Linear equation with two methods

Statement

Solve the linear equation, trying both methods (parametric formulae and auxiliary angle): $\sin x + \cos x = 1$

Solution

The parametric-formulae method. Set $t = \tan(x/2)$, with $\sin x = \dfrac{2t}{1+t^2}$ and $\cos x = \dfrac{1-t^2}{1+t^2}$: $\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = 1 \iff 2t + 1 - t^2 = 1 + t^2 \iff 2t - 2t^2 = 0 \iff 2t(1-t) = 0.$ Hence $t = 0$ or $t = 1$.

• $t = 0$: $\tan(x/2) = 0 \Rightarrow x/2 = k\pi \Rightarrow x = 2k\pi$;
• $t = 1$: $\tan(x/2) = 1 \Rightarrow x/2 = \pi/4 + k\pi \Rightarrow x = \pi/2 + 2k\pi$.

Checking the case $x=\pi$: $\sin\pi + \cos\pi = 0 - 1 = -1 \ne 1$: not a solution.

The auxiliary-angle method. Here $a=b=1$, so $\sqrt{a^2+b^2} = \sqrt{2}$. Dividing: $\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}} \iff \sin\left(x+\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}},$ having chosen $\varphi = \pi/4$ (with $\cos\varphi = \sin\varphi = 1/\sqrt2$). Whence $x+\dfrac{\pi}{4} = \dfrac{\pi}{4}+2k\pi$ or $x+\dfrac{\pi}{4} = \dfrac{3\pi}{4}+2k\pi$.

The two methods give the same result: $\boxed{\,x = 2k\pi \ \vee\ x = \frac{\pi}{2} + 2k\pi,\quad k\in\mathbb{Z}.}$

Links

Topics: Trigonometric equations
Concepts: Auxiliary angle · Linear trigonometric equation · Parametric formulae
Functions: Cosine · Sine · Tangent
Skills: Solving equations · Using formulae
Exercise type: Solving an equation