Question 6.6.5.11: Solving a trigonometric equation linear in Sine and Cosine S......

Solution A

Attempts to use available identities do not lead to equations that are easy to solve. (Try it yourself.) Therefore, given the form of this equation, square each side.

(\sin\theta+\cos\theta)^{2}=1             Square each side.

\sin^{2}\theta+2\sin\theta\cos\theta+\cos^{2}\theta=1                  Remove parentheses.

2\sin\theta\cos\theta=0              \sin^{2}\theta+\cos^{2}\theta=1

Setting each factor equal to zero leads to

The apparent solutions are

Because both sides of the original equation were squared, these apparent solutions must be checked to see whether any are extraneous.

\theta=0:\quad\sin0+\cos0=0+1=1              A solution

\theta=\pi:\quad\sin \pi+\cos \pi=0+(-1)=-1          Not a solution

\theta=\frac{\pi}{2}:\quad\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1+0=1                 A solution

\theta=\frac{3\pi}{2}:\quad\sin\frac{3\pi}{2}+\cos\frac{3\pi}{2}=-1+0=-1                      Not a solution

The values \theta = \pi and \theta={\frac{3\pi}{2}} are extraneous. The solution set is \left\{0,{\frac{\pi}{2}}\right\}.

Solution B

Start with the equation

and divide each side by \sqrt{2}. (The reason for this choice will become apparent shortly.) Then

The left side now resembles the formula for the sine of the sum of two angles, one of which is \theta. The other angle is unknown (call it Ø.) Then

\sin\left(\theta+\phi\right)\,=\,\sin\theta\cos\phi\,+\,\cos\theta\sin\phi\,=\,{\frac{1}{\sqrt{2}}}={\frac{\sqrt{2}}{2}}       (8)

where

The angle \phi is therefore \frac{\pi}{4}. As a result, equation (8) becomes

In the interval [0,2\pi), there are two angles whose sine is \frac{\sqrt{2}}{2}: \frac{\pi}{4} and \frac{3\pi}{4}. See Figure 29. As a result,

\theta+{\frac{\pi}{4}}={\frac{\pi}{4}}        or        \theta+{\frac{\pi}{4}}={\frac{3\pi}{4}}

\theta=0            or           \theta=\frac{\pi}{2}

The solution set is \left\{0,{\frac{\pi}{2}}\right\}.