Question #e8ab5

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Question #e8ab5

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Answer 1 · 2016-08-21T15:37:31

$cos (x + y) = \frac{a^{2} + b^{2}}{2} - 1$ $cos(x+y)=(a^2+b^2)/2-1$

Explanation:

First, recall what $cos (x + y)$ $cos(x+y)$ is:
$cos (x + y) = cos x cos y + sin x sin y$ $cos(x+y)=cosxcosy+sinxsiny$

Note that:
${(sin x + sin y)}^{2} = a^{2}$ $(sinx+siny)^2=a^2$
$\to {sin}^{2} x + 2 sin x sin y + {sin}^{2} y = a^{2}$ $->sin^2x+2sinxsiny+sin^2y=a^2$

And:
${(cos x + cos y)}^{2} = b^{2}$ $(cosx+cosy)^2=b^2$
$\to {cos}^{2} x + 2 cos x cos y + {cos}^{2} y = b^{2}$ $->cos^2x+2cosxcosy+cos^2y=b^2$

Now we have these two equations:
${sin}^{2} x + 2 sin x sin y + {sin}^{2} y = a^{2}$ $sin^2x+2sinxsiny+sin^2y=a^2$
${cos}^{2} x + 2 cos x cos y + {cos}^{2} y = b^{2}$ $cos^2x+2cosxcosy+cos^2y=b^2$

If we add them together, we have:
${sin}^{2} x + 2 sin x sin y + {sin}^{2} y + {cos}^{2} x + 2 cos x cos y + {cos}^{2} y = a^{2} + b^{2}$ $sin^2x+2sinxsiny+sin^2y+cos^2x+2cosxcosy+cos^2y=a^2+b^2$

Don't let the size of this equation throw you off. Look for identities and simplifications:
$({sin}^{2} x + {cos}^{2} x) + (2 sin x sin y + 2 cos x cos y) + ({cos}^{2} y + {sin}^{2} y) = a^{2} + b^{2}$ $(sin^2x+cos^2x)+(2sinxsiny+2cosxcosy)+(cos^2y+sin^2y)=a^2+b^2$

Since ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ (Pythagorean Identity) and ${cos}^{2} y + {sin}^{2} y = 1$ $cos^2y+sin^2y=1$ (Pythagorean Identity), we can simplify the equation to:
$1 + (2 sin x sin y + 2 cos x cos y) + 1 = a^{2} + b^{2}$ $1+(2sinxsiny+2cosxcosy)+1=a^2+b^2$
$\to (2 sin x sin y + 2 cos x cos y) + 2 = a^{2} + b^{2}$ $->(2sinxsiny+2cosxcosy)+2=a^2+b^2$

We can factor out a $2$ $2$ twice:
$2 (sin x sin y + cos x cos y) + 2 = a^{2} + b^{2}$ $2(sinxsiny+cosxcosy)+2=a^2+b^2$
$\to 2 ((sin x sin y + cos x cos y) + 1) = a^{2} + b^{2}$ $->2((sinxsiny+cosxcosy)+1)=a^2+b^2$

And divide:
$(sin x sin y + cos x cos y) + 1 = \frac{a^{2} + b^{2}}{2}$ $(sinxsiny+cosxcosy)+1=(a^2+b^2)/2$

And subtract:
$sin x sin y + cos x cos y = \frac{a^{2} + b^{2}}{2} - 1$ $sinxsiny+cosxcosy=(a^2+b^2)/2-1$

Finally, since $cos (x + y) = cos x cos y + sin x sin y$ $cos(x+y)=cosxcosy+sinxsiny$ , we have:
$cos (x + y) = \frac{a^{2} + b^{2}}{2} - 1$ $cos(x+y)=(a^2+b^2)/2-1$

Answer 2 · 2016-08-21T15:40:49

Given

$sinx+siny=a.......(1)$

$cosx+cosy=b.......(2)$

Squaring and adding (1) & (2)

$(cosx+cosy)^2+(sinx+siny)^2= a^2+b^2$

$=>2(cosxcosy+sinxsiny)+2=a^2+b^2$

$=>2cos(x-y)=a^2+b^2-2....(3)$

Squaring and Subtracting (1) from(2)

$(cosx+cosy)^2-(sinx+siny)^2= b^2-a^2$

$=>2cos(x+y)+cos^2x-sin^2x+cos^2y-sin^2y=b^2-a^2$

$=>2cos(x+y)+cos2x+cos2y=b^2-a^2$

$=>2cos(x+y)+2cos(x+y)cos(x-y)=b^2-a^2$

$=>cos(x+y)(2+2cos(x-y))=b^2-a^2$

( $"From (3) "2cos(x-y)=a^2+b^2-2$ )

$=>cos(x+y)(2+b^2+a^2-2)=b^2-a^2$

$=>cos(x+y)(b^2+a^2)=b^2-a^2$

$=>cos(x+y)=(b^2-a^2)/(b^2+a^2)$

Answer 3 · 2016-08-21T15:54:46

$cos(x+y)=(b^2-a^2)/(b^2+a^2)$ .

Explanation:

$sinx+siny=a rArr 2sin((x+y)/2)cos((x-y)/2)=a.........(1)$ .

$cosx+cosy=b rArr 2cos((x+y)/2)cos((x-y)/2)=b..........(2)$ .

Dividing $(1)$ by $(2)$ , we have, $tan((x+y)/2)=a/b$ .

Now, $cos(x+y)={1-tan^2((x+y)/2)}/{1+tan^2((x+y)/2)}$

$=(1-a^2/b^2)/(1+a^2/b^2)=(b^2-a^2)/(b^2+a^2)$ .

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