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I've come across a deceptively simple algebraic equation involving trig functions.

Solve the following for $\gamma$: $$a = -x \sin(\gamma) + z \cos(\gamma)$$ where $a$, $x$, $z$ are constants.

After realizing that I didn't know how to solve it, I plugged it into Wolfram Alpha. The step-by-step solutions mention something called Weierstrass ("tangent half-angle") Substitution. After researching this method, I've learned that most examples are methods for solving integrals.

My question is: Why does the Weierstrass Substitution work for the above equation?

Blue
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Ian Gullett
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  • "Why does it work?" It works because it works. "Why use it?" It can be a hassle to invoke a host of identities to manipulate a trig eqn into a solvable form. The Weierstrass Substitution cuts through the clutter to provide a purely-mechanical way of writing trig functions in terms of a common quantity, with an added benefit of converting what could be a convoluted trig eqn into a straightforward polynomial one. Polynomials are "easy" to solve, after all. The trade-off is that you might miss an insightful trig-specific approach; see, eg, here. – Blue Feb 11 '20 at 17:19

1 Answers1

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$$ a = -x \sin \gamma + z \cos \gamma $$ $$ \frac{a}{\sqrt{x^2 + z^2}} = \sin \gamma \frac{-x}{\sqrt{x^2 + z^2}} + \cos \gamma \frac{z}{\sqrt{x^2 + z^2}} $$ so take $\delta$ with $\tan \delta = \frac{-z}{x},$ $\cos \delta = \frac{-x}{\sqrt{x^2 + z^2}}$ and $\sin \delta = \frac{z}{\sqrt{x^2 + z^2}}.$ Now $\delta$ is solvable and $$ \frac{a}{\sqrt{x^2 + z^2}} = \sin (\gamma + \delta) $$

Will Jagy
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