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:

Answer: (3) cot α

Solution:

Given,

cos x + cos y + cos α = 0

cos x + cos y = -cos α

2 cos(x + y)/2 cos(x – y)/2 = -cos α….(i)

sin x + sin y + sin α = 0

sin x + sin y = -sin α

2 sin(x + y)/2 cos(x – y)/2 = -sin α….(ii)

Dividing (i) by (ii),

[2 cos(x + y)/2 cos(x – y)/2]/ [2 sin(x + y)/2 cos(x – y)/2] = (-cos α)/(-sin α)

cot(x + y)/2 = cot α

Answer 2

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:

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