Question

If `cos(t) = −7/19` and tan(t) < 0, find sin(t) and cos(-t). ?


Answer 1

`sin(t)=(2sqrt78)/361`
`cos(-t)=-7/19`

Explanation:

`tan(t)<0` simply gives us quadrant information - the tangent function is negative in quadrants II and IV, while the cosine function is negative in quadrants II and III. Therefore `t` must occur in quadrant II.

To find `sin(t)`, you can use the identity `sin^2t+cos^2t=1`. Inputting knowns:
`sin^2t+(-7/19)^2=1` Simplify:
`sin^2t=312/361`
`sin(t)=(sqrt312)/361` or `(2sqrt78)/361`
We know this is positive because sine is positive in QII.

Because cosine is an even function, `cos(-t)=cos(t)`, so `cos(-t)=-7/19`

Answer 2

`sint=(2sqrt78)/19and cos(-t)=-7/19`

Explanation:

Here,

`cost=-7/19=> color(blue)(II^(nd)Quadrant )orIII^(rd)Quadrant...to(1)`

`tant < 0=> color(blue)(II^(nd) Quadrant) or IV^(th)Quadrant...to(2)`

From `(1) and (2)`,we can say that

`pi/2<= t <= pi to color(blue)(II^(nd) Quadrant)=>(sine) > 0`

So,

`(i).sin(t)=+sqrt(1-cos^2t)=sqrt(1-49/361)=sqrt312/19= (2sqrt78)/19`

We know that,

`"cosine is "color(red)"even function"=>cos(-theta)=costheta`

`(ii).cos(-t)=cost=-7/19`

Note:

`pi/2<= t <= pi =>-pi/2 >= (-t) >= -pi`

`i.e.-pi <= (-t) <= -pi/2=>III^(rd) Quadrant=>` cosine `< 0`