Solving a trigonometric equation linear in Sine and Cosine Solve the equation:

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Solving a trigonometric equation linear in Sine and Cosine

Solve the equation: [latex]\sin heta+\cos heta=1,\quad0\leq heta\lt 2\pi[/latex]

Accepted Answer

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.

[latex]\sin heta+\cos heta=1[/latex]

[latex](\sin heta+\cos heta)^{2}=1[/latex] Square each side.

[latex]\sin^{2} heta+2\sin heta\cos heta+\cos^{2} heta=1[/latex] Remove parentheses.

[latex]2\sin heta\cos heta=0[/latex] [latex]\sin^{2} heta+\cos^{2} heta=1[/latex]

[latex]\sin heta \cos heta=0[/latex]

Setting each factor equal to zero leads to

[latex]\sin heta=0\;\;\;\mathrm{or}\;\;\;\cos heta=0[/latex]

The apparent solutions are

[latex]\displaystyle{ heta=0,\;\;\;\;\; heta=\pi,\;\;\;\;\;\; heta={\frac{\pi}{2}},\;\;\;\;\;\; heta={\frac{3\pi}{2}}}[/latex]

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

[latex] heta=0:\quad\sin0+\cos0=0+1=1[/latex] A solution

[latex] heta=\pi:\quad\sin \pi+\cos \pi=0+(-1)=-1[/latex] Not a solution

[latex] heta=\frac{\pi}{2}:\quad\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1+0=1[/latex] A solution

[latex] heta=\frac{3\pi}{2}:\quad\sin\frac{3\pi}{2}+\cos\frac{3\pi}{2}=-1+0=-1[/latex] Not a solution

The values [latex] heta = \pi[/latex] and [latex] heta={\frac{3\pi}{2}}[/latex] are extraneous. The solution set is [latex]\left\{0,{\frac{\pi}{2}} ight\}.[/latex]

Solution B

Start with the equation

[latex]\sin heta+\cos heta=1[/latex]

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

[latex]{\frac{1}{\sqrt{2}}}\sin heta+{\frac{1}{\sqrt{2}}}\cos heta={\frac{1}{\sqrt{2}}}[/latex]

The left side now resembles the formula for the sine of the sum of two angles, one of which is [latex] heta[/latex]. The other angle is unknown (call it Ø.) Then

[latex]\sin\left( heta+\phi ight)\,=\,\sin heta\cos\phi\,+\,\cos heta\sin\phi\,=\,{\frac{1}{\sqrt{2}}}={\frac{\sqrt{2}}{2}}[/latex] (8)

where

[latex]\cos\phi={\frac{1}{\sqrt{2}}}={\frac{\sqrt{2}}{2}}\qquad\sin\phi={\frac{1}{\sqrt{2}}}={\frac{\sqrt{2}}{2}}\qquad0\leq\phi\lt 2\pi[/latex]

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

[latex]\sin\left( heta+{\frac{\pi}{4}} ight)={\frac{\sqrt{2}}{2}}[/latex]

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

[latex] heta+{\frac{\pi}{4}}={\frac{\pi}{4}}[/latex] or [latex] heta+{\frac{\pi}{4}}={\frac{3\pi}{4}}[/latex]

[latex] heta=0[/latex] or [latex] heta=\frac{\pi}{2}[/latex]

The solution set is [latex]\left\{0,{\frac{\pi}{2}} ight\}.[/latex]

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

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