Solving a trigonometric equation linear in sin θ and cos θ Solve asin θ + bcos θ = c (9)

Question

Solving a trigonometric equation linear in [latex]\sin heta[/latex] and [latex]\cos heta[/latex]

Solve

[latex]a\sin heta+b\cos heta=c[/latex] (9)

where a, b, and c are constant and either [latex]a eq0[/latex] or [latex]b eq0[/latex].

Accepted Answer

Divide each side of equation (9) by [latex]{\sqrt{a^{2}+b^{2}}}\,.[/latex] Then

[latex]{\frac{a}{\sqrt{a^{2}+b^{2}}}}\sin heta+{\frac{b}{\sqrt{a^{2}+b^{2}}}}\cos heta={\frac{c}{\sqrt{a^{2}+b^{2}}}}[/latex] (10)

There is a unique angle [latex]\phi,0\,\leq\,\phi\lt 2\pi,[/latex] for which

[latex]\cos\phi={\frac{a}{\sqrt{a^{2}+b^{2}}}}\quad{\mathrm{and}}\quad\sin\phi={\frac{b}{\sqrt{a^{2}+b^{2}}}}[/latex] (11)

Figure 30 shows the situation for [latex]a\gt0[/latex] and [latex]b\gt0[/latex]. Equation (10) may be written as

[latex]\sin heta\cos\phi+\cos heta\sin\phi={\frac{c}{\sqrt{a^{2}+b^{2}}}}[/latex]

or, equivalently,

[latex]\sin\left( heta+\phi ight)\,={\frac{c}{\sqrt{a^{2}+b^{2}}}}[/latex] (12)

where [latex]\phi[/latex] satisfies equation (11).

If [latex]|c|\gt {\sqrt{a^{2}+b^{2}}},[/latex] then [latex]\sin\left( heta+\phi ight)\gt 1[/latex] or [latex]\sin\left( heta+\phi ight)\lt -1[/latex], and equation (12) has no solution.

If [latex]|c|\leq{\sqrt{a^{2}+b^{2}}},[/latex] then the solution of equation (12) are

[latex] heta+\,\phi=\sin^{-1}\frac{c}{\sqrt{a^{2}+b^{2}}}\;\;\;\mathrm{or}\;\;\; heta+\,\phi=\pi-\sin^{-1}\frac{c}{\sqrt{a^{2}+b^{2}}}[/latex]

Because the angle [latex]\phi[/latex] is determined by equations (11), these give the solutions to equation (9).

Question 6.6.5.12: Solving a trigonometric equation linear in sin θ and cos θ S......

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