soit ` x ` un élément de `[0 ,pi ] ` et ` x ne {pi}/2`

on pose `P(x) = 2(1-cos^2(x+{pi}/2)) -cos(pi-x)sin(pi+x)`

1) Simplifier `P(x)` puis Calculer `P(0) ` et `P({pi}/4)`

2)

a) Montrer que `P(x) = {2-tanx}/{1+tan^2x}`

b) Calculer `P({pi}/3)` et `P(pi)`

3) On suppose que `P(x)= 2 ` et `x` un élément de l’intervalle `[0 ,{pi}/2[`

a) Trouver la valeur de `tanx`

b) En déduire la valeur de `x`

Simplifier `P(x) = 2(1-cos^2(x+{pi}/2)) -cos(pi-x)sin(pi+x)`

Rappels

pour tout `x ` de `R`

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

2 `cos(pi- x) = -cosx`

3 `sin(pi +x) = -sinx`

4 `sin(x+{pi}/2)= cosx`



On a `P(x) = 2(1-cos^2(x+{pi}/2)) -cos(pi-x)sin(pi+x)`

Comme `cos^2x+sin^2x= 1 `

`=> 1-cos^2(x+{pi}/2) = sin^2(x+(pi)/2) `

` = cos^2x ` car et `sin(x+{pi}/2)=cosx`

Comme `cos(pi- x) = -cosx` et `sin(pi +x) = -sinx`

`=> cos(pi-x)sin(pi+x) = -cosx xx (-sinx ) = cosx xx sinx `

alors `P(x) = 2sin^2(x+{pi}/2) -[-(cosx)(-sinx)] `

`=> P(x) = 2cos^2x -sinxxxcosx `

`=> P(x) = cosx( 2cosx -sinx) `

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Calculons de `P(0)`

on a ` P(0) = cos0( 2cos0-sin0) `

or `cos(0) = 1 ` et `sin(0)=0`

alors `P(0) = 1xx(2xx1-0) = 1xx2=2`

`=> P(0)= 2 `

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Calculons `P((pi)/4)`

on a ` P((pi)/4) = cos((pi)/4)( 2cos((pi)/4)-sin((pi)/4)) `

or `cos((pi)/4) = sin((pi)/4) = (sqrt(2))/2 `

` => P((pi)/4) = (sqrt(2))/2( (2sqrt(2))/2-(sqrt(2))/2) = 2/4 =1/2 `


`P((pi)/4)= 1/2`

2) Montrer que `P(x) = {2-tanx}/{1+tan^2x} `

Rappels

` 1 +(tanx)^2 = 1+(sinx)^2/(cosx)^2 = (cos^2x+sin^2x)/(cosx)^2 = 1/(cosx)^2 `

`cos^2x= 1/{1+tan^2x}`



on a ` P(x) = 2cos^2x -sinxxxcosx `

` => P(x) = 2cos^2x - {sinx}/{cosx}xxcos^2x `

` => P(x) = 2cos^2x -tanx cos^2x `

` => P(x) = cos^2x(2 -tanx) `

or `cos^2x= 1/{1+tan^2x}`

` => P(x) = 1/{1+tan^2x}xx(2 -tanx) = {2-tanx}/{1+tan^2x} `

alors ` => P(x) = {2-tanx}/{1+tan^2x} `

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Calculons `P({pi}/3)`

on a ` P(x) = {2-tanx}/{1+tan^2x} `

alors ` P({pi}/3) = {2-tan({pi}/3)}/{1+tan^2({pi}/3)} `


on a `tan({pi}/3)= sqrt(3) ` alors `tan^2({pi}/3) = 3`

` => P({pi}/3) = {2-sqrt(3)}/{1+3} = {2-sqrt(3)}/4 `

`=> P((pi)/3)={2-sqrt(3)}/4 `

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Calculons `P(pi)`

on a ` P(pi) = {2-tan(pi)}/{1+tan^2(pi)} `

or `tan(pi)= 0 `

` => P(pi) = {2-0}/{1+0} = 2/1= 2 `

`=> P(pi)= 2 `

3) On suppose que `P(x)= 2 ` et `x` un élément de l’intervalle `[0 ,{pi}/2[`

a) Trouver la valeur de `tanx`

on a ` P(x) = {2-tan(x)}/{1+tan^2(x)} `

alors `P(x)= 2 `

` {2-tan(x)}/{1+tan^2(x)} = 2 `

`2-tanx = 2+2tan^2x`

`-tanx = 2tan^2x`

` tanx(2tanx+1) = 0 `

on a ` x in [0,{pi}/2[` alors `sinx >= 0 ` et `cosx > 0 `

alors `tanx >= 0 ` `=> 2tanx >= 0 `

`=> 2tanx +1 >= 1 > 0 `

` => tanx(2tanx+1) = 0 `

équivalente à `tanx = 0 `

`=> tanx = 0 `

3 b) En déduire la valeur de `x`

On a ` tanx = 0 `

`{sinx}/{cosx} = 0 `

`sinx= 0 `

or `x in [0,{pi}/2[`

alors ` x = 0 `