Unit 6, lesson 58

Rewriting Trigonometric Powers and Products

How double-angle, power-reduction and product-to-sum identities turn squares and products of sines and cosines into sums of single sines and cosines, and how to pick the identity that fits your goal.

  • Theory part 115
  • Practice part 116
  • Trigonometric Powers and Products

A product like sin⁡5xcos⁡3x\sin 5x \cos 3x is awkward to work with, but the same function written as half of sin⁡8x+sin⁡2x\sin 8x + \sin 2x is just two plain waves added together. Calculus prefers the second form, because a single sine or cosine of a multiple of xx is easy to integrate and easy to read. A handful of identities makes that rewrite possible, and choosing the right one is most of the skill.

Key terms

Double-angle identity

Definition For every real number uu,

sin⁡2u=2sin⁡ucos⁡u,cos⁡2u=cos⁡2u−sin⁡2u=2cos⁡2u−1=1−2sin⁡2u.\begin{aligned} \sin 2u &= 2\sin u \cos u, \\[0.6em] \cos 2u &= \cos^2 u - \sin^2 u \\[0.4em] &= 2\cos^2 u - 1 \\[0.4em] &= 1 - 2\sin^2 u. \end{aligned}

These are the angle addition formulas with A=B=uA = B = u. For all real AA and BB,

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B,sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B.\begin{aligned} \cos(A + B) &= \cos A \cos B \\[0.2em] &\quad - \sin A \sin B, \\[0.6em] \sin(A + B) &= \sin A \cos B \\[0.2em] &\quad + \cos A \sin B. \end{aligned}

To see why, rotate the plane counterclockwise about the origin by the angle AA. The rotation sends (1,0)(1, 0) to (cos⁡A,sin⁡A)(\cos A, \sin A) and (0,1)(0, 1) to (−sin⁡A,cos⁡A)(-\sin A, \cos A), and because it preserves sums and scalar multiples it sends (p,q)=p(1,0)+q(0,1)(p, q) = p(1, 0) + q(0, 1) to (pcos⁡A−qsin⁡A, psin⁡A+qcos⁡A)(p\cos A - q\sin A,\ p\sin A + q\cos A). The point (cos⁡B,sin⁡B)(\cos B, \sin B) at angle BB lands at angle A+BA + B, which is the point (cos⁡(A+B),sin⁡(A+B))(\cos(A + B), \sin(A + B)), and comparing coordinates gives both formulas. The last two forms of cos⁡2u\cos 2u come from replacing sin⁡2u\sin^2 u or cos⁡2u\cos^2 u using sin⁡2u+cos⁡2u=1\sin^2 u + \cos^2 u = 1.

In plain words Doubling the angle does not double the sine. It produces a product of sine and cosine, and for cosine it produces a difference of squares.

Example Suppose sin⁡θ=35\sin\theta = \frac{3}{5} with θ\theta in the second quadrant. Then cos⁡θ=−45\cos\theta = -\frac{4}{5}, and

sin⁡2θ=2⋅35⋅(−45)=−2425.\sin 2\theta = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = -\frac{24}{25}.

Squared trigonometric function

Definition The notation sin⁡2u\sin^2 u means (sin⁡u)2(\sin u)^2, the square of the output, and likewise for cos⁡2u\cos^2 u and other positive integer powers. It does not mean sin⁡(u2)\sin(u^2). By contrast sin⁡−1u\sin^{-1} u is the inverse sine arcsin⁡u\arcsin u, not a reciprocal, so the superscript convention applies only to positive powers.

In plain words The small exponent squares the value of the sine, not the angle.

Example At u=π6u = \frac{\pi}{6},

sin⁡2π6=(12)2=14,\sin^2 \frac{\pi}{6} = \left(\frac{1}{2}\right)^2 = \frac{1}{4},

while sin⁡(π236)≈0.2707\sin\left(\frac{\pi^2}{36}\right) \approx 0.2707. The two are different numbers.

Power reduction

Definition Solving the double-angle identities for the squares gives, for every real uu,

cos⁡2u=1+cos⁡2u2,sin⁡2u=1−cos⁡2u2.\begin{gathered} \cos^2 u = \frac{1 + \cos 2u}{2}, \\[0.6em] \sin^2 u = \frac{1 - \cos 2u}{2}. \end{gathered}

No square root and no sign choice is involved, because these formulas rewrite the squared function itself. A sign question appears only if you try to undo the square, since the root of the right side of the first formula is ∣cos⁡u∣\lvert \cos u \rvert, not cos⁡u\cos u.

In plain words A square of a sine or cosine becomes a constant plus a cosine at twice the angle. The power drops from two to one, and the frequency doubles.

Example A cosine square with a scaled angle becomes

cos⁡23x=1+cos⁡6x2=12+12cos⁡6x.\begin{aligned} \cos^2 3x &= \frac{1 + \cos 6x}{2} \\[0.6em] &= \frac{1}{2} + \frac{1}{2}\cos 6x. \end{aligned}

Angle grouping

Definition Every identity above holds for any real angle uu, so uu may be replaced by a whole expression such as 3x3x, x2\frac{x}{2} or x+1x + 1. The replacement must be made for every occurrence of uu at once, keeping the whole expression together as one angle. In particular 2u2u becomes twice the whole expression.

In plain words Treat the angle as a single block. Double the block, not one piece of it.

Example With u=x2u = \frac{x}{2} the angle 2u2u is xx, so

sin⁡2x2=1−cos⁡x2.\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}.

A common slip with cos⁡23x\cos^2 3x is to write 1+3cos⁡2x2\frac{1 + 3\cos 2x}{2} or 1+cos⁡5x2\frac{1 + \cos 5x}{2}. The angle 3x3x doubled is 6x6x, and the coefficient stays inside the cosine.

Product-to-sum

Definition Adding and subtracting the angle addition formulas for A+BA + B and A−BA - B gives, for all real AA and BB,

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B),2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B),2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B).\begin{aligned} 2\sin A \cos B &= \sin(A + B) \\[0.2em] &\quad + \sin(A - B), \\[0.6em] 2\cos A \cos B &= \cos(A - B) \\[0.2em] &\quad + \cos(A + B), \\[0.6em] 2\sin A \sin B &= \cos(A - B) \\[0.2em] &\quad - \cos(A + B). \end{aligned}

In plain words A product of two waves equals a sum of two waves, one at the sum of the angles and one at the difference.

Example For a mixed-frequency product, take A=2xA = 2x and B=5xB = 5x. Then A−B=−3xA - B = -3x, and since sine is odd, sin⁡(−3x)=−sin⁡3x\sin(-3x) = -\sin 3x. So

sin⁡2xcos⁡5x=sin⁡7x+sin⁡(−3x)2=sin⁡7x−sin⁡3x2.\begin{aligned} \sin 2x \cos 5x &= \frac{\sin 7x + \sin(-3x)}{2} \\[0.6em] &= \frac{\sin 7x - \sin 3x}{2}. \end{aligned}

The order matters in the first formula. Here the sine carries AA, so the difference is A−B=2x−5xA - B = 2x - 5x.

Purposeful rewrite

Definition A purposeful rewrite replaces an expression with an equivalent one chosen for a stated goal. A frequent goal in calculus is a sum of constants and single first powers sin⁡kx\sin kx or cos⁡kx\cos kx. The working rule is to reduce even powers of sine and cosine with power reduction, to split products of different angles with product-to-sum, and to stop as soon as the target form is reached.

In plain words Know what form you want before you start, and use the fewest identities that get you there.

Example Write sin⁡4x\sin^4 x with no powers. Square the power-reduction formula and reduce again.

sin⁡4x=(1−cos⁡2x2)2=1−2cos⁡2x+cos⁡22x4=14(1−2cos⁡2x+1+cos⁡4x2)=3−4cos⁡2x+cos⁡4x8.\begin{aligned} \sin^4 x &= \left(\frac{1 - \cos 2x}{2}\right)^2 \\[0.6em] &= \frac{1 - 2\cos 2x + \cos^2 2x}{4} \\[0.6em] &= \frac{1}{4}\bigg(1 - 2\cos 2x \\[0.4em] &\qquad + \frac{1 + \cos 4x}{2}\bigg) \\[0.6em] &= \frac{3 - 4\cos 2x + \cos 4x}{8}. \end{aligned}

The second step applied power reduction with the grouped angle u=2xu = 2x.

Equivalent expression

Definition Two expressions in xx are equivalent when they are defined for the same inputs and give equal values at every one of them. An identity proves equivalence. Checking a few numerical values can catch a mistake, but agreement at finitely many points is evidence, not proof.

In plain words A correct rewrite changes the look of an expression and nothing else. Plugging in a number is a quick test that tells you when something has gone wrong.

Example Test the result for sin⁡4x\sin^4 x at x=π4x = \frac{\pi}{4}. On the left, (22)4=14\left(\frac{\sqrt{2}}{2}\right)^4 = \frac{1}{4}. On the right,

3−4cos⁡π2+cos⁡π8=3−0−18=14.\begin{aligned} \frac{3 - 4\cos\frac{\pi}{2} + \cos\pi}{8} &= \frac{3 - 0 - 1}{8} \\[0.6em] &= \frac{1}{4}. \end{aligned}

They agree. In contrast, cos⁡x\cos x and 1+cos⁡2x2\sqrt{\frac{1 + \cos 2x}{2}} are not equivalent. At x=πx = \pi the first is −1-1 and the second is 11, because the root equals ∣cos⁡x∣\lvert \cos x \rvert.

Check yourself

  1. Rewrite sin⁡24x\sin^2 4x without a power.
Show the answer

With the grouped angle u=4xu = 4x, power reduction gives

sin⁡24x=1−cos⁡8x2.\sin^2 4x = \frac{1 - \cos 8x}{2}.
  1. Write cos⁡3xcos⁡x\cos 3x \cos x as a sum.
Show the answer

Using 2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2\cos A \cos B = \cos(A - B) + \cos(A + B) with A=3xA = 3x and B=xB = x,

cos⁡3xcos⁡x=cos⁡2x+cos⁡4x2.\cos 3x \cos x = \frac{\cos 2x + \cos 4x}{2}.
  1. Write sin⁡xcos⁡4x\sin x \cos 4x as a sum of sines.
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Here A=xA = x and B=4xB = 4x, so A−B=−3xA - B = -3x and sin⁡(−3x)=−sin⁡3x\sin(-3x) = -\sin 3x. The result is

sin⁡xcos⁡4x=sin⁡5x−sin⁡3x2.\sin x \cos 4x = \frac{\sin 5x - \sin 3x}{2}.

How this lesson connects