Question

solve the equation on the interval 0≤θ<2π:

sinθ + cosθ = -√(2)

Answer 1

$$\sin\theta+\cos\theta=-\sqrt2$$Okay, first of all we're going to square both sides.$$(\sin\theta+\cos\theta)^2=(-\sqrt2)^2\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=2$$Now, you should immediately see the Pythagorean identity, \(\sin^2\theta+\cos^2\theta=1\). Substitute to yield:$$1+2\sin\theta\cos\theta=2$$The second identity you should notice is the double-angle sine identity, \(2\sin\theta\cos\theta=\sin2\theta\); substitute to yield:$$1+\sin2\theta=2\\\sin2\theta=1\\2\theta=\sin^{-1}1$$We know from the unit circle that only one angle \(0\le\theta\lt2\pi\) satisfies \(\sin\theta=1\), namely \(\frac\pi2\). Now we have:$$2\theta=\frac\pi2\\\theta=\frac\pi4$$