Question

If tan x = -3/4 and pi/2<x<pi find values of cos x and sin x?

`tan x = -3/4`and`pi/2 <x<pi` find the value of `cos x` and `sin x`

Answer 1

`cosx=-4/5 and sinx=3/5`

Explanation:

Here,

`tanx=-3/4 < 0 and pi/2 < x < pi=>II^(nd)Quadrant`

So,

`color(red)(sinx > 0 and cosx < 0`

Now,

`sec^2x=1+tan^2x=1+9/16=25/16`

`=>cos^2x=1/sec^2x=16/25`

`=>color(blue)(cosx=-4/5to[becausecosx < 0]`

We know that,

`sin^2x=1-cos^2x=1-16/25=9/25`

`=>color(blue)(sinx=+3/5to[becausesinx > 0]`

Hence,

`cosx=-4/5 and sinx=3/5`

Answer 2

Second quadrant, positive sine and negative cosine,

`cos x= - 4/5`

`sin x= 3/5`

Explanation:

We're looking for `x` in the second quadrant, so we want a positive sine and negative cosine.

In general, the multivalued

`arctan (a/b)`

refers to a right triangle, opposite `a`, adjacent `b` so hypotenuse `sqrt{a^2+b^2}.` The sign on the square root is always ambiguous in these, so

`cos arctan(a/b) = pm b/sqrt{a^2+b^2}`

`sin arctan(a/b) = pm a/sqrt{a^2+b^2}`

Of course `3^2+4^2=5^2` so the hypotenuse is `5`.

`cos arctan({-3}/4) = pm 4/5`

`sin arctan({-3}/4) = pm 3/5`

We want a positive sine and negative cosine.

`cos x= - 4/5`

`sin x= 3/5`