7.3 Trigonometric Identities

7.3 Trigonometric Identities

An identity is an equation true for every value in the domain of the variable — unlike a conditional equation, which is true only for particular values. Identities are the algebra of trigonometry: they let you rewrite one function in terms of another, collapse a compound angle into known values, and reduce a messy expression to something you can evaluate by hand. Because Section 2 of the exam bans calculators, identity fluency is what makes exact-value questions solvable at all.

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1. The Fundamental Identities

Reciprocal and Quotient Identities

$\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta}$

$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}$

Pythagorean Identities

These come directly from the unit circle equation $x^2 + y^2 = 1$ with $x = \cos\theta$ and $y = \sin\theta$:

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

Dividing every term by $\cos^2\theta$ yields: $1 + \tan^2\theta = \sec^2\theta$

Dividing every term by $\sin^2\theta$ yields: $1 + \cot^2\theta = \csc^2\theta$

Memorize the first and derive the other two — that takes five seconds and eliminates sign errors.

Even-Odd Identities

$\cos(-\theta) = \cos\theta \quad (\text{even}); \qquad \sin(-\theta) = -\sin\theta, \quad \tan(-\theta) = -\tan\theta \quad (\text{odd})$

Cosine and secant are even; the other four are odd. This is why $\cos$ graphs are symmetric about the $y$-axis and $\sin$ graphs about the origin.

Co-Function Identities

$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta, \quad \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta, \quad \tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta$

Each function equals the co-function of the complementary angle — the reason $\sin 30^\circ = \cos 60^\circ = \frac{1}{2}$.

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2. Sum and Difference Formulas

\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 \\ \tan(\alpha \pm \beta) &= \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta} \end{aligned}$$ Note the **opposite sign** in the cosine formula: a sum inside produces a minus outside. Students lose more points to this single sign than to any other identity. #### Worked Example 7.3A: Exact value of a non-special angle Find the exact value of $\cos(105^\circ)$. *Solution*: 1. Decompose $105^\circ$ into special angles: $105^\circ = 45^\circ + 60^\circ$. 2. Apply the cosine sum formula: $$\cos(45^\circ + 60^\circ) = \cos 45^\circ \cos 60^\circ - \sin 45^\circ \sin 60^\circ$$ 3. Substitute exact values: $$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$$ Sanity check: $105^\circ$ lies in Quadrant II, where cosine is negative, and $\sqrt{2} < \sqrt{6}$ makes the result negative. ✔ --- ## 3. Double-Angle and Half-Angle Formulas ### Double-Angle $$\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}$$ The three forms of $\cos(2\theta)$ are interchangeable via $\sin^2\theta + \cos^2\theta = 1$. Choose whichever form matches the function already in your problem — that is the entire trick. ### Half-Angle $$\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{2}}, \quad \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos\theta}{2}}$$ The $\pm$ is **not** ambiguity — it is resolved by the quadrant containing $\frac{\theta}{2}$, which you must determine before choosing. #### Worked Example 7.3B: Double-angle with quadrant reasoning If $\sin\theta = \frac{3}{5}$ and $\theta$ lies in Quadrant II, find $\cos(2\theta)$. *Solution*: 1. Use the form of $\cos(2\theta)$ that needs only sine: $\cos(2\theta) = 1 - 2\sin^2\theta$. 2. Substitute: $1 - 2\left(\frac{3}{5}\right)^2 = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$. Because this form uses only $\sin\theta$, the quadrant never enters the computation — a genuine shortcut. Had you chosen $\cos(2\theta) = 2\cos^2\theta - 1$, you would first need $\cos\theta = -\frac{4}{5}$ (negative in Quadrant II), and squaring would erase the sign anyway: $2\left(\frac{16}{25}\right) - 1 = \frac{7}{25}$. ✔ Same answer, more work. --- ## 4. Verifying an Identity To *verify* an identity, transform one side until it matches the other. Never move terms across the equals sign as though solving — you are proving, not solving. **Strategy:** start with the more complicated side; convert everything to sines and cosines; combine fractions over a common denominator; look for a Pythagorean substitution. #### Worked Example 7.3C Verify that $\dfrac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta$. *Solution*: Work the left side and multiply by the conjugate. $$\frac{\cos\theta}{1 - \sin\theta} \cdot \frac{1 + \sin\theta}{1 + \sin\theta} = \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta}$$ The denominator is $\cos^2\theta$ by the Pythagorean identity: $$= \frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta$$ Verified. --- ## 5. Identity Reference Table | Family | Identity | Most common use | | :--- | :--- | :--- | | Pythagorean | $\sin^2\theta + \cos^2\theta = 1$ | Convert between sine and cosine | | Pythagorean | $1 + \tan^2\theta = \sec^2\theta$ | Simplify expressions in $\sec$ | | Sum/Difference | $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ | Exact values of $15^\circ$, $75^\circ$, $105^\circ$ | | Double-angle | $\sin(2\theta) = 2\sin\theta\cos\theta$ | Collapse a product into one function | | Double-angle | $\cos(2\theta) = 1 - 2\sin^2\theta$ | When only $\sin\theta$ is known | | Co-function | $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$ | Complementary-angle questions | --- ## 6. CLEP Exam Traps & Common Errors - **Trap 1: The cosine sum sign.** $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$. The inner plus becomes an outer minus. - **Trap 2: Distributing a function over addition.** $\sin(\alpha + \beta) \neq \sin\alpha + \sin\beta$. Trigonometric functions are not linear; test it with $\alpha = \beta = \frac{\pi}{2}$, where the left side is $0$ and the right side is $2$. - **Trap 3: Misreading $\sin^2\theta$.** It means $(\sin\theta)^2$, not $\sin(\theta^2)$ and not $\sin(\sin\theta)$. - **Trap 4: Dropping the half-angle sign decision.** The $\pm$ must be resolved from the quadrant of $\frac{\theta}{2}$, which is *not* the quadrant of $\theta$. If $\theta = 300^\circ$ (Quadrant IV), then $\frac{\theta}{2} = 150^\circ$ sits in Quadrant II. - **Trap 5: "Solving" a verification.** Cross-multiplying across the equals sign assumes what you are trying to prove. Transform one side only.