If α and β be two distinct real numbers such that (α-β) ≠ 2nπ for any intetger n, satisfying the equation a cos θ+b sin θ=c then prove that (i) cos (α+β)=a²-b²/a²+b² (ii) sin (α+β)=2ab/a²+b².

Question

If [latex] \alpha [/latex] and [latex] \beta [/latex] be two distinct real numbers such that [latex] (\alpha-\beta) eq 2 n \pi [/latex] for any intetger [latex] n [/latex], satisfying the equation [latex] a \cos heta+b \sin heta=c [/latex] then prove that

(i) [latex] \cos (\alpha+\beta)=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\quad [/latex](ii) [latex] \sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}} [/latex].

Accepted Answer

Since [latex] \alpha [/latex] and [latex] \beta [/latex] satisfy the equation [latex] a \cos heta+b \sin heta=c [/latex], we have

[latex]\begin{array}{l}a \cos \alpha+b \sin \alpha=c, \qquad \qquad ...(i)\a \cos \beta+b \sin \beta=c .\qquad \qquad ...(ii)\end{array}[/latex]

Subtracting (ii) from (i), we get

[latex]\begin{array}{l}a(\cos \alpha-\cos \beta)+b(\sin \alpha-\sin \beta)=0 \ \\Rightarrow-2 a \sin \left(\frac{\alpha+\beta}{2} ight) \sin \left(\frac{\alpha-\beta}{2} ight)+2 b \cos \left(\frac{\alpha+\beta}{2} ight) \sin \left(\frac{\alpha-\beta}{2} ight)=0 \ \\Rightarrow-2 \sin \left(\frac{\alpha-\beta}{2} ight)\left[a \sin \left(\frac{\alpha+\beta}{2} ight)-b \cos \left(\frac{\alpha+\beta}{2} ight) ight]=0 \ \\Rightarrow a \sin \left(\frac{\alpha+\beta}{2} ight)-b \cos \left(\frac{\alpha+\beta}{2} ight)=0\ \{\left[\begin{array}{l}\because(\alpha-\beta) eq 2 n \pi \Rightarrow\left(\frac{\alpha-\beta}{2} ight) eq n \pi \ \ herefore \sin \left(\frac{\alpha-\beta}{2} ight) eq \sin n \pi eq 0\end{array} ight]} \ \\Rightarrow an \left(\frac{\alpha+\beta}{2} ight)=\frac{b}{a} .\qquad \qquad ...(iii) \ \ ext { (i) } \cos (\alpha+\beta)=\frac{1- an ^{2}\left(\frac{\alpha+\beta}{2} ight)}{1+ an ^{2}\left(\frac{\alpha+\beta}{2} ight)} \quad\left[\because \cos x=\frac{1- an ^{2}\left(\frac{x}{2} ight)}{1+ an ^{2}\left(\frac{x}{2} ight)} ight] \ \=\frac{\left(1-\frac{b^{2}}{a^{2}} ight)}{\left(1+\frac{b^{2}}{a^{2}} ight)}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \quad[ ext { using (iii)]. } \ \ ext{ (ii) }\sin (\alpha+\beta)=\frac{2 an \left(\frac{\alpha+\beta}{2} ight)}{1+ an ^{2}\left(\frac{\alpha+\beta}{2} ight)}\left[\because \sin x=\frac{2 an \left(\frac{x}{2} ight)}{1+ an ^{2}\left(\frac{x}{2} ight)} ight] \ \=\frac{2\left(\frac{b}{a} ight)}{\left(1+\frac{b^{2}}{a^{2}} ight)}=\frac{2 a b}{a^{2}+b^{2}} \quad[ ext { using (iii)]. } \end{array}[/latex]

Question 15.5.6: If α and β be two distinct real numbers such that (α-β) ≠ 2n......

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