The Auxiliary Angle and Evaluating Expressions

What this move does

Think of $a\sin x+b\cos x$ as the height of an L-shaped path: walk $a$ along the angle $x$, then turn left $90^{\circ }$ and walk $b$. The vertical parts of the two legs are exactly $a\sin x$ and $b\cos x$. The L is rigid and turns as one piece, so its tip always lies on the circle of radius $R=\sqrt{a^2+b^2}$, and the hypotenuse always runs a fixed angle $\phi$ ahead of the first leg:

This is the auxiliary angle formula. Use it in three steps:

1. Take out $R$: $R=\sqrt{a^2+b^2}$. What is left in the brackets is $\frac{a}{R}\sin x+\frac{b}{R}\cos x$, and the squares of the two coefficients add up to $1$.
2. Recognize $\phi$: $\frac{a}{R}$ and $\frac{b}{R}$ are the cosine and sine of $\phi$, the direction angle of the point $(a,b)$. The quadrant is decided by $(a,b)$.
3. After combining, use the properties of sine: the extreme values $\pm R$, the period, the monotonic intervals and the axes of symmetry all come down to $\sin (x+\phi )$. If $x$ is restricted to an interval, find the range of $x+\phi$ first.

Work through one

Evaluating expressions: look at the angles, then pin down the range

The auxiliary angle “combines two terms into one”. Evaluation problems often go the other way: they split an angle into a sum or difference of known angles, and every formula in this chapter gets used. For example:

Given $\alpha \in \left(0,\frac{\pi }{2}\right)$ and $\sin \left(\alpha -\frac{\pi }{6}\right)=\frac{1}{3}$, find $\cos \alpha$.

• Look at the angles: $\alpha =\left(\alpha -\frac{\pi }{6}\right)+\frac{\pi }{6}$, a known angle plus a special angle.
• Pin down the range: $\alpha -\frac{\pi }{6}\in \left(-\frac{\pi }{6},\frac{\pi }{3}\right)$, and its sine is positive, so it lies in $\left(0,\frac{\pi }{3}\right)$ and its cosine is positive: $\cos \left(\alpha -\frac{\pi }{6}\right)=\frac{2\sqrt{2}}{3}$.
• Apply the formula: $\cos \alpha =\frac{2\sqrt{2}}{3}\cdot \frac{\sqrt{3}}{2}-\frac{1}{3}\cdot \frac{1}{2}=\frac{2\sqrt{6}-1}{6}$.

Other common ways to split angles are $2\alpha =(\alpha +\beta )+(\alpha -\beta )$ and $\frac{\pi }{4}+\alpha =\frac{\pi }{2}-\left(\frac{\pi }{4}-\alpha \right)$, and writing $1$ as ${\sin }^2\alpha +{\cos }^2\alpha$ or as $\tan \frac{\pi }{4}$. If only $\tan \frac{\alpha }{2}$ is left in the expression, the tangent half-angle substitution $\sin \alpha =\frac{2t}{1+t^2}$, $\cos \alpha =\frac{1-t^2}{1+t^2}$ ($t=\tan \frac{\alpha }{2}$) turns everything into a rational expression in $t$.

Practice