Question

I have to answer these equations but I don't know how to?

Answer 1

`tan(-x)=-0.5`
`sin(-x)=-0.7`
`cos(-x)=0.2`
`tan(pi+x)=-4`

Explanation:

Tangent and Sine are odd functions. In any odd function, `f(-x)=-f(x)`. Applying this to tangent, `tan(-x)=-tan(x)`, so if `tan(x)=0.5`, `tan(-x)=-0.5`. The same process yields us `sin(-x)=-0.7`.

Cosine is an even function. In an even function, `f(-x)=f(x)`. In other words, `cos(-x)=cos(x)`. If `cos(x)=0.2`, `cos(-x)=0.2`.

Tangent is a function with a period of `pi`. Therefore, every `pi`, tangent will be the same number. As such, `tan(pi+x)=tan(x)`, so `tan(x)=-4`

Answer 2

If `tan x = .5` then `tan(-x)=-tan x = -.5`

If `sin x = .7` then `sin(-x) = -sin x =- .7`

If `cos x = .2` then `cos(-x) = cos x = .2`

If `tan x = -4` then `tan(pi+x)=tan x= -4`

Explanation:

These are asking the basic question of what happens to a trig function when we negate its argument. Negating an angle means reflecting it in the `x` axis. This flips the sign of the sine, but leaves the cosine alone. So,

`cos (-x) = cos x`

`sin(-x) = -sin x`

`tan (-x) = {sin(-x)}/{cos(-x)} = -tan (x)`

When we add `pi` to an angle we flip the sign on both sine and cosine.

`cos(x + \pi) = - cos x`

`sin(x + \pi) = - sin x`

`tan(x+ \pi) = {cos(x+pi)}/{sin(x+pi)} = tan x`

With that as background, let's do the questions:

If `tan x = .5` then `tan(-x)=-tan x = -.5`

If `sin x = .7` then `sin(-x) = -sin x =- .7`

If `cos x = .2` then `cos(-x) = cos x = .2`

If `tan x = -4` then `tan(pi+x)=tan x= -4`