Trigonometric Identities
Master the fundamental equations that are always true
If you have made it this far in your study of trigonometry, you have already encountered several useful relationships between the trigonometric functions: the Pythagorean identity, the reciprocal identities, and the quotient identities. But here is an exciting truth: we have only scratched the surface. There is an entire universe of trigonometric identities, equations that are true for every valid input, and mastering them will transform how you work with trigonometric expressions.
You might wonder why we need so many identities. Think of them like a well-stocked toolbox. A carpenter does not just use one saw for everything; different tools are suited for different jobs. Similarly, some problems become trivial when you apply the right identity, while the same problem might seem impossible with the wrong approach. The sum and difference formulas let you break apart complicated angles. The double-angle formulas compress two angles into one. The half-angle formulas do the reverse. Each identity gives you a new way to see and simplify expressions.
By the end of this chapter, you will have a comprehensive toolkit of identities and, more importantly, you will know when and how to use each one.
Core Concepts
Review: The Fundamental Identities
Before we learn new identities, let us consolidate what we already know. These fundamental identities form the foundation for everything else.
The Pythagorean Identities:
From the unit circle, where any point $(\cos\theta, \sin\theta)$ satisfies $x^2 + y^2 = 1$:
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\cos^2\theta$ or $\sin^2\theta$ gives us two more forms:
$$\tan^2\theta + 1 = \sec^2\theta$$ $$1 + \cot^2\theta = \csc^2\theta$$
The Reciprocal Identities:
$$\csc\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta}$$
The Quotient Identities:
$$\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$$
These identities are your starting point for nearly every simplification and verification problem.
Cofunction Identities
Here is something interesting: the sine of an angle equals the cosine of its complement, and vice versa. This is where the “co” in cosine, cotangent, and cosecant comes from. The prefix means “complement.”
If two angles are complementary (they add up to $\frac{\pi}{2}$ or 90 degrees), their cofunctions are equal:
$$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$$ $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta$$ $$\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta$$ $$\cot\left(\frac{\pi}{2} - \theta\right) = \tan\theta$$ $$\sec\left(\frac{\pi}{2} - \theta\right) = \csc\theta$$ $$\csc\left(\frac{\pi}{2} - \theta\right) = \sec\theta$$
Why does this work? Consider a right triangle with acute angles $\theta$ and $\frac{\pi}{2} - \theta$. The side that is opposite to one angle is adjacent to the other. So the sine of one angle (opposite over hypotenuse) equals the cosine of the other (adjacent over hypotenuse).
This is why $\sin 30° = \cos 60° = \frac{1}{2}$ and $\sin 60° = \cos 30° = \frac{\sqrt{3}}{2}$. The angles are complements.
Sum and Difference Formulas
What if you need to find the sine or cosine of 75 degrees? It is not a special angle you have memorized. But notice that $75° = 45° + 30°$. If we had a formula for the sine of a sum, we could use what we know about 45 degrees and 30 degrees.
That is exactly what the sum and difference formulas give us:
Sine of a Sum or Difference: $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ $$\sin(A - B) = \sin A \cos B - \cos A \sin B$$
Cosine of a Sum or Difference: $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$
Tangent of a Sum or Difference: $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
Notice the patterns: for sine, the operations match (plus gives plus, minus gives minus). For cosine, the operations are opposite (plus gives minus, minus gives plus). The tangent formulas have the same pattern in the numerator as sine, but the denominator signs flip.
A helpful memory device: “Sine keeps the Sign, Cosine Changes.” For the sine formula, the sign between the two terms matches the sign in the argument. For cosine, the sign is the opposite.
Double-Angle Formulas
What happens when both angles in a sum formula are the same? We get the double-angle formulas:
Sine Double-Angle: $$\sin(2\theta) = 2\sin\theta\cos\theta$$
This comes directly from $\sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta$.
Cosine Double-Angle: $$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$
This formula has two alternative forms, obtained by using $\sin^2\theta + \cos^2\theta = 1$:
$$\cos(2\theta) = 2\cos^2\theta - 1$$ $$\cos(2\theta) = 1 - 2\sin^2\theta$$
Having three forms of the cosine double-angle formula is incredibly useful. You can choose the form that best fits your situation.
Tangent Double-Angle: $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$
Half-Angle Formulas
The half-angle formulas let you find trigonometric values for half of a known angle. They are derived by solving the double-angle formulas for the single angle.
Starting with $\cos(2\theta) = 2\cos^2\theta - 1$ and solving for $\cos\theta$:
$$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$$ $$\cos\theta = \pm\sqrt{\frac{1 + \cos(2\theta)}{2}}$$
Replacing $\theta$ with $\frac{\alpha}{2}$ (so $2\theta = \alpha$):
$$\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}$$
Similarly, from $\cos(2\theta) = 1 - 2\sin^2\theta$:
$$\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}$$
Tangent Half-Angle: $$\tan\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}}$$
There are also two alternative forms without the square root: $$\tan\frac{\alpha}{2} = \frac{\sin\alpha}{1 + \cos\alpha} = \frac{1 - \cos\alpha}{\sin\alpha}$$
Important: The $\pm$ sign in the half-angle formulas depends on the quadrant where $\frac{\alpha}{2}$ lies. You must determine this from the context of the problem.
Power-Reducing Formulas
Sometimes you need to rewrite powers of trigonometric functions as expressions involving only first powers. This is especially important in calculus when integrating. The power-reducing formulas come directly from rearranging the double-angle formulas:
$$\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$$
$$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$$
$$\tan^2\theta = \frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}$$
These formulas “reduce” a squared function to a non-squared one (at the cost of introducing a double angle). For higher powers, you apply these formulas repeatedly.
Verifying Identities: Strategies and Techniques
Verifying an identity means showing that the left side of an equation can be transformed into the right side (or vice versa, or both sides into the same thing). This is different from solving an equation. You are not finding values of the variable; you are proving the equation is true for all valid values.
Strategy 1: Work on the more complicated side.
Start with whichever side looks more complex and try to simplify it to match the other side.
Strategy 2: Convert everything to sines and cosines.
When you are stuck, replacing tangent, secant, cotangent, and cosecant with their sine and cosine definitions often reveals simplifications.
Strategy 3: Look for Pythagorean identity opportunities.
Whenever you see $\sin^2\theta + \cos^2\theta$, replace it with 1. Whenever you see $1 - \sin^2\theta$, replace it with $\cos^2\theta$. And so on.
Strategy 4: Factor when possible.
Expressions like $\sin^2\theta - \cos^2\theta$ can be factored as $(\sin\theta - \cos\theta)(\sin\theta + \cos\theta)$.
Strategy 5: Get common denominators.
When adding or subtracting fractions, combining them under a common denominator often reveals simplifications.
Strategy 6: Multiply by a clever form of 1.
If you see a troublesome expression like $\frac{1}{1 - \sin\theta}$, try multiplying by $\frac{1 + \sin\theta}{1 + \sin\theta}$. This creates a difference of squares in the denominator.
Important: Never move terms from one side to the other. That is solving, not verifying. In a verification, each side must be manipulated independently (or you can work from both sides toward a common middle).
Notation and Terminology
| Term | Symbol/Expression | Meaning |
|---|---|---|
| Identity | $\sin^2\theta + \cos^2\theta = 1$ | An equation true for all valid inputs |
| Cofunction | sin/cos, tan/cot, sec/csc | Pairs of functions related by complements |
| Sum formula | $\sin(A + B)$ | Formula for trig function of a sum of angles |
| Difference formula | $\cos(A - B)$ | Formula for trig function of a difference of angles |
| Double-angle formula | $\sin(2\theta)$ | Formula expressing a trig function of $2\theta$ in terms of $\theta$ |
| Half-angle formula | $\cos\frac{\alpha}{2}$ | Formula expressing a trig function of $\frac{\alpha}{2}$ in terms of $\alpha$ |
| Power-reducing formula | $\sin^2\theta = \ldots$ | Formula that rewrites powers of trig functions without the power |
| Verify/Prove | — | Show that both sides of an identity are equivalent |
Examples
Express $\sin 20°$ in terms of a trigonometric function of a different angle.
Solution: Using the cofunction identity: $$\sin\theta = \cos\left(90° - \theta\right)$$
We have: $$\sin 20° = \cos(90° - 20°) = \cos 70°$$
Answer: $\sin 20° = \cos 70°$
This shows why complementary angles have this special relationship. In a right triangle with a 20 degree angle, the side opposite 20 degrees is the same side that is adjacent to the 70 degree angle.
Find the exact value of $\sin 75°$.
Step 1: Recognize that $75° = 45° + 30°$.
Step 2: Apply the sine sum formula: $$\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30°$$
Step 3: Substitute the known values: $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$
Step 4: Simplify: $$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$
Answer: $\sin 75° = \frac{\sqrt{6} + \sqrt{2}}{4}$
If $\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant I, find $\sin(2\theta)$.
Step 1: First, find $\cos\theta$ using the Pythagorean identity: $$\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$ $$\cos\theta = \frac{4}{5} \quad \text{(positive because Quadrant I)}$$
Step 2: Apply the double-angle formula: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$$
Answer: $\sin(2\theta) = \frac{24}{25}$
Find the exact value of $\cos 22.5°$.
Step 1: Recognize that $22.5° = \frac{45°}{2}$.
Step 2: Apply the half-angle formula: $$\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}$$
With $\alpha = 45°$: $$\cos 22.5° = \pm\sqrt{\frac{1 + \cos 45°}{2}}$$
Step 3: Since $22.5°$ is in Quadrant I, cosine is positive.
Step 4: Substitute $\cos 45° = \frac{\sqrt{2}}{2}$: $$\cos 22.5° = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}}$$
Step 5: Simplify: $$\cos 22.5° = \frac{\sqrt{2 + \sqrt{2}}}{2}$$
Answer: $\cos 22.5° = \frac{\sqrt{2 + \sqrt{2}}}{2}$
Rewrite $\sin^4\theta$ in terms of first powers of cosine.
Step 1: Write $\sin^4\theta = (\sin^2\theta)^2$.
Step 2: Apply the power-reducing formula for $\sin^2\theta$: $$\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$$
So: $$\sin^4\theta = \left(\frac{1 - \cos(2\theta)}{2}\right)^2 = \frac{(1 - \cos(2\theta))^2}{4}$$
Step 3: Expand the numerator: $$= \frac{1 - 2\cos(2\theta) + \cos^2(2\theta)}{4}$$
Step 4: Apply the power-reducing formula to $\cos^2(2\theta)$: $$\cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2}$$
Step 5: Substitute: $$\sin^4\theta = \frac{1 - 2\cos(2\theta) + \frac{1 + \cos(4\theta)}{2}}{4}$$
Step 6: Simplify by getting a common denominator in the numerator: $$= \frac{\frac{2 - 4\cos(2\theta) + 1 + \cos(4\theta)}{2}}{4} = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}$$
Answer: $\sin^4\theta = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}$
Find the exact value of $\tan 15°$.
Step 1: Recognize that $15° = 45° - 30°$.
Step 2: Apply the tangent difference formula: $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
Step 3: Substitute $A = 45°$ and $B = 30°$: $$\tan 15° = \frac{\tan 45° - \tan 30°}{1 + \tan 45° \tan 30°} = \frac{1 - \frac{\sqrt{3}}{3}}{1 + 1 \cdot \frac{\sqrt{3}}{3}}$$
Step 4: Simplify the fractions: $$= \frac{\frac{3 - \sqrt{3}}{3}}{\frac{3 + \sqrt{3}}{3}} = \frac{3 - \sqrt{3}}{3 + \sqrt{3}}$$
Step 5: Rationalize by multiplying by $\frac{3 - \sqrt{3}}{3 - \sqrt{3}}$: $$= \frac{(3 - \sqrt{3})^2}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{9 - 6\sqrt{3} + 3}{9 - 3} = \frac{12 - 6\sqrt{3}}{6} = 2 - \sqrt{3}$$
Answer: $\tan 15° = 2 - \sqrt{3}$
Verify: $\frac{\sin(2\theta)}{\sin\theta} - \frac{\cos(2\theta)}{\cos\theta} = \sec\theta$
Step 1: Work on the left side. Replace $\sin(2\theta)$ and $\cos(2\theta)$ with their double-angle formulas: $$\frac{2\sin\theta\cos\theta}{\sin\theta} - \frac{\cos^2\theta - \sin^2\theta}{\cos\theta}$$
Step 2: Simplify the first fraction: $$2\cos\theta - \frac{\cos^2\theta - \sin^2\theta}{\cos\theta}$$
Step 3: Get a common denominator: $$\frac{2\cos^2\theta}{\cos\theta} - \frac{\cos^2\theta - \sin^2\theta}{\cos\theta} = \frac{2\cos^2\theta - \cos^2\theta + \sin^2\theta}{\cos\theta}$$
Step 4: Simplify the numerator: $$= \frac{\cos^2\theta + \sin^2\theta}{\cos\theta}$$
Step 5: Apply the Pythagorean identity: $$= \frac{1}{\cos\theta} = \sec\theta$$
Conclusion: The left side equals the right side, so the identity is verified.
Verify: $\frac{1 + \tan^2\theta}{1 - \tan^2\theta} = \frac{1}{\cos(2\theta)}$
Step 1: Work on the left side. Use the Pythagorean identity $1 + \tan^2\theta = \sec^2\theta$: $$\frac{\sec^2\theta}{1 - \tan^2\theta}$$
Step 2: Convert to sines and cosines: $$\frac{\frac{1}{\cos^2\theta}}{1 - \frac{\sin^2\theta}{\cos^2\theta}} = \frac{\frac{1}{\cos^2\theta}}{\frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta}}$$
Step 3: Simplify the complex fraction (dividing is the same as multiplying by the reciprocal): $$= \frac{1}{\cos^2\theta} \cdot \frac{\cos^2\theta}{\cos^2\theta - \sin^2\theta} = \frac{1}{\cos^2\theta - \sin^2\theta}$$
Step 4: Recognize the denominator as a double-angle formula: $$\cos^2\theta - \sin^2\theta = \cos(2\theta)$$
So: $$= \frac{1}{\cos(2\theta)}$$
Conclusion: The left side equals the right side, so the identity is verified.
Solve $\cos(2\theta) + \cos\theta = 0$ for $0 \leq \theta < 2\pi$.
Step 1: Replace $\cos(2\theta)$ using the double-angle formula $\cos(2\theta) = 2\cos^2\theta - 1$: $$2\cos^2\theta - 1 + \cos\theta = 0$$
Step 2: Rearrange: $$2\cos^2\theta + \cos\theta - 1 = 0$$
Step 3: This is a quadratic in $\cos\theta$. Let $u = \cos\theta$: $$2u^2 + u - 1 = 0$$
Step 4: Factor: $$(2u - 1)(u + 1) = 0$$
Step 5: Solve: $$u = \frac{1}{2} \quad \text{or} \quad u = -1$$ $$\cos\theta = \frac{1}{2} \quad \text{or} \quad \cos\theta = -1$$
Step 6: Find $\theta$:
- If $\cos\theta = \frac{1}{2}$: $\theta = \frac{\pi}{3}$ or $\theta = \frac{5\pi}{3}$
- If $\cos\theta = -1$: $\theta = \pi$
Answer: $\theta = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$
Key Properties and Rules
Complete Summary of Major Identities
Pythagorean Identities: $$\sin^2\theta + \cos^2\theta = 1$$ $$\tan^2\theta + 1 = \sec^2\theta$$ $$1 + \cot^2\theta = \csc^2\theta$$
Cofunction Identities (where $\theta$ is in radians): $$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta \qquad \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta$$ $$\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta \qquad \cot\left(\frac{\pi}{2} - \theta\right) = \tan\theta$$ $$\sec\left(\frac{\pi}{2} - \theta\right) = \csc\theta \qquad \csc\left(\frac{\pi}{2} - \theta\right) = \sec\theta$$
Sum and Difference Formulas: $$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$ $$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$ $$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$
Double-Angle Formulas: $$\sin(2\theta) = 2\sin\theta\cos\theta$$ $$\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$ $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$
Half-Angle Formulas: $$\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}$$ $$\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}$$ $$\tan\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} = \frac{\sin\alpha}{1 + \cos\alpha} = \frac{1 - \cos\alpha}{\sin\alpha}$$
Power-Reducing Formulas: $$\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$$ $$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$$ $$\tan^2\theta = \frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}$$
Memory Tips
- Sum formulas: “Sine keeps the Sign, Cosine Changes.” The sign between terms in sine matches the input; in cosine, it is opposite.
- Double-angle for sine: It is just the sum formula with $A = B = \theta$, so $2\sin\theta\cos\theta$.
- Double-angle for cosine: Remember one form, derive the others using $\sin^2\theta + \cos^2\theta = 1$.
- Half-angle formulas: Solve the double-angle formulas for the single-angle term.
- Power-reducing: Same as half-angle squared formulas, just written differently.
Real-World Applications
Signal Processing and Electronics
When electrical engineers analyze circuits with alternating current (AC), they deal with signals that vary sinusoidally. The power delivered by such a signal involves $\sin^2$ or $\cos^2$ terms. Using the power-reducing formula, engineers can express instantaneous power in terms of the double frequency: $$P = V_0 I_0 \sin^2(\omega t) = \frac{V_0 I_0}{2}(1 - \cos(2\omega t))$$
This reveals that power oscillates at twice the frequency of the voltage and current, a fundamental insight in electrical engineering.
Physics: Wave Interference
When two waves meet, they combine according to the sum of their displacements. If two sound waves or light waves with slightly different frequencies interfere, the result involves expressions like $\sin(A) + \sin(B)$. The sum-to-product identities (derived from sum and difference formulas) let physicists predict the “beat frequency” you hear when two similar musical notes are played together: $$\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
Computer Graphics: Rotation
When rotating an object in a video game or 3D modeling software, the new coordinates after rotating by angle $\theta$ are given by: $$x’ = x\cos\theta - y\sin\theta$$ $$y’ = x\sin\theta + y\cos\theta$$
If you need to rotate by $2\theta$, you could apply this twice, or you could use the double-angle formulas to compute the rotation directly. Game engines use these identities constantly to optimize performance.
Navigation and GPS
Aircraft navigation systems use great-circle routes, the shortest paths on a sphere. Calculating these paths requires extensive use of trigonometric identities, particularly the sum and difference formulas for cosine, to determine distances and bearings between two points on Earth’s surface.
Acoustics and Music
The sound of a musical instrument is not a pure sine wave but a combination of a fundamental frequency and its harmonics (multiples of the fundamental). Analyzing this sound mathematically involves expressing $\sin(n\theta)$ for various integers $n$ in terms of powers of $\sin\theta$ and $\cos\theta$, which requires the identities in this chapter.
Optics and Photography
Polarizing filters on cameras and sunglasses work by blocking light waves vibrating in certain directions. The intensity of light passing through a polarizer at angle $\theta$ follows Malus’s law: $I = I_0 \cos^2\theta$. Using the power-reducing formula, this can be rewritten as: $$I = \frac{I_0}{2}(1 + \cos(2\theta))$$
This form is often more useful for calculations in optical systems.
Self-Test Problems
Problem 1: Use a cofunction identity to find an equivalent expression for $\cos 35°$ involving a different trigonometric function.
Show Answer
Using the cofunction identity $\cos\theta = \sin(90° - \theta)$: $$\cos 35° = \sin(90° - 35°) = \sin 55°$$
Problem 2: Find the exact value of $\cos 105°$ using a sum or difference formula.
Show Answer
Recognize that $105° = 60° + 45°$.
Using the cosine sum formula: $$\cos 105° = \cos(60° + 45°) = \cos 60° \cos 45° - \sin 60° \sin 45°$$ $$= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$$ $$= \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$$
Problem 3: If $\cos\theta = -\frac{4}{5}$ and $\theta$ is in Quadrant II, find $\cos(2\theta)$.
Show Answer
Using the double-angle formula $\cos(2\theta) = 2\cos^2\theta - 1$: $$\cos(2\theta) = 2\left(-\frac{4}{5}\right)^2 - 1 = 2 \cdot \frac{16}{25} - 1 = \frac{32}{25} - \frac{25}{25} = \frac{7}{25}$$
Problem 4: Find the exact value of $\sin 15°$ using a half-angle formula.
Show Answer
Since $15° = \frac{30°}{2}$, use the half-angle formula with $\alpha = 30°$: $$\sin 15° = \sin\frac{30°}{2} = \sqrt{\frac{1 - \cos 30°}{2}}$$
(Positive since $15°$ is in Quadrant I.)
$$= \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$$
Note: This can also be written as $\frac{\sqrt{6} - \sqrt{2}}{4}$ using the difference formula for $\sin(45° - 30°)$.
Problem 5: Verify the identity: $\frac{\sin(2\theta)}{1 + \cos(2\theta)} = \tan\theta$
Show Answer
Work on the left side.
Step 1: Replace with double-angle formulas. Use $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\cos(2\theta) = 2\cos^2\theta - 1$: $$\frac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta}$$
Step 2: Simplify: $$= \frac{\sin\theta}{\cos\theta} = \tan\theta$$
The left side equals the right side, so the identity is verified.
Problem 6: Write $\cos^2(3\theta)$ without any powers using power-reducing formulas.
Show Answer
Using the power-reducing formula $\cos^2\phi = \frac{1 + \cos(2\phi)}{2}$ with $\phi = 3\theta$: $$\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}$$
Problem 7: Verify: $\tan\theta + \cot\theta = 2\csc(2\theta)$
Show Answer
Work on the left side.
Step 1: Convert to sines and cosines: $$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}$$
Step 2: Get a common denominator: $$= \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}$$
Step 3: Apply the Pythagorean identity to the numerator: $$= \frac{1}{\sin\theta\cos\theta}$$
Step 4: Recognize that $\sin(2\theta) = 2\sin\theta\cos\theta$, so $\sin\theta\cos\theta = \frac{\sin(2\theta)}{2}$: $$= \frac{1}{\frac{\sin(2\theta)}{2}} = \frac{2}{\sin(2\theta)} = 2\csc(2\theta)$$
The left side equals the right side, so the identity is verified.
Summary
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Cofunction identities relate the trigonometric function of an angle to its cofunction of the complementary angle. The sine of any angle equals the cosine of its complement, and this pattern extends to all six functions.
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Sum and difference formulas express $\sin(A \pm B)$, $\cos(A \pm B)$, and $\tan(A \pm B)$ in terms of the trig functions of $A$ and $B$ separately. These let you find exact values for angles like 75 degrees (45 degrees plus 30 degrees) or 15 degrees (45 degrees minus 30 degrees).
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Double-angle formulas are special cases of sum formulas where both angles are the same. The cosine double-angle formula has three equivalent forms, giving you flexibility in choosing which to use.
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Half-angle formulas let you find trig values for half of a known angle. The plus-or-minus sign must be determined from the quadrant where the half-angle lies.
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Power-reducing formulas rewrite squared trig functions without the exponent, useful for integration in calculus and for simplifying expressions.
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Verifying identities requires showing both sides of an equation are equivalent without moving terms across the equals sign. Key strategies include working on the more complicated side, converting to sines and cosines, looking for Pythagorean identity opportunities, factoring, getting common denominators, and multiplying by clever forms of 1.
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These identities have wide applications in physics (waves, optics, mechanics), engineering (signal processing, circuit analysis), computer graphics (rotations, animations), and navigation (GPS, flight paths).
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The more you practice with these identities, the more you will develop intuition for which identity to apply in a given situation. Like learning a new language, fluency comes with regular use.