Chapter 6: Trigonometric Identities & Equations
Trigonometric identities are equations that hold true for all values of the variable where both sides are defined. They form the backbone of trigonometric manipulation and are essential for simplifying expressions, evaluating integrals in calculus, and solving trigonometric equations. In this chapter, we systematically develop all major families of identities—from the fundamental Pythagorean identities through sum and difference formulas, double and half angle formulas, and beyond—and then apply them to verify identities and solve equations.
In This Chapter
6.1 Fundamental Identities
Every trigonometric identity ultimately traces back to a small set of fundamental relationships. These are the building blocks from which all other identities are derived. Mastering them is not optional—they must become second nature.
The first identity follows directly from the unit circle definition: if $(\cos\theta, \sin\theta)$ is a point on the unit circle $x^2 + y^2 = 1$, then $\cos^2\theta + \sin^2\theta = 1$. Dividing both sides by $\cos^2\theta$ gives the second identity, and dividing by $\sin^2\theta$ gives the third.
Simplify $\sec^2\theta - \tan^2\theta$.
Solution. From the second Pythagorean identity, $1 + \tan^2\theta = \sec^2\theta$, which rearranges to:
$$\sec^2\theta - \tan^2\theta = 1.$$No matter what value of $\theta$ we substitute (where the expression is defined), the result is always $1$.
Simplify $\dfrac{\cot\theta}{\csc\theta}$.
Solution. Convert everything to sine and cosine:
$$\frac{\cot\theta}{\csc\theta} = \frac{\frac{\cos\theta}{\sin\theta}}{\frac{1}{\sin\theta}} = \frac{\cos\theta}{\sin\theta} \cdot \sin\theta = \cos\theta.$$6.2 Sum and Difference Formulas
The sum and difference formulas allow us to find the exact trigonometric values of angles that are sums or differences of known angles (such as $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$). They are also essential for deriving the double angle and half angle formulas.
Note the sign pattern carefully: in the cosine formula, the sign in the result is opposite to the sign in the argument. In the sine and tangent formulas, the sign in the result matches the sign in the argument.
Find the exact value of $\sin(75^\circ)$.
Solution. Write $75^\circ = 45^\circ + 30^\circ$ and apply the sine sum formula:
$$\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$Find the exact value of $\cos(15^\circ)$.
Solution. Write $15^\circ = 45^\circ - 30^\circ$:
$$\cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$Notice that $\sin(75^\circ) = \cos(15^\circ)$, which confirms the cofunction identity since $75^\circ + 15^\circ = 90^\circ$.
Find the exact value of $\tan\left(\frac{7\pi}{12}\right)$.
Solution. Write $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$:
$$\tan\left(\frac{7\pi}{12}\right) = \frac{\tan\frac{\pi}{3} + \tan\frac{\pi}{4}}{1 - \tan\frac{\pi}{3}\tan\frac{\pi}{4}} = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}.$$Rationalize by multiplying by $\frac{1 + \sqrt{3}}{1 + \sqrt{3}}$:
$$= \frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})} = \frac{(\sqrt{3}+1)^2}{1-3} = \frac{3 + 2\sqrt{3} + 1}{-2} = \frac{4 + 2\sqrt{3}}{-2} = -(2+\sqrt{3}).$$6.3 Double Angle Formulas
The double angle formulas are special cases of the sum formulas with $B = A$. They arise frequently in calculus, physics, and engineering, and are among the most commonly tested identities.
The cosine double angle formula has three equivalent forms. Choose the form that best suits the problem: if you know $\cos A$, use $2\cos^2 A - 1$; if you know $\sin A$, use $1 - 2\sin^2 A$; if you know both, use $\cos^2 A - \sin^2 A$.
Derivation. Set $B = A$ in the sum formulas. For sine: $\sin(A + A) = \sin A\cos A + \cos A\sin A = 2\sin A\cos A$. For cosine: $\cos(A + A) = \cos A\cos A - \sin A\sin A = \cos^2 A - \sin^2 A$. Substituting $\sin^2 A = 1 - \cos^2 A$ gives $2\cos^2 A - 1$; substituting $\cos^2 A = 1 - \sin^2 A$ gives $1 - 2\sin^2 A$.
If $\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant II, find $\sin(2\theta)$ and $\cos(2\theta)$.
Solution. Since $\theta$ is in QII, $\cos\theta < 0$. From the Pythagorean identity:
$$\cos\theta = -\sqrt{1 - \sin^2\theta} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}.$$Now apply the double angle formulas:
$$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = -\frac{24}{25}$$ $$\cos(2\theta) = \cos^2\theta - \sin^2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.$$6.4 Half Angle Formulas
The half angle formulas express trigonometric functions of $\frac{A}{2}$ in terms of trigonometric functions of $A$. They are derived directly from the double angle formulas for cosine by solving for $\sin\frac{A}{2}$ or $\cos\frac{A}{2}$.
The $\pm$ sign in the sine and cosine formulas is determined by the quadrant in which $\frac{A}{2}$ lies. The tangent half angle formula has no ambiguity because the two equivalent forms automatically give the correct sign.
Derivation. Start from $\cos(2\alpha) = 1 - 2\sin^2\alpha$. Let $\alpha = \frac{A}{2}$, so $\cos A = 1 - 2\sin^2\frac{A}{2}$. Solving: $\sin^2\frac{A}{2} = \frac{1 - \cos A}{2}$, hence $\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}$. The cosine formula follows similarly from $\cos(2\alpha) = 2\cos^2\alpha - 1$.
Find the exact value of $\cos(22.5^\circ)$.
Solution. Since $22.5^\circ = \frac{45^\circ}{2}$, apply the half angle formula with $A = 45^\circ$:
$$\cos(22.5^\circ) = +\sqrt{\frac{1 + \cos 45^\circ}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}.$$We chose the positive root because $22.5^\circ$ is in Quadrant I, where cosine is positive.
6.5 Power-Reducing Formulas
Power-reducing formulas express $\sin^2\theta$ and $\cos^2\theta$ (and higher powers) in terms of first powers of cosine at double the angle. These are indispensable in calculus when integrating even powers of sine and cosine.
These follow immediately from the double angle formulas for cosine. From $\cos(2\theta) = 1 - 2\sin^2\theta$, solving for $\sin^2\theta$ gives the first formula. From $\cos(2\theta) = 2\cos^2\theta - 1$, solving for $\cos^2\theta$ gives the second. Dividing the first by the second gives the third.
Rewrite $\sin^4\theta$ without any power greater than 1.
Solution. Apply the power-reducing formula twice:
$$\sin^4\theta = \left(\sin^2\theta\right)^2 = \left(\frac{1 - \cos(2\theta)}{2}\right)^2 = \frac{1 - 2\cos(2\theta) + \cos^2(2\theta)}{4}.$$Now reduce $\cos^2(2\theta)$ using the same technique with angle $2\theta$:
$$\cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2}.$$Substituting:
$$\sin^4\theta = \frac{1 - 2\cos(2\theta) + \frac{1 + \cos(4\theta)}{2}}{4} = \frac{2 - 4\cos(2\theta) + 1 + \cos(4\theta)}{8} = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}.$$6.6 Product-to-Sum and Sum-to-Product Formulas
Product-to-sum formulas convert a product of two trigonometric functions into a sum (or difference), while sum-to-product formulas do the reverse. These are derived by adding or subtracting the sum and difference formulas.
Derivation sketch. For the product-to-sum formulas, add the identities $\sin(A+B) = \sin A\cos B + \cos A\sin B$ and $\sin(A-B) = \sin A\cos B - \cos A\sin B$ to get $\sin(A+B) + \sin(A-B) = 2\sin A\cos B$, which gives the first formula after dividing by 2. The sum-to-product formulas follow by substituting $A = \frac{C+D}{2}$ and $B = \frac{C-D}{2}$.
Express $\sin(5x)\cos(3x)$ as a sum.
Solution. Apply the product-to-sum formula with $A = 5x$ and $B = 3x$:
$$\sin(5x)\cos(3x) = \frac{1}{2}[\sin(5x+3x) + \sin(5x-3x)] = \frac{1}{2}[\sin(8x) + \sin(2x)].$$Express $\cos(4\theta) - \cos(2\theta)$ as a product.
Solution. Apply the sum-to-product formula with $C = 4\theta$ and $D = 2\theta$:
$$\cos(4\theta) - \cos(2\theta) = -2\sin\frac{4\theta+2\theta}{2}\sin\frac{4\theta-2\theta}{2} = -2\sin(3\theta)\sin(\theta).$$6.7 Verifying Trigonometric Identities
Verifying an identity means showing that the left side and right side of an equation are equal for all values in the domain. Unlike solving an equation, you do not perform the same operation on both sides. Instead, you transform one side (or both sides independently) until they match.
- Work with the more complicated side. Transform it to match the simpler side.
- Convert to sines and cosines. When in doubt, rewrite everything in terms of $\sin\theta$ and $\cos\theta$.
- Factor and simplify. Look for common factors, difference of squares ($a^2 - b^2$), or perfect squares.
- Multiply by a conjugate. Expressions like $1 - \sin\theta$ or $1 + \cos\theta$ often benefit from conjugate multiplication.
- Use Pythagorean substitutions. Replace $\sin^2\theta$ with $1 - \cos^2\theta$ (or vice versa) to create opportunities for simplification.
Verify: $\tan\theta + \cot\theta = \sec\theta\csc\theta$.
Solution. Work with the left side:
$$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} = \sec\theta\csc\theta. \quad \checkmark$$Verify: $\dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$.
Solution. Start with the left side and multiply numerator and denominator by the conjugate $1 - \cos\theta$:
$$\frac{\sin\theta}{1 + \cos\theta} \cdot \frac{1 - \cos\theta}{1 - \cos\theta} = \frac{\sin\theta(1 - \cos\theta)}{1 - \cos^2\theta} = \frac{\sin\theta(1 - \cos\theta)}{\sin^2\theta} = \frac{1 - \cos\theta}{\sin\theta}. \quad \checkmark$$Verify: $\cos^4\theta - \sin^4\theta = \cos(2\theta)$.
Solution. Factor the left side as a difference of squares:
$$\cos^4\theta - \sin^4\theta = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta) = 1 \cdot (\cos^2\theta - \sin^2\theta) = \cos(2\theta). \quad \checkmark$$6.8 Solving Trigonometric Equations
Unlike identities (which are true for all valid inputs), trigonometric equations are true only for specific values. Solving them requires combining algebraic techniques with knowledge of the unit circle and the periodic nature of trig functions.
Linear Trigonometric Equations
A linear trigonometric equation involves a single trig function to the first power. The strategy is to isolate the trig function, then use inverse functions and the unit circle to find all solutions.
Solve $2\sin\theta - 1 = 0$ for $0 \le \theta < 2\pi$.
Solution. Isolate $\sin\theta$:
$$\sin\theta = \frac{1}{2}.$$On the unit circle, $\sin\theta = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ (QI) and $\theta = \frac{5\pi}{6}$ (QII).
The solution set is $\theta \in \left\{\dfrac{\pi}{6},\, \dfrac{5\pi}{6}\right\}$.
For the general solution (all real numbers): $\theta = \frac{\pi}{6} + 2n\pi$ or $\theta = \frac{5\pi}{6} + 2n\pi$, where $n$ is any integer.
Quadratic Trigonometric Equations
When a trig equation involves squared trig functions, treat the trig function as a variable and solve the resulting quadratic. Factor or use the quadratic formula, then solve for the angle.
Solve $2\cos^2\theta - \cos\theta - 1 = 0$ for $0 \le \theta < 2\pi$.
Solution. Let $u = \cos\theta$. The equation becomes $2u^2 - u - 1 = 0$, which factors as:
$$(2u + 1)(u - 1) = 0.$$So $u = -\frac{1}{2}$ or $u = 1$, meaning $\cos\theta = -\frac{1}{2}$ or $\cos\theta = 1$.
- $\cos\theta = -\frac{1}{2}$: $\theta = \frac{2\pi}{3}$ or $\theta = \frac{4\pi}{3}$.
- $\cos\theta = 1$: $\theta = 0$.
Solution set: $\theta \in \left\{0,\, \dfrac{2\pi}{3},\, \dfrac{4\pi}{3}\right\}$.
Equations with Multiple Angles
When the argument of the trig function is a multiple of $\theta$ (such as $2\theta$ or $3\theta$), solve for the multiple angle first, then divide to find $\theta$. Be careful to account for extra solutions that arise from the expanded period.
Solve $\sin(2\theta) = \frac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$.
Solution. Let $u = 2\theta$. Since $0 \le \theta < 2\pi$, we need $0 \le u < 4\pi$.
Solve $\sin u = \frac{\sqrt{3}}{2}$. In the interval $[0, 4\pi)$, the solutions are:
$$u = \frac{\pi}{3},\quad \frac{2\pi}{3},\quad \frac{\pi}{3} + 2\pi = \frac{7\pi}{3},\quad \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}.$$Divide each by 2 to recover $\theta$:
$$\theta = \frac{\pi}{6},\quad \frac{\pi}{3},\quad \frac{7\pi}{6},\quad \frac{4\pi}{3}.$$Solve $\sin(2\theta) = \cos\theta$ for $0 \le \theta < 2\pi$.
Solution. Replace $\sin(2\theta)$ with $2\sin\theta\cos\theta$:
$$2\sin\theta\cos\theta = \cos\theta.$$Move all terms to one side and factor:
$$2\sin\theta\cos\theta - \cos\theta = 0$$ $$\cos\theta(2\sin\theta - 1) = 0.$$Set each factor to zero:
- $\cos\theta = 0$: $\theta = \frac{\pi}{2}$ or $\theta = \frac{3\pi}{2}$.
- $\sin\theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$.
Solution set: $\theta \in \left\{\dfrac{\pi}{6},\, \dfrac{\pi}{2},\, \dfrac{5\pi}{6},\, \dfrac{3\pi}{2}\right\}$.
6.9 Practice Problems
Work through the following problems to solidify your understanding. Each covers a technique from this chapter. Try each problem on your own before revealing the solution.
Find the exact value of $\sin(105^\circ)$.
Show Solution
Write $105^\circ = 60^\circ + 45^\circ$ and apply the sine sum formula:
$$\sin(105^\circ) = \sin 60^\circ\cos 45^\circ + \cos 60^\circ\sin 45^\circ = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$Verify the identity: $\dfrac{1 - \cos(2\theta)}{\sin(2\theta)} = \tan\theta$.
Show Solution
Apply double angle formulas to the left side: $1 - \cos(2\theta) = 2\sin^2\theta$ and $\sin(2\theta) = 2\sin\theta\cos\theta$:
$$\frac{1 - \cos(2\theta)}{\sin(2\theta)} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta. \quad \checkmark$$If $\cos\alpha = -\frac{5}{13}$ and $\alpha$ is in Quadrant III, find $\sin(2\alpha)$.
Show Solution
In QIII, $\sin\alpha < 0$. From the Pythagorean identity:
$$\sin\alpha = -\sqrt{1 - \cos^2\alpha} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}.$$Apply the double angle formula:
$$\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2\left(-\frac{12}{13}\right)\left(-\frac{5}{13}\right) = \frac{120}{169}.$$Find the exact value of $\tan(22.5^\circ)$ using a half angle formula.
Show Solution
Use the half angle formula for tangent with $A = 45^\circ$:
$$\tan(22.5^\circ) = \tan\frac{45^\circ}{2} = \frac{1 - \cos 45^\circ}{\sin 45^\circ} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\frac{2 - \sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}}.$$Rationalize:
$$= \frac{(2 - \sqrt{2})\sqrt{2}}{2} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1.$$Solve $2\sin^2\theta + \sin\theta - 1 = 0$ for $0 \le \theta < 2\pi$.
Show Solution
Let $u = \sin\theta$. Factor the quadratic $2u^2 + u - 1 = 0$:
$$(2u - 1)(u + 1) = 0.$$So $\sin\theta = \frac{1}{2}$ or $\sin\theta = -1$.
- $\sin\theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$.
- $\sin\theta = -1$: $\theta = \frac{3\pi}{2}$.
Solution set: $\theta \in \left\{\dfrac{\pi}{6},\, \dfrac{5\pi}{6},\, \dfrac{3\pi}{2}\right\}$.
Verify: $\sec\theta - \cos\theta = \sin\theta\tan\theta$.
Show Solution
Work with the left side, converting to sine and cosine:
$$\sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} = \sin\theta \cdot \frac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta. \quad \checkmark$$Express $\cos(3\theta)\cos(\theta)$ as a sum of trig functions.
Show Solution
Apply the product-to-sum formula with $A = 3\theta$ and $B = \theta$:
$$\cos(3\theta)\cos(\theta) = \frac{1}{2}[\cos(3\theta - \theta) + \cos(3\theta + \theta)] = \frac{1}{2}[\cos(2\theta) + \cos(4\theta)].$$Solve $\cos(2\theta) + \cos\theta = 0$ for $0 \le \theta < 2\pi$.
Show Solution
Replace $\cos(2\theta)$ with $2\cos^2\theta - 1$:
$$2\cos^2\theta - 1 + \cos\theta = 0 \quad \Rightarrow \quad 2\cos^2\theta + \cos\theta - 1 = 0.$$Factor: $(2\cos\theta - 1)(\cos\theta + 1) = 0$.
- $\cos\theta = \frac{1}{2}$: $\theta = \frac{\pi}{3}$ or $\theta = \frac{5\pi}{3}$.
- $\cos\theta = -1$: $\theta = \pi$.
Solution set: $\theta \in \left\{\dfrac{\pi}{3},\, \pi,\, \dfrac{5\pi}{3}\right\}$.
Rewrite $\cos^2\theta\sin^2\theta$ using only first powers of cosine functions (no squared terms).
Show Solution
Use the power-reducing formulas:
$$\cos^2\theta\sin^2\theta = \frac{1 + \cos(2\theta)}{2} \cdot \frac{1 - \cos(2\theta)}{2} = \frac{1 - \cos^2(2\theta)}{4}.$$Apply the power-reducing formula again to $\cos^2(2\theta)$:
$$= \frac{1 - \frac{1 + \cos(4\theta)}{2}}{4} = \frac{\frac{2 - 1 - \cos(4\theta)}{2}}{4} = \frac{1 - \cos(4\theta)}{8}.$$Alternatively, note that $\cos^2\theta\sin^2\theta = \frac{1}{4}\sin^2(2\theta) = \frac{1}{4} \cdot \frac{1 - \cos(4\theta)}{2} = \frac{1 - \cos(4\theta)}{8}$.
Solve $\tan^2\theta - 3 = 0$ for $0 \le \theta < 2\pi$.
Show Solution
Isolate $\tan^2\theta$:
$$\tan^2\theta = 3 \quad \Rightarrow \quad \tan\theta = \pm\sqrt{3}.$$- $\tan\theta = \sqrt{3}$: $\theta = \frac{\pi}{3}$ (QI) or $\theta = \frac{4\pi}{3}$ (QIII).
- $\tan\theta = -\sqrt{3}$: $\theta = \frac{2\pi}{3}$ (QII) or $\theta = \frac{5\pi}{3}$ (QIV).
Solution set: $\theta \in \left\{\dfrac{\pi}{3},\, \dfrac{2\pi}{3},\, \dfrac{4\pi}{3},\, \dfrac{5\pi}{3}\right\}$.
Verify: $\dfrac{\sin(A+B)}{\sin(A-B)} = \dfrac{\tan A + \tan B}{\tan A - \tan B}$.
Show Solution
Work with the right side. Write $\tan A = \frac{\sin A}{\cos A}$ and $\tan B = \frac{\sin B}{\cos B}$:
$$\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}} = \frac{\frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B}}{\frac{\sin A\cos B - \cos A\sin B}{\cos A\cos B}}.$$The $\cos A\cos B$ denominators cancel:
$$= \frac{\sin A\cos B + \cos A\sin B}{\sin A\cos B - \cos A\sin B} = \frac{\sin(A+B)}{\sin(A-B)}. \quad \checkmark$$Solve $\sin\theta + \cos\theta = 1$ for $0 \le \theta < 2\pi$.
Show Solution
Square both sides (we will check for extraneous solutions):
$$\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1$$ $$1 + \sin(2\theta) = 1$$ $$\sin(2\theta) = 0.$$So $2\theta = 0, \pi, 2\pi, 3\pi$, giving $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$.
Check each in the original equation $\sin\theta + \cos\theta = 1$:
- $\theta = 0$: $0 + 1 = 1$. Valid.
- $\theta = \frac{\pi}{2}$: $1 + 0 = 1$. Valid.
- $\theta = \pi$: $0 + (-1) = -1 \ne 1$. Extraneous.
- $\theta = \frac{3\pi}{2}$: $-1 + 0 = -1 \ne 1$. Extraneous.
Solution set: $\theta \in \left\{0,\, \dfrac{\pi}{2}\right\}$.
Chapter Summary
In this chapter, we developed a comprehensive toolkit for working with trigonometric expressions and equations:
- Fundamental Identities (Pythagorean, reciprocal, quotient, even-odd, cofunction) form the foundation for all trigonometric manipulation.
- Sum and Difference Formulas express $\sin(A \pm B)$, $\cos(A \pm B)$, and $\tan(A \pm B)$ in terms of functions of $A$ and $B$ individually, enabling exact evaluation of non-standard angles.
- Double Angle Formulas are the special case $B = A$ and appear throughout calculus and physics.
- Half Angle Formulas allow computation of trig values at half of known angles, derived by inverting the double angle formulas.
- Power-Reducing Formulas eliminate even powers of sine and cosine, which is essential for integration in calculus.
- Product-to-Sum and Sum-to-Product Formulas convert between products and sums of trig functions, useful in signal processing and Fourier analysis.
- Verifying Identities requires transforming one side of an equation to match the other, using strategies like converting to sine/cosine, factoring, and conjugate multiplication.
- Solving Trig Equations combines algebraic techniques (factoring, quadratic formula) with unit circle knowledge, and requires careful attention to domain and periodicity.
Mastery of these identities and techniques prepares you for calculus, where they are used extensively in differentiation, integration, and series expansion. In the next chapter, we explore vectors and polar coordinates, which provide alternative ways to represent points and curves in the plane.
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