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MATH
[1251]

Examples from Trigonometric and geometric applications

All examples from Trigonometric and geometric applications.

Expand sin⁡3θ\sin^3\theta

Using the exponential form,

sin⁡3θ=(eiθ−e−iθ2i)3=e3iθ−3eiθ+3e−iθ−e−3iθ(2i)3=3sin⁡θ−sin⁡(3θ)4.\begin{aligned} \sin^3\theta &=\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^3\\ &=\frac{e^{3i\theta}-3e^{i\theta}+3e^{-i\theta}-e^{-3i\theta}}{(2i)^3}\\ &=\frac{3\sin\theta-\sin(3\theta)}4. \end{aligned}

For the last step, group opposite exponents and use eikθ−e−ikθ=2isin⁡(kθ)e^{ik\theta}-e^{-ik\theta}=2i\sin(k\theta), together with (2i)3=−8i(2i)^3=-8i. This step turns the exponential calculation back into a real trigonometric identity.

For the reverse direction, the imaginary part of

(cos⁡θ+isin⁡θ)3(\cos\theta+i\sin\theta)^3

gives

sin⁡(3θ)=3cos⁡2θsin⁡θ−sin⁡3θ.\sin(3\theta)=3\cos^2\theta\sin\theta-\sin^3\theta.

Using cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta recovers the same cubic identity.

Either way, the cube of one oscillation becomes a combination of oscillations at frequencies 11 and 33.

Interpret a combined locus

Describe

S={z∈C:∣z−(1+i)∣≤2,0≤Arg⁡(z−(1+i))<π2}.S=\left\{z\in\mathbb C:|z-(1+i)|\le2,\quad 0\le\operatorname{Arg}(z-(1+i))<\frac\pi2\right\}.

Writing w=z−(1+i)w=z-(1+i) moves the centre to the origin. The modulus condition restricts ww to the closed disc of radius 22; the argument condition keeps only directions from the positive real axis up to, but not including, the positive imaginary axis. Thus SS is the quarter-disc of radius 22 lying above and to the right of 1+i1+i, with its boundary only partly included:

  • the horizontal radial edge is included, except for the centre;
  • the circular arc is included, except for its upper endpoint;
  • the vertical radial edge is excluded.

The centre is omitted because Arg⁡(0)\operatorname{Arg}(0) is undefined, a detail the disc condition alone would miss.

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