1.1, 1.5 - Trigonometric identities and equations

1.1, 1.5 - Trigonometric identities and equations

A calculator gives one inverse-trigonometric angle, but an equation may have several roots in its interval. This lesson combines the quadratic methods of 1.1 with the identities and repeating graphs of 1.5. You will simplify and prove identities, find every permitted angle, and turn mixed trigonometric equations into quadratics without losing solutions.

Two basic identities

The point (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta) lies on a circle of radius 11. Pythagoras therefore gives cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1. The ratio of its vertical and horizontal coordinates gives tangent wherever the horizontal coordinate is nonzero.

Two basic trigonometric identities

sin⁡2θ+cos⁡2θ≡1for every real angle θ,\sin^2\theta+\cos^2\theta\equiv1 \quad\text{for every real angle }\theta, tan⁡θ≡sin⁡θcos⁡θwhere cos⁡θ≠0.\tan\theta\equiv\frac{\sin\theta}{\cos\theta} \quad\text{where }\cos\theta\ne0.

Both identities are supplied in MF19. Here sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2, not sin⁡(θ2)\sin(\theta^2).

An identity is true for every input on which its expressions are defined. An equation is a condition whose particular solutions must be found. To prove an identity, transform one side into the other using valid algebra; checking a few angles cannot prove it for all angles.

Rearranging the first identity gives sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta and cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta. Recovering a ratio itself from its square needs the correct sign. If cos⁡θ=−3/5\cos\theta=-3/5 in quadrant II, then sin⁡2θ=16/25\sin^2\theta=16/25, and the quadrant chooses sin⁡θ=4/5\sin\theta=4/5, not −4/5-4/5.

Worked example 1: simplifying a quotient

Question. Simplify (1−sin⁡2θ)/cos⁡θ(1-\sin^2\theta)/\cos\theta, where cos⁡θ≠0\cos\theta\ne0.

Plan. Replace the numerator by a cosine square, then cancel the nonzero cosine factor.

1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θ=cos⁡θ.\frac{1-\sin^2\theta}{\cos\theta} =\frac{\cos^2\theta}{\cos\theta} =\cos\theta.

Answer. cos⁡θ\cos\theta, on the original domain cos⁡θ≠0\cos\theta\ne0.

Check. At θ=60∘\theta=60^\circ, the original gives (1−3/4)/(1/2)=1/2(1-3/4)/(1/2)=1/2, matching cosine. The algebra above, rather than this one numerical check, establishes the identity.

The simplified expression is defined at more angles than the original quotient. Simplifying cannot supply values at inputs where the original denominator was zero.

Worked example 2: combining both identities

Question. Prove that sin⁡θtan⁡θ+cos⁡θ≡1/cos⁡θ\sin\theta\tan\theta+\cos\theta\equiv1/\cos\theta where cos⁡θ≠0\cos\theta\ne0.

Plan. Replace tangent by its sine–cosine quotient, put both terms over one denominator, then use Pythagoras.

sin⁡θtan⁡θ+cos⁡θ=sin⁡2θcos⁡θ+cos⁡2θcos⁡θ=sin⁡2θ+cos⁡2θcos⁡θ=1cos⁡θ.\begin{aligned} \sin\theta\tan\theta+\cos\theta &=\frac{\sin^2\theta}{\cos\theta}+\frac{\cos^2\theta}{\cos\theta}\\ &=\frac{\sin^2\theta+\cos^2\theta}{\cos\theta}\\ &=\frac1{\cos\theta}. \end{aligned}

Answer. The left side equals the right side for every angle with cos⁡θ≠0\cos\theta\ne0, as required.

Check. Every division uses the stated nonzero denominator; neither side is defined when cos⁡θ=0\cos\theta=0. For a numerical check, at 45∘45^\circ both sides equal 2\sqrt2.

In a proof, each line must follow from the previous one. Starting by assuming the desired identity and merely rearranging it into a true statement does not by itself show the original claim unless every step is reversible and that reasoning is made clear. Working from one side usually makes the logic easiest to see.

All roots in an interval

A principal inverse value gives one angle with a requested ratio. A graph or the unit circle gives the other angles, and the stated interval decides which to keep. First isolate the trigonometric function; keep the degree or radian unit consistent throughout.

For a positive ratio with acute reference angle α\alpha, the quadrant signs give the following positions during one turn. The negative-ratio rows use the same positive magnitude and reference angle.

ConditionAngles in 0∘≤θ<360∘0^\circ\le\theta<360^\circ
sin⁡θ=sin⁡α>0\sin\theta=\sin\alpha>0α, 180∘−α\alpha,\ 180^\circ-\alpha
sin⁡θ=−sin⁡α<0\sin\theta=-\sin\alpha<0180∘+α, 360∘−α180^\circ+\alpha,\ 360^\circ-\alpha
cos⁡θ=cos⁡α>0\cos\theta=\cos\alpha>0α, 360∘−α\alpha,\ 360^\circ-\alpha
cos⁡θ=−cos⁡α<0\cos\theta=-\cos\alpha<0180∘−α, 180∘+α180^\circ-\alpha,\ 180^\circ+\alpha
tan⁡θ=tan⁡α>0\tan\theta=\tan\alpha>0α, 180∘+α\alpha,\ 180^\circ+\alpha
tan⁡θ=−tan⁡α<0\tan\theta=-\tan\alpha<0180∘−α, 360∘−α180^\circ-\alpha,\ 360^\circ-\alpha

In radians, replace 180∘180^\circ by π\pi and 360∘360^\circ by 2π2\pi. For intervals longer than one cycle, repeat the appropriate sine/cosine period or tangent period until the whole interval has been covered. For negative intervals, subtract complete periods as needed. List actual roots in the requested interval.

The table uses an acute reference angle. Handle axis values directly: in [0∘,360∘][0^\circ,360^\circ], sine zero occurs at 0∘,180∘,360∘0^\circ,180^\circ,360^\circ, cosine 11 at 0∘,360∘0^\circ,360^\circ, and cosine −1-1 at 180∘180^\circ. Sine 11 occurs only at 90∘90^\circ, and sine −1-1 only at 270∘270^\circ. Tangent zero occurs at 0∘,180∘,360∘0^\circ,180^\circ,360^\circ. A sine or cosine value outside [−1,1][-1,1] is impossible.

Worked example 3: three equation families

Question. Solve each equation for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ: sin⁡θ=−3/2\sin\theta=-\sqrt3/2, cos⁡θ=1/2\cos\theta=1/2, and tan⁡θ=−1\tan\theta=-1.

Plan. Find a positive acute reference angle for each magnitude, then choose the quadrants with the required sign.

For sine, the reference angle is 60∘60^\circ. Sine is negative in quadrants III and IV, giving 180∘+60∘=240∘180^\circ+60^\circ=240^\circ and 360∘−60∘=300∘360^\circ-60^\circ=300^\circ.

For cosine, the reference angle is 60∘60^\circ. Cosine is positive in quadrants I and IV, giving 60∘60^\circ and 300∘300^\circ.

For tangent, the reference angle is 45∘45^\circ. Tangent is negative in quadrants II and IV, giving 135∘135^\circ and 315∘315^\circ.

Answer. Sine equation: 240∘,300∘240^\circ,300^\circ. Cosine equation: 60∘,300∘60^\circ,300^\circ. Tangent equation: 135∘,315∘135^\circ,315^\circ.

Check. The two sine angles have equal negative vertical coordinates; the cosine angles have equal positive horizontal coordinates. The tangent roots differ by 180∘180^\circ, its period. Neither endpoint satisfies any of these three equations.

When the inner angle differs from the variable, transform the interval as well as the equation. Solving only within one turn of the inner angle can discard valid roots.

Worked example 4: a shifted, doubled angle

Question. Solve sin⁡(2x+π/6)=1/2\sin(2x+\pi/6)=1/2 for 0≤x≤π0\le x\le\pi.

Plan. Call the inner angle u=2x+π/6u=2x+\pi/6. Find every permitted uu, then undo the linear transformation.

0≤x≤π⟹π6≤u≤13π6.0\le x\le\pi \quad\Longrightarrow\quad \frac\pi6\le u\le\frac{13\pi}{6}.

Within this interval, sine is 1/21/2 at

u=π6,5π6,13π6.u=\frac\pi6,\quad\frac{5\pi}6,\quad\frac{13\pi}6.

Using x=(u−π/6)/2x=(u-\pi/6)/2 gives

x=0,π3,π.x=0,\quad\frac\pi3,\quad\pi.

Answer. x=0,π/3,πx=0,\pi/3,\pi.

Check. Their inner angles are respectively π/6,5π/6,13π/6\pi/6,5\pi/6,13\pi/6, all with sine 1/21/2. Both interval endpoints satisfy the equation and are included by the weak inequalities.

For an interval crossing zero, include the matching negative-angle positions too.

Worked example 5: four roots across a negative interval

Question. Solve 2sin⁡2x+1=02\sin2x+1=0 for −π<x<π-\pi<x<\pi.

Plan. Isolate sine and use u=2xu=2x, whose interval is −2π<u<2π-2\pi<u<2\pi.

sin⁡u=−12.\sin u=-\frac12.

In the positive turn, u=7π/6u=7\pi/6 or 11π/611\pi/6. Subtract 2π2\pi to obtain the corresponding negative values −5π/6-5\pi/6 and −π/6-\pi/6. All four lie in the transformed interval. Dividing by 22,

x=−5π12,−π12,7π12,11π12.x=-\frac{5\pi}{12},\quad-\frac\pi{12},\quad\frac{7\pi}{12},\quad\frac{11\pi}{12}.

Answer. The four values above.

Check. Doubling each answer gives one of the four listed angles with sine −1/2-1/2. The inner interval spans two complete sine periods, so its two roots per period account for all four. The excluded endpoints would have sine zero and are not roots anyway.

For a non-exact ratio, find a reference angle with the calculator and keep its full precision until the final answers. For example, sin⁡θ=0.4\sin\theta=0.4 on [0∘,360∘][0^\circ,360^\circ] has reference angle sin⁡−1(0.4)=23.578…∘\sin^{-1}(0.4)=23.578\ldots^\circ, giving 23.6∘23.6^\circ and 156.4∘156.4^\circ to one decimal place. Apply the same quadrant reasoning used for exact values.

Watch out

A doubled input doubles the interval that must be searched. Write the inner-angle interval before finding roots, then substitute each candidate into the original equation and check the original endpoints.

Use tangent's half-turn period in this interval.

Quadratics in a trig function

A quadratic in sin⁡θ\sin\theta or cos⁡θ\cos\theta can be solved like a quadratic in any other repeated expression. Use one letter for the ratio, solve for that letter, reject impossible ratio values, then find all angles for each remaining value. Do not confuse a value of the ratio with an angle.

If both sine and cosine appear but one is squared, the identity sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta, or its cosine counterpart, may make the equation a quadratic in just one function. This is the substitution idea from 1.1, with an extra restriction: sine and cosine values must lie in [−1,1][-1,1].

Worked example 6: two valid sine values

Question. Solve 2sin⁡2θ+sin⁡θ−1=02\sin^2\theta+\sin\theta-1=0 for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

Plan. Substitute s=sin⁡θs=\sin\theta, factor the quadratic and solve both resulting sine equations.

2s2+s−1=(2s−1)(s+1)=0,s=12 or s=−1.2s^2+s-1=(2s-1)(s+1)=0, \qquad s=\frac12\text{ or }s=-1.

Both values are possible. For sin⁡θ=1/2\sin\theta=1/2, θ=30∘,150∘\theta=30^\circ,150^\circ. For sin⁡θ=−1\sin\theta=-1, θ=270∘\theta=270^\circ.

Answer. θ=30∘,150∘,270∘\theta=30^\circ,150^\circ,270^\circ.

Check. With s=1/2s=1/2, the original left side is 1/2+1/2−1=01/2+1/2-1=0; with s=−1s=-1, it is 2−1−1=02-1-1=0. The sine graph gives two angles for the first value and one for the second, so there are three distinct roots.

A mixed equation can produce an algebraic root that no angle can attain.

Worked example 7: an impossible cosine value

Question. Solve 2sin⁡2x=3cos⁡x2\sin^2x=3\cos x for 0≤x≤2π0\le x\le2\pi.

Plan. Replace the sine square by 1−cos⁡2x1-\cos^2x, then solve the quadratic in c=cos⁡xc=\cos x.

2(1−c2)=3c⟹2c2+3c−2=0.2(1-c^2)=3c \quad\Longrightarrow\quad 2c^2+3c-2=0. (2c−1)(c+2)=0⟹c=12 or c=−2.(2c-1)(c+2)=0 \quad\Longrightarrow\quad c=\frac12\text{ or }c=-2.

Reject c=−2c=-2, since cosine cannot be below −1-1. The remaining equation cos⁡x=1/2\cos x=1/2 gives x=π/3,5π/3x=\pi/3,5\pi/3.

Answer. x=π/3x=\pi/3 and x=5π/3x=5\pi/3.

Check. At either angle, sin⁡2x=3/4\sin^2x=3/4 and cos⁡x=1/2\cos x=1/2, so both sides of the original equation equal 3/23/2. Neither endpoint satisfies the equation.

There is another danger: division by a trigonometric factor can remove valid solutions when that factor is zero. Factor first and use the zero-product rule instead.

Worked example 8: roots that division would lose

Question. Solve sin⁡θcos⁡θ=2sin⁡θ\sin\theta\cos\theta=2\sin\theta for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

Plan. Bring both terms to one side and factor without dividing by sin⁡θ\sin\theta.

sin⁡θ(cos⁡θ−2)=0.\sin\theta(\cos\theta-2)=0.

Thus sin⁡θ=0\sin\theta=0 or cos⁡θ=2\cos\theta=2. The second is impossible, while the first gives all three axis roots in the closed interval.

Answer. θ=0∘,180∘,360∘\theta=0^\circ,180^\circ,360^\circ.

Check. At each angle, both sides of the original equation are zero. Dividing by sin⁡θ\sin\theta at the start would have left only the impossible equation cos⁡θ=2\cos\theta=2, incorrectly losing every root.

The identities used here preserve equality on their stated domains. If a rearrangement involves a denominator or a division by a variable expression, keep track of where it is valid and check any excluded possibilities separately.

Quick Recap

  • Use the unit-circle identities with their denominator conditions; a proof needs valid algebra for all permitted inputs.
  • Solve for the inner angle over its transformed interval, then return to the original variable.
  • A quadratic substitution finds ratio values first; each valid value may produce several angles.
  • Factor rather than divide by a possibly zero ratio, and check boundaries and the original equation.