Question

Question #e8ab5

Answer 1

`cos(x+y)=(a^2+b^2)/2-1`

Explanation:

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

Note that:
`(sinx+siny)^2=a^2`
`->sin^2x+2sinxsiny+sin^2y=a^2`

And:
`(cosx+cosy)^2=b^2`
`->cos^2x+2cosxcosy+cos^2y=b^2`

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

If we add them together, we have:
`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^2x+cos^2x)+(2sinxsiny+2cosxcosy)+(cos^2y+sin^2y)=a^2+b^2`

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

We can factor out a `2` twice:
`2(sinxsiny+cosxcosy)+2=a^2+b^2`
`->2((sinxsiny+cosxcosy)+1)=a^2+b^2`

And divide:
`(sinxsiny+cosxcosy)+1=(a^2+b^2)/2`

And subtract:
`sinxsiny+cosxcosy=(a^2+b^2)/2-1`

Finally, since `cos(x+y)=cosxcosy+sinxsiny`, we have:
`cos(x+y)=(a^2+b^2)/2-1`

Answer 2

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

`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)`.

Enjoy Maths.!