Question

$if, \: \: \:\:\: \cos \: x \: + \cos \: y = a \\ \: \: \: and \:\: \:\: \sin \: x + \sin \: y \: = b \\ then \: prove \: that, \: \: \\ \cos(x + y) = \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

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Answer 1

ANSWER:

Given:

• cos x + cos y = a
• sin x + sin y = b

To Prove:

• cos(x + y) = (a² - b²)/(a² + b²)

Proof:

$\text{We are given that,}\\\\:\longrightarrow\cos x+\cos y=a\\\\\text{Squaring both sides,}\\\\:\implies(\cos x+\cos y)^2=(a)^2\\\\:\implies\cos^2x+\cos^2y+2\cos x\cos y=a^2- - - -(1)\\\\\text{And, we are also given that,}\\\\:\longrightarrow\sin x+\sin y=b\\\\\text{Squaring both sides,}\\\\:\implies(\sin x+\sin y)^2=(b)^2\\\\:\implies\sin^2x+\sin^2y+2\sin x\sin y=b^2- - - -(2)\\\\\text{Now, we'll add (1) and (2)}\\\\:\implies\cos^2x+\cos^2y+2\cos x\cos y+\sin^2x+\sin^2y+2\sin x\sin y=a^2+b^2\\\\:\implies(\sin^2x+\cos^2x)+(\sin^2y+\cos^2y)+2(\cos x\cos y+\sin x\sin y)=a^2+b^2$

$\text{We know that,}\\\\:\hookrightarrow\sin^2\theta+\cos^2\theta=1\\\\:\hookrightarrow\text{And,}\:\cos\theta\cos\phi+\sin\theta\sin\phi=cos(\theta-\phi)\\\\\text{So,}\\\\:\implies(\sin^2x+\cos^2x)+(\sin^2y+\cos^2y)+2(\cos x\cos y+\sin x\sin y)=a^2+b^2\\\\:\implies1+1+2[\cos(x-y)]=a^2+b^2\\\\:\implies2+2\cos(x-y)=a^2+b^2\\\\:\implies2[1+\cos(x-y)]=a^2+b^2- - - -(3)\\\\\text{Now, we'll subtract (2) from (1)}\\\\:\implies\cos^2x+\cos^2y+2\cos x\cos y-(\sin^2x+\sin^2y+2\sin x\sin y)=a^2-b^2\\\\:\implies\cos^2x+\cos^2y+2\cos x\cos y-\sin^2x-\sin^2y-2\sin x\sin y=a^2-b^2 \\\\:\implies(\cos^2x-\sin^2x)+(\cos^2y-\sin^2y)+2(\cos x\cos y-\sin x\sin y)=a^2-b^2$

$\text{We know that,}\\\\:\hookrightarrow\cos^2\theta-\sin^2\theta=\cos2\theta\\\\:\hookrightarrow\text{And,}\:\cos\theta\cos\phi-\sin\theta\sin\phi=cos(\theta+\phi)\\\\\text{So,}\\\\:\implies(\cos^2x-\sin^2x)+(\cos^2y-\sin^2y)+2(\cos x\cos y-\sin x\sin y)=a^2-b^2\\\\:\implies\cos2x+\cos2y+2[\cos(x+y)]=a^2-b^2\\\\\text{We know that,}\\\\:\hookrightarrow\cos\theta+\cos\phi=2\times\cos\left(\frac{\theta+\phi}{2}\right)\times\cos\left(\frac{\theta-\phi}{2}\right)\\\\\text{So,}\\\\:\implies(\cos2x+\cos2y)+2[\cos(x+y)]=a^2-b^2\\\\:\implies2\times\cos\left(\frac{2x+2y}{2}\right)\times\cos\left(\frac{2x-2y}{2}\right)+2[\cos(x+y)]=a^2-b^2\\\\:\implies2\times\cos(x+y)\times\cos(x-y)+2[\cos(x+y)]=a^2-b^2$

$\text{Taking 2\cos(x+y) common,}\\\\:\implies2\cos(x+y)\bigg[\cos(x-y)+1\bigg]=a^2-b^2\\\\:\implies2\cos(x+y)\bigg[1+\cos(x-y)\bigg]=a^2-b^2\\\\:\implies\cos(x+y)\times2\bigg[1+\cos(x-y)\bigg]=a^2-b^2 - - - -(4)\\\\\text{Putting (3) in (4),}\\\\:\implies\cos(x+y)\times(a^2+b^2)=a^2-b^2\\\\\text{Transposing a^2+b^2 to RHS,}\\\\\bf{:\implies\cos(x+y)=\dfrac{a^2-b^2}{a^2+b^2}}\\\\\text{\bf{HENCE PROVED!!!}}$

Formulae Used:

$:\hookrightarrow1)\:\sin^2\theta+\cos^2\theta=1\\\\:\hookrightarrow2)\:\cos\theta\cos\phi+\sin\theta\sin\phi=cos(\theta-\phi)\\\\:\hookrightarrow3)\:\cos^2\theta-\sin^2\theta=\cos2\theta\\\\:\hookrightarrow4)\:\cos\theta\cos\phi-\sin\theta\sin\phi=cos(\theta+\phi)\\\\:\hookrightarrow5)\:\cos\theta+\cos\phi=2\times\cos\left(\frac{\theta+\phi}{2}\right)\times\cos\left(\frac{\theta-\phi}{2}\right)$

Answer 2

Answer:

$Given that,</p><p></p><p>→cosx+coy=a....[1]</p><p></p><p>→sinx+siny=b....[2]</p><p></p><p>Squaring and adding [1] and [2], we get,</p><p></p><p>→cos2x+2cosxcosy+cos2y+sin2x+2sinxsiny+sin2y=a2+b2</p><p></p><p>→2+2(cosxcosy+sinxsiny)=a2+b2</p><p></p><p>→2(1+cos(x−y))=a2+b2</p><p></p><p>→cos(x−y)=a2+b22−1</p><p></p><p>Dividing equation [1] by [2], we get,</p><p></p><p>→cosx+cosysinx+siny=ab</p><p></p><p>→2cos(x+y2)cos(x−y2)2sin(x+y2)cos(x−y2)=ab</p><p></p><p>→cot(x+y2)=ab</p><p></p><p>→tan(x+y2)=ba</p><p></p><p>→x+y2=tan−1(ba)</p><p></p><p>→x+y=2tan−1(ba)</p><p></p><p>As, 2tan−1x=sin−1(2x1+x2),we have,</p><p></p><p>→x+y=sin−1⎛⎜ ⎜⎝2⋅(ba)1+(ba)2⎞⎟ ⎟⎠=sin−1(2aba2+b2)</p><p></p><p>→sin(x+y)=2aba2+b2</p><p></p><p>Now,</p><p></p><p>LHS=sin2x+sin2y</p><p></p><p>=2sin(x+y)⋅cos(x−y)</p><p></p><p>=2[2aba2+b2][a2+b22−1]</p><p></p><p>=2ab[2a2+b2⋅a2+b22−2a2+b2]</p><p></p><p>=2ab[1−2a2+b2]=RHS</p><p></p><p>$

Step-by-step explanation:

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