tg x - tg x sin^2 x and cos x = -4/5

Asked on by elmee

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Since the request of the problem is vague, the only thing you can do, under the given conditions, is to evaluate the value of the expression `tan x - tan x*sin^2 x`  for `cos x = -4/5`  .

You should factor out `tan x ` such that:

`tan x - tan x*sin^2 x = tan x(1 - sin^2 x)`

You should remember the basic formula of trigonometry such that:

`sin^2 x + cos^2 x = 1 => cos^2 x = 1 - sin^2 x`

Hence, substituting `cos^2 x`  for `1 - sin^2 x`  yields:

`tan x(1 - sin^2 x) = tan x*cos^2 x`

You should remember that `tan x = sin x/cos x`  such that:

`sin x/cos x*cos^2 x = sin x*cos x `

Notice that the problem provides the values of cos x, hence, using the basic formula of trigonometry yields:

`sin x = +-sqrt(1 - 16/25) => sin x = +-sqrt(9/25) => sin x = +-3/5`

Notice that the value of cosine is negative, hence, the angle x is either in the quadrant 2, where sine is positive, or in quadrant 3, where sine is negative.

`tan x - tan x*sin^2 x = +-(3/4)*(4/5) = +-12/25`

Hence, evaluating the expression, under the given conditions, yields `tan x - tan x*sin^2 x =+-12/25` .

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llong4 | Student | (Level 1) eNoter

Posted on

Your question is unclear.

So let's assum that cos x=-4/5 and need to find out the value of tg x-tg x sin^2x.

sin^2 x=1-cos^2 x=1(-4/5)^2=9/25,

therefore sin x=-3/5 or 3/5.

If sin x=-3/5,  

then tg x=sin x/cos x=-3/5/(-4/5)=3/4.

Hence,tg x - tg x sin^2 x=3/4-3/4*(9/16)=48/64-27/64=21/64.



If sin x=3/5,

then tg x=sin x/cos x=3/5/(-4/5)=-3/4.

Hence,tg x - tg x sin^2 x=-3/4-(-3/4)*(9/16)

                                     = -48/64+27/64=-21/64.


Hope that helps.

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