6.9 - Trigonometric identities

6.9 - Trigonometric identities

Use two identities to simplify trigonometric expressions, prove equalities and turn equations into familiar algebra. All angles in this lesson are measured in degrees.

An equality that survives a changing angle

Choose any angle, add its sine and cosine, and square the result. Now subtract its cosine from its sine, square again, and add the two squares. Can the total change when the angle changes? Keep that question in mind as we build the tools to explain it.

Trigonometric identity

A trigonometric identity is an equality that holds for every angle for which both sides are defined. An equation may instead be true only for particular angles: sin⁡θ=12\sin\theta=\frac12, for example, is true at 30∘30^\circ but false at 90∘90^\circ.

An identity comes from the meaning of the ratios. On the unit circle, the point (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta) is always a distance 11 from the origin. Pythagoras therefore gives cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1, wherever the point lies. Tangent is the vertical coordinate divided by the horizontal coordinate, provided that horizontal coordinate is non-zero.

The two trigonometric identities

sin⁡2θ+cos⁡2θ=1andtan⁡θ=sin⁡θcos⁡θ\sin^2\theta+\cos^2\theta=1 \qquad\text{and}\qquad \tan\theta=\frac{\sin\theta}{\cos\theta}

The first holds for every angle. The second requires cos⁡θ≠0\cos\theta\ne0, because division by zero is undefined.
On the formulae sheet.

The notation sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2. It does not mean sin⁡(θ2)\sin(\theta^2). Likewise, tan⁡2θ=sin⁡2θcos⁡2θ\tan^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta} wherever cos⁡θ≠0\cos\theta\ne0; this follows by squaring the supplied quotient identity.

Rearranging the first identity gives sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta or cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta. These rearrangements are derived from the supplied identity. They let you replace a squared ratio with an expression in the other squared ratio.

Watch out

Wrong: cos⁡θ=1−sin⁡θ\cos\theta=1-\sin\theta. Right: cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta. At 30∘30^\circ, the wrong statement would give 32=12\frac{\sqrt3}{2}=\frac12, which is false. Keep the squares attached to the ratios.

Check the structure of an identity before using it. A missing square changes its meaning completely.

Simplifying an expression with the identities

Look at the form you want to reach. An answer involving tan⁡2θ\tan^2\theta suggests making the quotient sin⁡2θ/cos⁡2θ\sin^2\theta/\cos^2\theta. First simplify the denominator into one kind of ratio; then cancel common factors.

Worked example 1: making a tangent squared

Question. Simplify 12sin⁡2θ1+cos⁡2θ−sin⁡2θ\dfrac{12\sin^2\theta}{1+\cos^2\theta-\sin^2\theta} into the form ktan⁡2θk\tan^2\theta.

Plan. Replace 1−sin⁡2θ1-\sin^2\theta by cos⁡2θ\cos^2\theta, then use the quotient identity.

1+cos⁡2θ−sin⁡2θ=cos⁡2θ+cos⁡2θuse 1−sin⁡2θ=cos⁡2θ=2cos⁡2θcollect like terms12sin⁡2θ1+cos⁡2θ−sin⁡2θ=12sin⁡2θ2cos⁡2θreplace the denominator=6tan⁡2θuse the squared quotient identity\begin{aligned} 1+\cos^2\theta-\sin^2\theta &=\cos^2\theta+\cos^2\theta && \text{use }1-\sin^2\theta=\cos^2\theta\\ &=2\cos^2\theta && \text{collect like terms}\\ \frac{12\sin^2\theta}{1+\cos^2\theta-\sin^2\theta} &=\frac{12\sin^2\theta}{2\cos^2\theta} && \text{replace the denominator}\\ &=6\tan^2\theta && \text{use the squared quotient identity} \end{aligned}

Why this step? Converting the denominator to cosine squared creates exactly the quotient needed for tangent squared.

Answer. 6tan⁡2θ6\tan^2\theta, for cos⁡θ≠0\cos\theta\ne0. The original denominator is zero at the excluded angles too.

Check. At 45∘45^\circ, the original is 12(12)/(1+12−12)=612(\frac12)/(1+\frac12-\frac12)=6, matching 6tan⁡245∘=66\tan^245^\circ=6.

Worked example 2 changes only the coefficient of cos⁡2θ\cos^2\theta in the denominator from 11 to 33. The same replacement now leaves a different multiple of cosine squared.

Worked example 2: changing one coefficient

Question. Simplify 12sin⁡2θ1+3cos⁡2θ−sin⁡2θ\dfrac{12\sin^2\theta}{1+3\cos^2\theta-\sin^2\theta} into the form ktan⁡2θk\tan^2\theta.

Plan. Combine 1−sin⁡2θ1-\sin^2\theta with the existing cosine term.

1+3cos⁡2θ−sin⁡2θ=cos⁡2θ+3cos⁡2θuse the identity=4cos⁡2θcollect the cosine terms12sin⁡2θ1+3cos⁡2θ−sin⁡2θ=12sin⁡2θ4cos⁡2θreplace the denominator=3tan⁡2θsimplify the common factor\begin{aligned} 1+3\cos^2\theta-\sin^2\theta &=\cos^2\theta+3\cos^2\theta && \text{use the identity}\\ &=4\cos^2\theta && \text{collect the cosine terms}\\ \frac{12\sin^2\theta}{1+3\cos^2\theta-\sin^2\theta} &=\frac{12\sin^2\theta}{4\cos^2\theta} && \text{replace the denominator}\\ &=3\tan^2\theta && \text{simplify the common factor} \end{aligned}

Why this step? The identity replaces only 1−sin⁡2θ1-\sin^2\theta; the original 3cos⁡2θ3\cos^2\theta must still be included.

Answer. 3tan⁡2θ3\tan^2\theta, for cos⁡θ≠0\cos\theta\ne0.

Check. At 45∘45^\circ, the original gives 6/(1+32−12)=36/(1+\frac32-\frac12)=3, matching the simplified answer.

These are the same fraction rules you use in algebra. You may cancel a common factor of the whole numerator and denominator, but you may not cancel one term across a sum. Original exclusions remain in force after simplification.

Watch out

Wrong: cancel sin⁡2θ\sin^2\theta directly from 12sin⁡2θ/(1+cos⁡2θ−sin⁡2θ)12\sin^2\theta/(1+\cos^2\theta-\sin^2\theta). The denominator is a sum, not a product containing that factor. Right: simplify the whole denominator first.

Change the coefficient again and look for the same quotient.

Proving an identity for every allowed angle

A few matching numerical values are useful checks, but they do not prove an identity. A proof needs algebra that works for an arbitrary angle. Start with the more complicated side and transform it into the other side, keeping every equality justified.

Continue this lesson.

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