Question

Show that if $x$, $y$, and $z$ are consecutive terms of an arithmetic sequence, with $x \leq y \leq z$, and both sides of the equation are defined, then?

Answer 1

Answer:

x=a−h,y=a,z=a+h then you have  the following trigonometric identity

sin(a−h)=sinacosh−cosasinh

sin(a+h)=sinacosh+cosasinh

cos(a−h)=cosacosh+sinasinh

cos(a+h)=cosacosh−sinasinh

$\frac{sin(a-h)+sina+sin(a+h)}{cos(a-h)+cosa+cos(a+h)}$

$\frac{sina(2cosh+1)}{cosa(2cosh+1)}$

sina/cosa=tana

recall that a=y

sina/cosa=tany

Step-by-step explanation:

Show that if $x$, $y$, and $z$ are consecutive terms of an arithmetic sequence, with $x \leq y \leq z$, and both sides of the equation are defined, then?

the complete part of the question will be what follows

$\frac{sinx+siny+sinz}{cosx+cosy+cosz} =tany$

This are trigonometric identity,

We take the terms of the Linear Sequence as

x=a−h,y=a,z=a+h then you have  the following trigonometric identity

sin(a−h)=sinacosh−cosasinh

sin(a+h)=sinacosh+cosasinh

cos(a−h)=cosacosh+sinasinh

cos(a+h)=cosacosh−sinasinh

$\frac{sin(a-h)+sina+sin(a+h)}{cos(a-h)+cosa+cos(a+h)}$

$\frac{sina(2cosh+1)}{cosa(2cosh+1)}$

sina/cosa=tana

recall that a=y

sina/cosa=tany