Trigonometric Identities — Free Trigonometry Review Games.

This is the fundamental Pythagorean identity.

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

Third Pythagorean identity: 1 + cot^2(x) = csc^2(x).

Sine is an odd function: sin(-x) = -sin(x).

Cosine is an even function: cos(-x) = cos(x).

The sine sum formula.

The cosine sum formula has a minus between the products.

The double-angle sine formula.

All three forms of cos(2x) are equivalent. 1-2sin^2 = 2cos^2-1.

sin(x) * (1/sin(x)) = 1.

\(\sin(x/2) = \pm \sqrt{\frac{1 - \cos(x)}{2}}\), sign depends on the quadrant.

1 - cos^2(x) = sin^2(x), so sin^2(x)/sin(x) = sin(x).

The tangent difference formula.

Factor as difference of squares: (cos^2-sin^2)(cos^2+sin^2) = cos(2x)*1 = cos(2x).

\(\cos^2(x) = \frac{1}{2}\), \(\cos(x) = \pm \frac{\sqrt{2}}{2}\). Solutions: \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\).

The difference formula for sine is derived by substituting \(-B\) into the sum formula, giving \(\sin A\cos B - \cos A\sin B\) since cosine is even and sine is odd. The choice \"\(\sin A\cos B + \cos A\sin B\)\" is actually the sum formula \(\sin(A+B)\), not the difference. Always keep the sum and difference formulas paired but distinguish the sign carefully before applying them.

Because cosine is an even function, replacing \(B\) with \(-B\) in \(\cos(A+B)=\cos A\cos B-\sin A\sin B\) flips only the sine term's sign, giving \(\cos A\cos B+\sin A\sin B\). The choice \"\(\cos A\cos B - \sin A\sin B\)\" is instead the sum formula for cosine, so it belongs to \(\cos(A+B)\). Memorizing that cosine differences use a plus sign while sine differences use a minus sign prevents this common mix-up.

Dividing \(\sin(A+B)\) by \(\cos(A+B)\) and dividing every term by \(\cos A\cos B\) produces \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\). The choice \"\(\dfrac{\tan A+\tan B}{1+\tan A\tan B}\)\" incorrectly uses a plus sign in the denominator, which is the pattern for \(\tan(A-B)\) instead. Remember that the denominator sign is opposite the numerator sign in the tangent sum/difference formulas.

Substituting \(\sin^2(x) = 1-\cos^2(x)\) into \(\cos(2x)=\cos^2(x)-\sin^2(x)\) yields \(\cos(2x)=2\cos^2(x)-1\). The choice \"\(1-2\cos^2(x)\)\" reverses the sign pattern and is actually the wrong-variable version that would apply if you incorrectly substituted for cosine instead of sine. There are three equivalent double-angle forms for cosine, so always check which variable (sine or cosine only) the question asks for.

The double-angle formula for tangent comes from applying \(\tan(A+B)\) with \(A=B=x\), giving \(\tan(2x)=\dfrac{2\tan x}{1-\tan^2 x}\). The choice \"\(\dfrac{2\tan x}{1+\tan^2 x}\)\" wrongly uses a plus sign, confusing this identity with an unrelated Weierstrass-type substitution. Deriving double-angle formulas from the sum formulas by setting \(A=B\) is a reliable way to avoid sign errors.

The half-angle formula for cosine comes from solving \(\cos x = 2\cos^2(x/2)-1\) for \(\cos(x/2)\), giving \(\pm\sqrt{\dfrac{1+\cos x}{2}}\). The choice \"\(\pm\sqrt{\dfrac{1-\cos x}{2}}\)\" is actually the half-angle formula for sine, not cosine, so it's the wrong identity. The sign of a half-angle result must always be chosen based on which quadrant \(x/2\) lies in, not assumed positive.

Tangent is an odd function because it equals \(\sin x/\cos x\), and sine is odd while cosine is even, so \(\tan(-x) = \dfrac{-\sin x}{\cos x} = -\tan x\). The choice \"\(\tan x\)\" would only be true if tangent were an even function like cosine, which it is not. Knowing the even/odd classification of each trig function lets you quickly evaluate negative-angle expressions.

Rearranging the Pythagorean identity \(1+\tan^2(x)=\sec^2(x)\) gives \(\sec^2(x)-\tan^2(x)=1\) for all \(x\) where both functions are defined. The choice \"\(0\)\" would only hold if secant and tangent squared were equal, which contradicts the identity itself. Recognizing this identity instantly (rather than expanding) saves time on simplification problems.

The Pythagorean identity \(1+\cot^2(x)=\csc^2(x)\) rearranges directly to \(\csc^2(x)-\cot^2(x)=1\). The choice \"\(-1\)\" would require cotangent squared to exceed cosecant squared, which never happens under this identity. All three Pythagorean identities are just algebraic rearrangements of \(\sin^2x+\cos^2x=1\) after dividing by sine or cosine squared.

Solving \(\cos(2x)=1-2\sin^2(x)\) for \(\sin^2(x)\) gives \(\sin^2(x)=\dfrac{1-\cos(2x)}{2}\). The choice \"\(\dfrac{1+\cos(2x)}{2}\)\" is the power-reduction formula for \(\cos^2(x)\) instead, so it names the wrong squared function. Power-reduction formulas are essential for rewriting even powers of sine and cosine before integrating or solving equations.

Solving \(\cos(2x)=2\cos^2(x)-1\) for \(\cos^2(x)\) yields \(\cos^2(x)=\dfrac{1+\cos(2x)}{2}\). The choice \"\(\dfrac{1-\cos(2x)}{2}\)\" is the corresponding formula for \(\sin^2(x)\), not cosine, so it answers the wrong question. Keeping the plus sign with cosine and the minus sign with sine in these companion formulas avoids a very common exam error.

Since \(\sin(2x)=2\sin x\cos x\), dividing both sides by 2 shows \(\sin x\cos x = \dfrac{1}{2}\sin(2x)\). The choice \"\(\sin(2x)\)\" omits the necessary factor of \(\dfrac{1}{2}\), doubling the actual value. Whenever a product \(\sin x\cos x\) appears, immediately recognize it as a scaled double-angle expression.

One standard tangent half-angle identity is \(\tan(x/2)=\dfrac{\sin x}{1+\cos x}\), derived by multiplying the radical half-angle form by a conjugate to eliminate the square root. The choice \"\(\dfrac{1+\cos x}{\sin x}\)\" is actually the reciprocal of the correct identity and equals \(\cot(x/2)\), not \(\tan(x/2)\). This sign-free tangent half-angle form is especially useful because it avoids the ambiguity of choosing a \(\pm\) sign.

Writing \(75^\circ = 45^\circ+30^\circ\) and applying \(\sin(A+B)=\sin A\cos B+\cos A\sin B\) gives \(\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6+\sqrt2}{4}\). The choice \"\(\dfrac{\sqrt6-\sqrt2}{4}\)\" is the value for \(\sin(15^\circ)\), the difference angle rather than the sum. Breaking non-standard angles into sums or differences of \(30^\circ\), \(45^\circ\), and \(60^\circ\) is a core exact-value strategy.

Writing \(15^\circ=45^\circ-30^\circ\) and applying \(\cos(A-B)=\cos A\cos B+\sin A\sin B\) gives \(\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6+\sqrt2}{4}\). The choice \"\(\dfrac{\sqrt6-\sqrt2}{4}\)\" would result from mistakenly using a minus sign, which belongs to the sine difference formula instead. Always confirm whether the sum/difference formula for the specific function you're evaluating uses a plus or minus sign.

Writing \(105^\circ=60^\circ+45^\circ\) and applying \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\) with \(\tan60^\circ=\sqrt3\) and \(\tan45^\circ=1\) gives \(\dfrac{\sqrt3+1}{1-\sqrt3}\), which simplifies to \(-(2+\sqrt3)\). The choice \"\(2+\sqrt3\)\" ignores that \(105^\circ\) lies in the second quadrant where tangent is negative. Always check the sign of the result against the quadrant of the resulting angle.

With \(\cos x = 4/5\) (positive in Quadrant I), the double-angle formula \(\sin(2x)=2\sin x\cos x = 2(3/5)(4/5)=24/25\). The choice \"\(7/25\)\" is actually the value of \(\cos(2x)\) for this triangle, not \(\sin(2x)\), so it answers a different quantity. Always find the missing side (or cosine) from the Pythagorean identity before applying a double-angle formula.

Using \(\cos(2x)=2\cos^2(x)-1 = 2(16/25)-1 = 32/25-25/25 = 7/25\). The choice \"\(24/25\)\" is actually \(\sin(2x)\) for this same triangle, so it names the wrong double-angle output. Choosing the correct double-angle formula variant (in terms of cosine only) avoids needing the sine value at all.

Applying \(\cos(2x)=1-2\sin^2(x)=1-2(1/9)=1-2/9=7/9\). The choice \"\(8/9\)\" comes from an arithmetic slip forgetting to double \(\sin^2(x)\) before subtracting from 1. This form of the double-angle formula is ideal here since it requires only \(\sin x\), with no need to determine the sign of \(\cos x\).

Substituting \(\sin(2x)=2\sin x\cos x\) gives \(\dfrac{2\sin x\cos x}{2\sin x}=\cos x\) after canceling the common factor of \(2\sin x\). The choice \"\(1\)\" would only be correct if the numerator and denominator were identical, which they are not once the double-angle expansion is applied. Expanding double-angle expressions before canceling is essential to simplify correctly rather than canceling terms that only look similar.

Using \(1-\cos(2x)=2\sin^2x\) and \(\sin(2x)=2\sin x\cos x\), the expression becomes \(\dfrac{2\sin^2x}{2\sin x\cos x}=\dfrac{\sin x}{\cos x}=\tan x\). The choice \"\(\cot x\)\" is the reciprocal result and would arise from mistakenly swapping the numerator and denominator's simplified forms. This is a classic identity where converting both numerator and denominator into double-angle equivalents reveals a large common factor.

Applying 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 adding \(\pi\) reflects the sine graph, producing a sign flip rather than leaving the value unchanged. Recognizing that shifting by \(\pi\) always negates both sine and cosine speeds up simplification of phase-shifted expressions.

Applying the difference formula, \(\cos(x-\pi/2)=\cos x\cos(\pi/2)+\sin x\sin(\pi/2) = \cos x(0)+\sin x(1)=\sin x\). The choice \"\(-\sin x\)\" would result from a sign error in the difference formula's second term, which should add rather than subtract here. Co-function shifts of \(\pi/2\) are common enough that memorizing the resulting sign pattern saves significant computation time.

Since \(\pi/12 = 15^\circ = 45^\circ-30^\circ\), applying \(\sin(A-B)=\sin A\cos B-\cos A\sin B\) gives \(\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6-\sqrt2}{4}\). The choice \"\(\dfrac{\sqrt6+\sqrt2}{4}\)\" is the value for \(\sin(75^\circ)\), the sum angle instead of the difference. Both a half-angle formula from \(30^\circ\) and a difference formula from \(45^\circ-30^\circ\) will give this same correct value, offering a useful cross-check.

Since \(\pi/8\) is half of \(\pi/4\), using \(\cos(x/2)=\sqrt{\dfrac{1+\cos x}{2}}\) with \(x=\pi/4\) gives \(\sqrt{\dfrac{1+\frac{\sqrt2}{2}}{2}}=\dfrac{\sqrt{2+\sqrt2}}{2}\). The choice \"\(\dfrac{\sqrt{2-\sqrt2}}{2}\)\" is instead the value of \(\sin(\pi/8)\), using the minus-sign version of the half-angle formula. Since \(\pi/8\) lies in Quadrant I, the positive root is chosen without ambiguity.

Applying \(\tan(2x)=\dfrac{2\tan x}{1-\tan^2 x}=\dfrac{2(2)}{1-4}=\dfrac{4}{-3}=-\dfrac{4}{3}\). The choice \"\(4/3\)\" drops the negative sign that arises because the denominator \(1-\tan^2x\) is negative when \(|\tan x|>1\). Careful sign tracking in the denominator of the double-angle tangent formula is essential whenever \(\tan x\) exceeds 1 in magnitude.

Expanding gives \(\sin^2x+2\sin x\cos x+\cos^2x = 1+2\sin x\cos x\), and since \(2\sin x\cos x=\sin(2x)\), the result is \(1+\sin(2x)\). The choice \"\(1-\sin(2x)\)\" would arise from expanding \((\sin x-\cos x)^2\) instead, a different expression with a minus sign inside the parentheses. Recognizing the cross term \(2\sin x\cos x\) as a disguised double-angle expression is key to simplifying squared sum expressions.

This expression is exactly the power-reduction identity \(\cos^2(x)=\dfrac{1+\cos(2x)}{2}\), so it simplifies directly to \(\cos^2 x\). The choice \"\(\sin^2 x\)\" corresponds to the version with a minus sign in the numerator, \(\dfrac{1-\cos(2x)}{2}\), not the plus-sign version given here. Recognizing these power-reduction patterns on sight avoids unnecessary algebraic derivation during a timed exam.

Rewriting as \(2\sin x\cos x-\sin x=0\) factors to \(\sin x(2\cos x-1)=0\), giving \(\sin x=0\) at \(x=0,\pi\) and \(\cos x=1/2\) at \(x=\pi/3,5\pi/3\), so the complete solution set is \(\{0,\pi/3,\pi,5\pi/3\}\). The choice \"\(\{0,\pi\}\)\" only captures the solutions from the \(\sin x=0\) factor and omits the second factor entirely. Whenever a double-angle equation is set equal to a single-angle trig function, factor out the common term rather than dividing it away, since dividing can lose valid solutions.

This expression is precisely the sign-free half-angle identity for tangent, \(\tan(x/2)=\dfrac{\sin x}{1+\cos x}\). The choice \"\(\cot(x/2)\)\" is the reciprocal form, \(\dfrac{1+\cos x}{\sin x}\), which inverts numerator and denominator. This alternate tangent half-angle formula is valuable because it avoids radical signs and quadrant checks altogether.

Since \(x\) is in Quadrant II, \(x/2\) lies between \(\pi/4\) and \(\pi/2\), so sine is positive there, and \(\sin(x/2)=\sqrt{\dfrac{1-\cos x}{2}}=\sqrt{\dfrac{1+1/2}{2}}=\sqrt{3/4}=\dfrac{\sqrt3}{2}\). The choice \"\(-\sqrt3/2\)\" uses the correct magnitude but the wrong sign, since \(x/2\) in this quadrant range never produces a negative sine value. Determining the quadrant of the half-angle, not just the original angle, is the step students most often skip when applying half-angle formulas.

Expanding both terms with the sum and difference formulas gives \((\cos x\cos y-\sin x\sin y)+(\cos x\cos y+\sin x\sin y)\), and the sine terms cancel, leaving \(2\cos x\cos y\). The choice \"\(2\sin x\sin y\)\" is what remains if you instead subtracted the two expansions rather than adding them. This sum-to-product-style cancellation pattern is a useful trick for simplifying sums of related sum/difference expressions quickly.

Starting from \(\sin(2x)=2\sin x\cos x\) and dividing numerator and denominator by \(\cos^2x\) using \(1+\tan^2x=\sec^2x\) produces \(\sin(2x)=\dfrac{2\tan x}{1+\tan^2x}\) after multiplying through by \(\cos^2x\). The choice \"\(\dfrac{2\tan x}{1-\tan^2x}\)\" is instead the tangent double-angle formula for \(\tan(2x)\), not sine. Deriving a Weierstrass-style substitution from Pythagorean and double-angle identities together is a synthesis skill worth practicing for harder proofs.

Rewriting \(2\sin x\cos x-\cos x=0\) factors to \(\cos x(2\sin x-1)=0\), so \(\cos x=0\) gives \(x=\pi/2,3\pi/2\) and \(\sin x=1/2\) gives \(x=\pi/6,5\pi/6\), for a complete set of four solutions. The choice \"\(\{\pi/2,3\pi/2\}\)\" only accounts for the \(\cos x=0\) factor and misses the second factor's solutions entirely. Factoring rather than dividing by \(\cos x\) is essential here because dividing would silently discard the \(\cos x=0\) solutions.

Combining over a common denominator gives \(\dfrac{(1-\cos x)^2+\sin^2x}{\sin x(1-\cos x)}\), and expanding the numerator with the Pythagorean identity simplifies it to \(2-2\cos x\), which cancels one factor of \((1-\cos x)\) to leave \(\dfrac{2}{\sin x}=2\csc x\). The choice \"\(\csc x\)\" drops the factor of 2 that survives after the cancellation step, an easy arithmetic slip in a multi-step simplification. When combining fractions with conjugate-like denominators, always fully expand the numerator using Pythagorean identities before attempting to cancel.

With \(\cos x=12/13\) and \(\sin y=4/5\), the sum formula gives \(\sin(x+y)=\sin x\cos y+\cos x\sin y=(5/13)(3/5)+(12/13)(4/5)=15/65+48/65=63/65\). The choice \"\(33/65\)\" would result from subtracting the two products instead of adding them, which is the pattern for \(\sin(x-y)\), not \(\sin(x+y)\). This problem requires two Pythagorean-identity steps to find the missing cosine and sine before the sum formula can even be applied.

Combining \(\tan x+\cot x = \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x} = \dfrac{\sin^2x+\cos^2x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x}\), and since \(\sin x\cos x=\dfrac{1}{2}\sin(2x)\), this becomes \(\dfrac{2}{\sin(2x)}=2\csc(2x)\). The choice \"\(\csc(2x)\)\" omits the factor of 2 that arises from rewriting \(\sin x\cos x\) as half of \(\sin(2x)\). This identity elegantly links a Pythagorean-identity combination step directly to a double-angle result.

Since \((\sin^2x+\cos^2x)^2=1\) expands to \(\sin^4x+2\sin^2x\cos^2x+\cos^4x=1\), isolating the fourth-power terms gives \(\cos^4x+\sin^4x=1-2\sin^2x\cos^2x\). The choice \"\(1+2\sin^2(x)\cos^2(x)\)\" reverses the necessary sign, which would only be correct if the cross term were subtracted from the original squared sum rather than added. Squaring the Pythagorean identity itself is a powerful technique for handling fourth-power trig expressions.

Since \(\pi/12=45^\circ-30^\circ\), applying \(\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\) gives \(\dfrac{1-\frac{\sqrt3}{3}}{1+\frac{\sqrt3}{3}}\), which rationalizes to \(2-\sqrt3\). The choice \"\(2+\sqrt3\)\" is instead the value of \(\tan(75^\circ)\), the sum-angle version rather than the difference. Rationalizing the compound fraction that results from a tangent difference formula is a necessary algebraic step students often skip, leaving an unsimplified answer.

In Quadrant III, \(\cos x=-3/5\), and since \(x\) is between \(\pi\) and \(3\pi/2\), the half-angle \(x/2\) falls in Quadrant II where cosine is negative, so \(\cos(x/2)=-\sqrt{\dfrac{1+\cos x}{2}}=-\sqrt{\dfrac{1-3/5}{2}}=-\sqrt{1/5}=-\dfrac{\sqrt5}{5}\). The choice \"\(\sqrt5/5\)\" has the correct magnitude but the wrong sign, since it ignores that \(x/2\) lands in a quadrant where cosine is negative. Determining the sign of a half-angle result always requires locating the half-angle's own quadrant, not just the original angle's quadrant.

Substituting \(\sin(2x)=2\sin x\cos x\) and \(1+\cos(2x)=2\cos^2x\) gives \(\dfrac{2\sin x\cos x}{2\cos^2x}=\dfrac{\sin x}{\cos x}=\tan x\). The choice \"\(\cot x\)\" would arise from mistakenly using \(1-\cos(2x)=2\sin^2x\) in the denominator instead of the correct plus-sign identity. This expression is actually the tangent half-angle identity \(\tan(x)\) in disguise, since it equals \(\tan\left(\frac{2x}{2}\right)\).

Substituting \(\cos(2x)=2\cos^2x-1\) transforms the equation into \(2\cos^2x-1=2-5\cos x\), which rearranges to \(2\cos^2x+5\cos x-3=0\) and factors as \((2\cos x-1)(\cos x+3)=0\); since \(\cos x=-3\) is impossible, only \(\cos x=1/2\) applies, giving \(x=\pi/3,5\pi/3\). The choice \"\(\{\pi/6,11\pi/6\}\)\" corresponds to \(\cos x=\sqrt3/2\), a value that never arises from correctly solving this factored quadratic. Substituting a double-angle identity to convert a mixed-angle equation into a solvable quadratic in a single trig function is a key multi-step exam strategy.

Expanding the numerator gives \(\cos A\cos B+\sin A\sin B\), and dividing each term by \(\sin A\sin B\) produces \(\dfrac{\cos A\cos B}{\sin A\sin B}+1=\cot A\cot B+1\). The choice \"\(\cot A\cot B - 1\)\" would result from dividing \(\cos(A+B)\) instead, which has a minus sign between its two product terms. Dividing a sum or difference formula through by a matching product of sines or cosines is a standard technique for generating cotangent- or tangent-based identities.

Since \(\sin x=\sqrt{1-1/16}=\dfrac{\sqrt{15}}{4}\) in Quadrant I, the double-angle formula gives \(\sin(2x)=2\sin x\cos x=2\cdot\dfrac{\sqrt{15}}{4}\cdot\dfrac{1}{4}=\dfrac{\sqrt{15}}{8}\). The choice \"\(15/16\)\" incorrectly squares the sine value instead of multiplying it by cosine and doubling, confusing this with a different power-reduction calculation. This problem combines a Pythagorean-identity step to find the missing sine with a direct double-angle substitution.

Expanding the numerator gives \(\sin x\cos y+\cos x\sin y\), and dividing each term by \(\cos x\cos y\) produces \(\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}=\tan x+\tan y\). The choice \"\(\tan x-\tan y\)\" would only result from expanding \(\sin(x-y)\) instead, which has a minus sign between its terms. This technique of dividing a sum formula by a matching cosine product is the standard way to build tangent-sum identities from scratch during a proof.