every $x$. The other three are equations: $\sin x = \cos x$ holds at $45°$, $\sin x + \cos x = 1$ at $0°$ and $90°$, $\tan x = 1$ at $45°$, but none of them holds for every angle.

Answer: 1

$(1 + 3) \cdot \frac{1}{4} = 4 \cdot \frac{1}{4} = 1$, exactly the right side. This test agrees with the proof $(1 + \tan^2 x)\cos^2 x = \sec^2 x \cos^2 x = 1$.

Answer: $\tan 30° \cos 30° = \sin 30° = \frac{1}{2}$

1. What is the right side $\sin 30°$? 1/2
2. What is the left side $\tan 30° \cdot \cos 30° = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}$? 1/2
3. What is left side minus right side? 0

Answer: Divide both sides by $\cos x$

Dividing both sides by $\cos x$ treats the identity like an equation that is already true. The other moves change only one side and keep its value: a quotient identity, multiplying by a form of 1, cancelling a common factor.

Answer: -2

Left side: $\sin 180° = 0$. Right side: $2\sin 90° = 2$. The difference $0 - 2 = -2$ is not $0$, so $x = 90°$ is a counterexample and the claim is not an identity.

Answer: $\sin x \cot x = \sin x \cdot \frac{\cos x}{\sin x} = \cos x$

1. Rewrite $\cot x$ with sine and cosine. Which expression is it? $\dfrac{\cos x}{\sin x}$
2. The left side is now $\sin x \cdot \frac{\cos x}{\sin x}$. What is left after cancelling? $\cos x$
3. Check at $x = 60°$: what value must both sides give ($\cos 60°$)? 1/2

Answer: Multiply numerator and denominator by $1 - \cos x$

Multiplying by the conjugate $\frac{1 - \cos x}{1 - \cos x}$ gives $\frac{\sin x\,(1 - \cos x)}{1 - \cos^2 x} = \frac{\sin x\,(1 - \cos x)}{\sin^2 x} = \frac{1 - \cos x}{\sin x}$. Squaring or adding to both sides breaks the one-side rule.

Answer: 3/2

$\frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} = 1.5$. The left side gives the same: $\sec 60° - \cos 60° = 2 - 0.5 = 1.5$.

Answer: $\cos x\,(\sec x - \cos x) = 1 - \cos^2 x = \sin^2 x$

1. Distributing gives $\cos x \sec x - \cos^2 x$. What number is $\cos x \sec x$? 1
2. So the left side is $1 - \cos^2 x$. Which expression is this equal to? $\sin^2 x$
3. Check at $x = 30°$: what is $\sin^2 30°$? 1/4

Answer: $\dfrac{2}{1 - \cos^2 x} = \dfrac{2}{\sin^2 x} = 2\csc^2 x$

1. With the common denominator the numerator is $(1 + \cos x) + (1 - \cos x)$. What number is it? 2
2. Which expression is the denominator $(1 - \cos x)(1 + \cos x)$ equal to? $\sin^2 x$
3. So the left side is $\frac{2}{\sin^2 x} = 2\csc^2 x$. What is its value at $x = 30°$? 8

Answer: $45°$

At $0°$, $90°$ and $360°$ the sum is exactly 1. At $45°$ it is $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414 \neq 1$, so $45°$ is a counterexample.

Answer: 3

$\frac{1 - \sin^2 x}{1 - \cos^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$. At $30°$: $\cot 30° = \sqrt{3}$, so $\cot^2 30° = 3$. Directly: $\frac{1 - \frac{1}{4}}{1 - \frac{3}{4}} = \frac{3/4}{1/4} = 3$.

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

1. At $x = 45°$, $\sin x + \cos x = \sqrt{2}$. What is the left side $(\sqrt{2})^2$? 2
2. The right side is 1. What is left side minus right side at $45°$? 1
3. Expanding gives $(\sin x + \cos x)^2 = 1 + 2\sin x\cos x$. What is $2\sin 45° \cos 45°$? 1

Answer: $2\tan x \sec x$

$\frac{(1 + \sin x) - (1 - \sin x)}{1 - \sin^2 x} = \frac{2\sin x}{\cos^2 x} = 2 \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = 2\tan x \sec x$. Check at $60°$: $\frac{1}{1 - 0.866} - \frac{1}{1.866} \approx 7.46 - 0.54 = 6.93$ and $2\tan 60° \sec 60° = 2 \cdot \sqrt{3} \cdot 2 \approx 6.93$. The choice $2\sec^2 x$ belongs to the sum of the two fractions, not the difference.

Answer: 2

$\sec^2 x - \tan^2 x = 1$, so the left side is $\sec^2 x + \tan^2 x = (1 + \tan^2 x) + \tan^2 x = 1 + 2\tan^2 x$. So $A = 2$. Check at $45°$: $4 - 1 = 3 = 1 + 2 \cdot 1$.

Answer: $\dfrac{\cos x\,(1 - \sin x)}{1 - \sin^2 x} = \dfrac{\cos x\,(1 - \sin x)}{\cos^2 x} = \dfrac{1 - \sin x}{\cos x}$

1. Which form of 1 should you multiply the left side by so that you can simplify the denominator with a Pythagorean identity? $\dfrac{1 - \sin x}{1 - \sin x}$
2. Which expression is the new denominator $(1 + \sin x)(1 - \sin x)$ equal to? $\cos^2 x$
3. The left side is now $\frac{\cos x\,(1 - \sin x)}{\cos^2 x}$. What do you get after cancelling one $\cos x$? $\dfrac{1 - \sin x}{\cos x}$
4. Check at $x = 0°$: what value do both sides give? 1

Answer: $\cot^2 x - \cos^2 x = \cos^2 x \cdot \dfrac{1 - \sin^2 x}{\sin^2 x} = \cos^2 x \cdot \dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x \cos^2 x$

1. Write $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$ and factor $\cos^2 x$ out of the left side. Which expression do you get? $\cos^2 x\left(\dfrac{1}{\sin^2 x} - 1\right)$
2. With a common denominator, $\frac{1}{\sin^2 x} - 1 = \frac{1 - \sin^2 x}{\sin^2 x}$. Which expression is the numerator $1 - \sin^2 x$ equal to? $\cos^2 x$
3. The left side is now $\cos^2 x \cdot \frac{\cos^2 x}{\sin^2 x}$. Which expression is this? $\cot^2 x \cos^2 x$
4. Check at $x = 45°$: what value do both sides give? 1/2

Answer: $(\sin^2 x - \cos^2 x) \cdot 1 = 1 - 2\cos^2 x$

1. Factor the left side. Which factorization is correct? $(\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$
2. What number is the second factor $\sin^2 x + \cos^2 x$? 1
3. Replace $\sin^2 x$ by $1 - \cos^2 x$ in $\sin^2 x - \cos^2 x$. You get $1 + k\cos^2 x$. What is $k$? -2
4. Check at $x = 60°$: what is $\sin^4 60° - \cos^4 60° = \frac{9}{16} - \frac{1}{16}$? 1/2