Show that (sin x+siny+sinz)/(cosx+cosy+cosz)=tg y if 2y=x+z?

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You need to consider the given condition `2y = x + z` , such that:

`2y = x + z => y = (x+z)/2`

Taking sine and cosine function both sides, yields:

`sin y = sin ((x+z)/2)`

`cos y = cos ((x+z)/2)`

You need to convert the summation `sin x + sin z` into a product, such that:

`sin x + sin z = 2 sin((x + z)/2)cos((x - z)/2)`

You need to convert the summation `cos x + cos z` into a product, such that:

`cos x + cos z = 2cos ((x+z)/2)cos ((x-z)/2)`

Replacing `2 sin((x + z)/2)cos((x - z)/2)` for `sin x + sin z` and `2cos ((x+z)/2)cos ((x-z)/2)` for `cos x + cos z` yields:

`(sin x+siny+sinz)/(cosx+cosy+cosz) = (2 sin((x + z)/2)cos((x - z)/2)+siny)/(2cos ((x+z)/2)cos ((x-z)/2)+cosy)`

Replacing `sin ((x+z)/2)` for `sin y` and `cos ((x+z)/2)` for `cos y` yields:

`(sin x+siny+sinz)/(cosx+cosy+cosz) = (2 sin((x + z)/2)cos((x - z)/2)+sin ((x+z)/2))/(2cos ((x+z)/2)cos ((x-z)/2)+cos ((x+z)/2))`

Factoring out `sin ((x+z)/2)` to numerator and `cos ((x+z)/2)` to denominator, yields:

`(sin x+siny+sinz)/(cosx+cosy+cosz) = (sin ((x+z)/2)(2cos((x - z)/2) + 1))/(cos ((x+z)/2)(2cos ((x-z)/2) + 1))`

Reducing duplicate factors yields:

`(sin x+siny+sinz)/(cosx+cosy+cosz) = (sin ((x+z)/2))/(cos ((x+z)/2))`

`(sin x+siny+sinz)/(cosx+cosy+cosz) = tan((x+z)/2)`

Replacing `y` for `((x+z)/2)` yields:

`(sin x+siny+sinz)/(cosx+cosy+cosz) = tan y`

**Hence, testing if the given identity holds, under the given condition `y = ((x+z)/2)` , yields that `(sin x+siny+sinz)/(cosx+cosy+cosz) = tan y` is valid.**

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