Analytic Trigonometry — Free Pre-Calculus Review Games.

The fundamental Pythagorean identity is sin^2(x) + cos^2(x) = 1.

By definition, tan(x) = sin(x)/cos(x).

Secant is the reciprocal of cosine: sec(x) = 1/cos(x).

\(\cos(x) = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5\) (positive in Q1).

Cosecant is the reciprocal of sine: csc(x) = 1/sin(x).

The sum formula for sine: sin(A+B) = sin(A)cos(B) + cos(A)sin(B).

cos(A-B) = cos(A)cos(B) + sin(A)sin(B).

The double-angle formula: sin(2x) = 2*sin(x)*cos(x).

One form of the double-angle formula: cos(2x) = cos^2(x) - sin^2(x).

From sin^2+cos^2=1: cos^2(x) = 1-sin^2(x).

tan(x)*cos(x) = (sin(x)/cos(x))*cos(x) = sin(x).

\(\sin(45)\cos(30)+\cos(45)\sin(30) = (\sqrt{2}/2)(\sqrt{3}/2)+(\sqrt{2}/2)(1/2) = (\sqrt{6}+\sqrt{2})/4\).

Pythagorean identity: 1 + tan^2(x) = sec^2(x).

sin(x) = 1/2. In [0,2pi): x = pi/6 and 5*pi/6.

Using sin^2=1-cos^2: cos(2x) = cos^2-sin^2 = 2cos^2(x)-1.

The sum formula for cosine is derived from the unit circle and angle rotation, giving \(\cos(A+B)=\cos A\cos B-\sin A\sin B\). The choice "\(\cos A\cos B + \sin A\sin B\)" is actually the formula for \(\cos(A-B)\), not \(\cos(A+B)\), so it is incorrect here. Students should memorize that cosine sum and difference formulas swap the sign compared to the operation in the angle.

The difference formula for sine keeps the same sign as the operation, so \(\sin(A-B)=\sin A\cos B-\cos A\sin B\). The option "\(\sin A\cos B + \cos A\sin B\)" is the formula for \(\sin(A+B)\), so it does not apply to a difference of angles. A helpful rule: for sine, the sign inside the formula matches the sign in the angle expression.

The tangent sum formula comes from dividing the sine sum formula by the cosine sum formula and simplifying, yielding \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\). The choice "\(\dfrac{\tan A-\tan B}{1+\tan A\tan B}\)" is actually the formula for \(\tan(A-B)\), so it belongs to the wrong operation. Remember that the denominator sign is always opposite the numerator's operation sign in tangent sum/difference formulas.

Complementary angles satisfy \(\sin(90^\circ-x)=\cos(x)\) because sine of an angle equals cosine of its complement on the unit circle. The choice "\(\sin(x)\)" ignores the complementary relationship entirely and would only be true if no subtraction occurred. Cofunction identities are essential shortcuts for rewriting expressions involving complementary angles.

By the cofunction relationship, \(\cos(90^\circ-x)=\sin(x)\) since cosine of an angle equals sine of its complement. The choice "\(-\sin(x)\)" incorrectly introduces a negative sign that has no basis in the complementary angle derivation. This pairing of sine and cosine as cofunctions is a foundational tool for simplifying trigonometric expressions.

Cosine is an even function because the unit circle x-coordinate is symmetric about the x-axis, so \(\cos(-x)=\cos(x)\). The option "\(\cos(-x)=-\cos(x)\)" describes odd symmetry, which applies to sine, not cosine. Knowing which trig functions are even versus odd helps simplify expressions with negative angles quickly.

Sine is an odd function because the unit circle y-coordinate flips sign when the angle is negated, giving \(\sin(-x)=-\sin(x)\). The choice "\(\sin(-x)=\sin(x)\)" describes even symmetry, which is a property of cosine, not sine. Recognizing even/odd symmetry is a quick way to simplify negative-angle expressions on exams.

The tangent double-angle formula comes from applying the sum formula with \(A=B=x\), giving \(\tan(2x)=\dfrac{2\tan(x)}{1-\tan^2(x)}\). The option "\(\dfrac{2\tan(x)}{1+\tan^2(x)}\)" incorrectly uses a plus sign in the denominator, which does not follow from the derivation. Students should recall that this formula is undefined whenever \(\tan(x)=\pm1\), since the denominator becomes zero.

Substituting \(\cos^2(x)=1-\sin^2(x)\) into \(\cos(2x)=\cos^2(x)-\sin^2(x)\) produces \(\cos(2x)=1-2\sin^2(x)\). The choice "\(2\sin^2(x)-1\)" reverses the sign and actually equals \(-\cos(2x)\), not \(\cos(2x)\). There are three equivalent forms of the cosine double-angle formula, and choosing the right one depends on which function's value is known.

Cotangent is defined as the reciprocal of tangent, so \(\cot(x)=\dfrac{\cos(x)}{\sin(x)}\). The choice "\(\dfrac{\sin(x)}{\cos(x)}\)" is actually the definition of \(\tan(x)\), the reciprocal relationship of cotangent. Keeping reciprocal and quotient identities straight is essential for simplifying complex trigonometric expressions.

Dividing \(\sin^2(x)+\cos^2(x)=1\) by \(\sin^2(x)\) produces \(1+\cot^2(x)=\csc^2(x)\), a standard Pythagorean identity. The choice "\(1+\tan^2(x)=\csc^2(x)\)" mixes tangent into an identity that actually pairs with secant, not cosecant. There are three Pythagorean identities total, and each pairs specific reciprocal functions together.

The Pythagorean identity \(1+\tan^2(x)=\sec^2(x)\) rearranges directly to \(\sec^2(x)-\tan^2(x)=1\). The choice "\(-1\)" reverses the sign, which would only occur if the terms were subtracted in the opposite order. This identity is frequently used to simplify or verify more complex trigonometric expressions involving secant and tangent.

The tangent difference formula is \(\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\), derived from dividing the sine and cosine difference formulas. The choice "\(\dfrac{\tan A+\tan B}{1-\tan A\tan B}\)" is actually the formula for \(\tan(A+B)\), so it applies to the wrong operation. Always check that the denominator sign is opposite the numerator's sign when working with tangent sum and difference formulas.

This expression is one of the standard forms of the double-angle formula, since \(\cos(2x)=2\cos^2(x)-1\) comes from substituting \(\sin^2(x)=1-\cos^2(x)\) into \(\cos^2(x)-\sin^2(x)\). The choice "\(\sin(2x)\)" confuses the cosine double-angle form with the sine double-angle form, \(2\sin(x)\cos(x)\). Recognizing all three equivalent forms of \(\cos(2x)\) allows quick simplification depending on which function's value is given.

Using \(\cos(2x)=1-2\sin^2(x)\) with \(\sin(x)=\frac{1}{3}\) gives \(1-2\left(\frac{1}{9}\right)=\frac{7}{9}\). The choice "\(-\frac{7}{9}\)" results from a sign error when distributing the negative through the subtraction. This formula is the fastest way to find \(\cos(2x)\) when only \(\sin(x)\) is known, without needing to first solve for \(\cos(x)\).

This expression is exactly the right-hand side of the sine sum formula, so it simplifies to \(\sin(x+y)\). The choice "\(\sin(x-y)\)" would require a minus sign between the terms instead of a plus sign, which does not match the given expression. Recognizing sum and difference formula patterns in reverse is a key skill for quickly condensing expanded expressions.

This expression matches the cosine sum formula exactly, so it condenses to \(\cos(x+y)\). The choice "\(\cos(x-y)\)" would instead require a plus sign between the two product terms, not a minus sign. Being able to recognize this pattern in reverse helps condense expanded trigonometric expressions back into single-angle form.

Substituting \(\sin(2x)=2\sin(x)\cos(x)\) into the numerator gives \(\dfrac{2\sin(x)\cos(x)}{2\sin(x)}\), and the \(2\sin(x)\) terms cancel to leave \(\cos(x)\). The choice "\(2\cos(x)\)" comes from failing to fully cancel the factor of 2 in both numerator and denominator. This kind of cancellation is common when verifying identities that involve double-angle expressions divided by single-angle terms.

The Pythagorean identity gives \(1-\cos^2(x)=\sin^2(x)\), so the expression becomes \(\dfrac{\sin^2(x)}{\sin(x)}=\sin(x)\). The choice "\(\tan(x)\)" incorrectly assumes a division by cosine occurs somewhere, which never appears in this simplification. Substituting Pythagorean identities before simplifying fractions is a common first step in verifying trigonometric identities.

Rewriting in terms of sine and cosine, \(\tan(x)\csc(x)=\dfrac{\sin(x)}{\cos(x)}\cdot\dfrac{1}{\sin(x)}=\dfrac{1}{\cos(x)}=\sec(x)\). The choice "\(\cot(x)\)" would result from mistakenly canceling cosine instead of sine during simplification. Converting all functions to sine and cosine first is a reliable strategy for simplifying products of trigonometric functions.

Using \(\cos(60^\circ+45^\circ)=\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ=\frac{1}{2}\cdot\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}\) gives \(\frac{\sqrt2-\sqrt6}{4}\). The choice "\(\frac{\sqrt2+\sqrt6}{4}\)" incorrectly adds instead of subtracting the two product terms, violating the cosine sum formula's structure. This method allows exact trigonometric values for angles not on the standard unit circle by decomposing them into known special angles.

Using \(\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac{1}{2}\) gives \(\frac{\sqrt6+\sqrt2}{4}\). The choice "\(\frac{\sqrt6-\sqrt2}{4}\)" incorrectly subtracts the terms, which would only be correct for a sum formula with mismatched signs. This technique of splitting non-standard angles into sums or differences of \(30^\circ\), \(45^\circ\), and \(60^\circ\) is essential for finding exact trig values.

In Quadrant II, sine is positive, so \(\sin(x)=\frac{3}{5}\), and \(\sin(2x)=2\sin(x)\cos(x)=2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)=-\frac{24}{25}\). The choice "\(\frac{24}{25}\)" ignores the negative sign of cosine in Quadrant II, producing an incorrect positive result. Determining the correct sign of each function based on the quadrant is essential before applying double-angle formulas.

Expanding with the sum formula, \(\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)=\sin(x)(-1)+\cos(x)(0)=-\sin(x)\). The choice "\(\sin(x)\)" ignores that \(\cos(\pi)=-1\) flips the sign of the sine term. Knowing exact values of trig functions at multiples of \(\pi\) makes these sum-formula simplifications quick and reliable.

Using the difference formula, \(\cos\left(x-\frac{\pi}{2}\right)=\cos(x)\cos\left(\frac{\pi}{2}\right)+\sin(x)\sin\left(\frac{\pi}{2}\right)=\cos(x)(0)+\sin(x)(1)=\sin(x)\). The choice "\(-\sin(x)\)" incorrectly introduces a negative sign, which would only occur if the angle order were reversed. This result matches the cofunction identity and shows how sum/difference formulas can verify cofunction relationships.

Expanding gives \(\sin^2(x)+2\sin(x)\cos(x)+\cos^2(x)=1+2\sin(x)\cos(x)\), and since \(\sin(2x)=2\sin(x)\cos(x)\), this equals \(1+\sin(2x)\). The choice "\(1-\sin(2x)\)" would only result from squaring \((\sin(x)-\cos(x))\) instead, not the sum. This combination of the Pythagorean identity and the double-angle formula is a common technique for simplifying squared binomials of sine and cosine.

Since \(1-\cos(2x)=2\sin^2(x)\) and \(\sin(2x)=2\sin(x)\cos(x)\), the ratio simplifies to \(\dfrac{2\sin^2(x)}{2\sin(x)\cos(x)}=\dfrac{\sin(x)}{\cos(x)}=\tan(x)\). The choice "\(\cot(x)\)" results from mistakenly inverting the final ratio of sine over cosine. Substituting double-angle formulas before simplifying fractions is a reliable strategy for verifying more complex identities.

Since \(\sin(2x)=2\sin(x)\cos(x)\), dividing both sides by 2 gives \(\sin(x)\cos(x)=\dfrac{1}{2}\sin(2x)\). The choice "\(\sin(2x)\)" omits the necessary factor of one-half, overstating the value by a factor of two. This rewritten form is frequently used to simplify integrals and expressions involving products of sine and cosine.

Factoring as a difference of squares gives \((\cos^2(x)-\sin^2(x))(\cos^2(x)+\sin^2(x))\), and since the second factor equals 1 by the Pythagorean identity, this reduces to \(\cos^2(x)-\sin^2(x)=\cos(2x)\). The choice "\(1\)" incorrectly assumes both factors simplify to constants, ignoring that only one factor equals 1. Factoring as a difference of squares before applying identities is a powerful technique for simplifying higher-power trigonometric expressions.

With \(\cos(A)=\frac{4}{5}\) and \(\sin(B)=\frac{12}{13}\), the sum formula gives \(\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}\). The choice "\(\dfrac{16}{65}\)" results from subtracting the two product terms instead of adding them, which applies the wrong formula. Finding the missing cosine or sine value using the Pythagorean identity is a required first step before applying sum and difference formulas.

Expanding with the sum formula, \(\cos(x+\pi)=\cos(x)\cos(\pi)-\sin(x)\sin(\pi)=\cos(x)(-1)-\sin(x)(0)=-\cos(x)\). The choice "\(\cos(x)\)" ignores that \(\cos(\pi)=-1\) negates the entire expression. This confirms that adding \(\pi\) to any angle reflects the point through the origin, flipping the sign of cosine.

Using the difference formula, \(\sin(\pi-x)=\sin(\pi)\cos(x)-\cos(\pi)\sin(x)=0-(-1)\sin(x)=\sin(x)\). The choice "\(-\sin(x)\)" incorrectly keeps the negative sign that should cancel when \(\cos(\pi)=-1\) is multiplied through. This identity shows that supplementary angles share the same sine value, a fact often used to find reference angles.

Substituting \(\sin(2x)=2\sin(x)\cos(x)\) and \(1+\cos(2x)=2\cos^2(x)\) gives \(\dfrac{2\sin(x)\cos(x)}{2\cos^2(x)}=\dfrac{\sin(x)}{\cos(x)}=\tan(x)\). The choice "\(\cot(x)\)" results from inverting the final simplified ratio incorrectly. This identity is a common verification exercise that requires applying both double-angle formulas simultaneously before canceling common factors.

Substituting \(\cos(2x)=2\cos^2(x)-1\) gives \(2\cos^2(x)-\cos(x)-1=0\), which factors as \((2\cos(x)+1)(\cos(x)-1)=0\), yielding \(\cos(x)=1\) or \(\cos(x)=-\frac{1}{2}\), so \(x=0,\frac{2\pi}{3},\frac{4\pi}{3}\). The choice "\(x=0,\pi\)" omits the solutions from \(\cos(x)=-\frac{1}{2}\) entirely, missing two valid angles. Converting a double-angle equation into a quadratic in a single trig function is a key strategy for solving these equations.

Using \(\tan(45^\circ+30^\circ)=\dfrac{1+\frac{1}{\sqrt3}}{1-\frac{1}{\sqrt3}}\) and rationalizing gives \(\dfrac{\sqrt3+1}{\sqrt3-1}\cdot\dfrac{\sqrt3+1}{\sqrt3+1}=\dfrac{4+2\sqrt3}{2}=2+\sqrt3\). The choice "\(2-\sqrt3\)" results from an error in the rationalization step, flipping the sign of the middle term. Multi-step exact value problems like this require careful algebraic simplification after applying the sum formula.

Rewriting with a common denominator, \(\tan(x)+\cot(x)=\dfrac{\sin(x)}{\cos(x)}+\dfrac{\cos(x)}{\sin(x)}=\dfrac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}=\dfrac{1}{\sin(x)\cos(x)}=\sec(x)\csc(x)\). The choice "\(\sec(x)+\csc(x)\)" incorrectly assumes the sum can be split apart without combining over a common denominator first. Combining fractions using a common denominator followed by a Pythagorean identity substitution is a core technique for verifying identities.

In Quadrant II, \(\sin(x)=\frac{3}{5}\) and \(\cos(x)=-\frac{4}{5}\) based on the reference triangle from \(\tan(x)=-\frac{3}{4}\), so \(\sin(2x)=2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)=-\frac{24}{25}\). The choice "\(\dfrac{7}{25}\)" incorrectly resembles a cosine double-angle value rather than applying the sine double-angle formula. This problem requires first reconstructing sine and cosine from a given tangent ratio and quadrant before applying the double-angle formula.

Substituting \(\sin(2x)=2\sin(x)\cos(x)\) gives \(2\sin(x)\cos(x)-\sin(x)=0\), which factors as \(\sin(x)(2\cos(x)-1)=0\), so \(\sin(x)=0\) giving \(x=0,\pi\), or \(\cos(x)=\frac{1}{2}\) giving \(x=\frac{\pi}{3},\frac{5\pi}{3}\). The choice "\(x=0,\pi\)" only captures the solutions from the \(\sin(x)=0\) factor and misses the two solutions from \(\cos(x)=\frac{1}{2}\). Factoring out a common trig term rather than dividing by it is essential to avoid losing solutions when solving double-angle equations.

Since \(\cos(4x)=1-2\sin^2(2x)\) and \(\sin(2x)=2\sin(x)\cos(x)\), substituting gives \(1-2(2\sin(x)\cos(x))^2=1-8\sin^2(x)\cos^2(x)\). The choice "\(8\sin^2(x)\cos^2(x)-1\)" reverses the sign of the entire expression, which does not match the correctly derived identity. Applying a double-angle formula twice, once for \(2x\) and once for \(4x\), is a common technique for expressing higher multiple angles.

With \(\cos(A)=\frac{15}{17}\) and \(\sin(B)=\frac{4}{5}\), the difference formula gives $\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)=\frac{15}{17}\left(-\frac{3}{5}\right)+\frac{8}{17}\left(\frac{4}{5}\right)=-\frac{45}{85}+\frac{32}{85}=-\frac{13}{85}$. The choice "\(\dfrac{13}{85}\)" drops the negative sign that results from the negative cosine value in Quadrant II. Careful quadrant analysis to determine correct signs for both angles is critical before applying sum and difference formulas.

Recognizing \(2\cos^2(x)-1=\cos(2x)\), the equation becomes \(\cos(2x)=0\), so \(2x=\frac{\pi}{2}+k\pi\), giving \(x=\frac{\pi}{4}+\frac{k\pi}{2}\) for \(k=0,1,2,3\), resulting in \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). The choice "\(x=\frac{\pi}{4},\frac{5\pi}{4}\)" only captures half of the valid solutions, missing two angles from the full period of \(2x\). Recognizing a double-angle formula hidden inside an equation before solving often simplifies the algebra significantly.

This expression matches the sine sum formula with \(A=2x\) and \(B=x\), so it simplifies to \(\sin(2x+x)=\sin(3x)\). The choice "\(3\sin(x)\)" incorrectly treats the angle addition as simple multiplication rather than applying the trigonometric sum formula. Recognizing that sum formulas can combine a double-angle term with a single-angle term is useful for deriving triple-angle expressions.

Using \(\cos(2x)=1-2\sin^2(x)\), solving \(\frac{7}{25}=1-2\sin^2(x)\) gives \(\sin^2(x)=\frac{9}{25}\), and since \(x\) is in Quadrant I, \(\sin(x)=\frac{3}{5}\). The choice "\(\dfrac{9}{25}\)" stops at the squared value without taking the square root to solve for \(\sin(x)\) itself. Working backward from a double-angle value to find the original single-angle function requires solving an algebraic equation and checking the quadrant for the correct sign.

Multiplying numerator and denominator by \((1+\sin(x))\) gives \(\dfrac{\cos(x)(1+\sin(x))}{1-\sin^2(x)}=\dfrac{\cos(x)(1+\sin(x))}{\cos^2(x)}=\dfrac{1+\sin(x)}{\cos(x)}=\sec(x)+\tan(x)\). The choice "\(\sec(x)-\tan(x)\)" would result from a sign error when splitting the final fraction into two separate terms. Multiplying by a conjugate is a powerful technique for verifying identities that contain a difference in the denominator.

Using reference triangles, \(\sin(A)=\frac{1}{\sqrt5}\), \(\cos(A)=\frac{2}{\sqrt5}\), \(\sin(B)=\frac{1}{\sqrt{10}}\), \(\cos(B)=\frac{3}{\sqrt{10}}\), so \(\sin(A+B)=\frac{1}{\sqrt5}\cdot\frac{3}{\sqrt{10}}+\frac{2}{\sqrt5}\cdot\frac{1}{\sqrt{10}}=\frac{5}{\sqrt{50}}=\frac{\sqrt2}{2}\). The choice "\(\dfrac{1}{2}\)" is a plausible-looking simplified fraction but does not follow from the correct radical simplification of \(\frac{5}{\sqrt{50}}\). Constructing right triangles from given tangent values is a necessary first step before applying sum formulas when sine and cosine are not directly given.