# If sin α + sin β = a and cos α + cos β = b, show that

Question:

If sin α + sin β = a and cos α + cos β = b, show that

(i) $\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$

(ii) $\cos (\alpha+\beta)=\frac{b^{2}-a^{2}}{b^{2}+a^{2}}$

Solution:

(i) $a^{2}+b^{2}=(\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}$

$\Rightarrow a^{2}+b^{2}=\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta+\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta$

$\Rightarrow a^{2}+b^{2}=\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \beta+\cos ^{2} \beta+2(\sin \alpha \sin \beta+\cos \alpha \cos \beta)$

$\Rightarrow a^{2}+b^{2}=2+2 \cos (\alpha-\beta) \quad \ldots(1)$

Now,

$b^{2}-a^{2}=(\cos \alpha+\cos \beta)^{2}-(\sin \alpha+\sin \beta)^{2}$

$\Rightarrow b^{2}-a^{2}=\cos ^{2} \alpha+\cos ^{2} \beta-\sin ^{2} \alpha-\sin ^{2} \beta+2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta$

$\Rightarrow b^{2}-a^{2}=\left(\cos ^{2} \alpha-\sin ^{2} \beta\right)+\left(\cos ^{2} \beta-\sin ^{2} \alpha\right)-2 \cos (\alpha+\beta)$

$\Rightarrow b^{2}-a^{2}=2 \cos (\alpha+\beta) \cos (\alpha-\beta)+2 \cos (\alpha-\beta)$

$\Rightarrow b^{2}-a^{2}=\cos (\alpha+\beta)(2+2 \cos (\alpha-\beta)) \ldots(2)$

From (1) and (2), we have

$b^{2}-a^{2}=\cos (\alpha+\beta)\left(a^{2}+b^{2}\right)$

$\Rightarrow \frac{b^{2}-a^{2}}{a^{2}+b^{2}}=\cos (\alpha+\beta)$

$\Rightarrow \sin (\alpha+\beta)=\sqrt{1-\cos ^{2}(\alpha+\beta)}$

$\Rightarrow \sin (\alpha+\beta)=\sqrt{1-\left(\frac{b^{2}-a^{2}}{b^{2}+a^{2}}\right)^{2}}=\sqrt{\frac{b^{4}+a^{4}-b^{4}-a^{4}+4 a^{2} b^{2}}{\left(b^{2}+a^{2}\right)^{2}}}$

$\Rightarrow \sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$

(ii) $a^{2}+b^{2}=(\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}$

$=\sin ^{2} \alpha+\sin ^{2} \beta+\cos ^{2} \alpha+\cos ^{2} \beta+2 \sin \alpha \sin \beta+2 \cos \alpha \cos \beta$

$=2+2 \cos (\alpha-\beta)$

$\Rightarrow b^{2}-a^{2}=(\cos \alpha+\cos \beta)^{2}-(\sin \alpha+\sin \beta)^{2}$

$\Rightarrow b^{2}-a^{2}=\cos ^{2} \alpha+\cos ^{2} \beta-\sin ^{2} \alpha-\sin ^{2} \beta+2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta$

$\Rightarrow b^{2}-a^{2}=\left(\cos ^{2} \alpha-\sin ^{2} \beta\right)+\left(\cos ^{2} \beta-\sin ^{2} \alpha\right)-2 \cos (\alpha+\beta)$

$\Rightarrow b^{2}-a^{2}=2 \cos (\alpha+\beta) \cos (\alpha-\beta)+2 \cos (\alpha-\beta)$

$\Rightarrow b^{2}-a^{2}=\cos (\alpha+\beta)(2+2 \cos (\alpha-\beta))$

$\Rightarrow b^{2}-a^{2}=\cos (\alpha+\beta)\left(a^{2}+b^{2}\right)$

$\frac{b^{2}-a^{2}}{a^{2}+b^{2}}=\cos (\alpha+\beta)$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}