# knowing that: sen(3π/2+x)=1/3 and x E ]π,2π[, calculate the exact value of cos (π+x) - 2cos (π/2+x)π-Pi

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You need to use the following formulas to evaluate `cos (pi+x` ) and `cos(pi/2+x) ` such that:

`cos(pi+x) = cos pi*cos x - sin pi*sin x`

Using `cos pi = -1` and `sin pi = 0 ` yields:

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

`cos(pi/2+x) = cos (pi/2)*cos x - sin (pi/2)*sin x`

Using `cos (pi/2) =0 ` and `sin (pi/2) =1` yields:

`cos(pi/2+x) = -sin x`

Hence, `cos(pi+x) - 2cos(pi/2+x) = -cos x + 2 sinx`

The probem provides the information `sin(3pi/2 + x) = 1/3` and `sin(3pi/2 + x) = sin 3pi/2*cos x + sin x*cos 3pi/2`

Using `cos 3pi/2 =0` and `sin 3pi/2 =-1` yields:

`sin(3pi/2 + x) = -cos x`

Hence, `-cos x = 1/3 =gt cos x = -1/3 `

Using Pythagorean's theorem yields:

`sin x = +-sqrt(1 - cos^2 x)`

`sin x = +-sqrt(1 - 1/9) =gt sin x = +-sqrt(8/9)`

`sin x = +-2sqrt2/3`

The problem provides the information that `x in [pi,2pi]` , hence, you need to use only the negative value for sin x, because the values of sine function are negative in quadrants 3 and 4.

Hence, `sin x = -2sqrt2/3.`

You need to substitute `-2sqrt2/3` for `sin x` and `-1/3` for `cos x` in expression `cos(pi+x) - 2cos(pi/2+x) = -cos x + 2 sinx ` such that:

`cos(pi+x) - 2cos(pi/2+x) = -(-1/3) + 2*(-2sqrt2/3)`

`cos(pi+x) - 2cos(pi/2+x) = 1/3 - 4sqrt2/3`

**Hence, evaluating the expression `cos(pi+x) - 2cos(pi/2+x) ` under given conditions yields `cos(pi+x) - 2cos(pi/2+x) = 1/3 - 4sqrt2/3.` **

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