1) Calculer `sinx` et `tanx` sachant que ` cosx=4/5` et ` 0 < x < pi/2`

2) Calculer `sinx` et `cosx` sachant que `tanx=1/3` ` -pi < x < -pi/2`

1) Calculer `sinx` et `tanx` sachant que ` cosx=4/5` sachant que ` 0 < x < pi/2`

Rappels

1 `a^2 = b^2 <=> [ a = b text{ ou } a = -b ] `

2 si ` a >= 0 ` et ` b >=0` alors `[ a^2= b^2 <=> a = b ] `







Puisque ` x in ]0, (pi)/2[ ` alors ` sinx > 0 ` et ` cosx > 0 `

On a sait que `sin^2x + cos^2x = 1 `

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

` sin^2x = 1 -(4/5)^2 = (25 -16)/(25)= 9/(25) `

` sinx = sqrt(9/(25)) = 3/5 ` car ` sinx >= 0 `

`=> sinx = 3/5 `

On a `tanx = (sinx)/(cosx)= (3/5)/(4/5) = 3/4 `

`=> tanx = 3/4 `

2) Calculer `sinx` et `cosx` sachant que `tanx=1/3` ` -pi < x < -pi/2`

Rappels

si `a >= 0` et ` b >= 0 ` alors `[ a^2 = b^2 <=> a = b ] `

si `a <= 0` et ` b <= 0 ` alors `[ a^2 = b^2 <=> a = b ] `



on a ` x in ]-pi , -(pi)/2[`





alors ` sinx < 0 ` et ` cosx < 0 `

On a ` tanx = 1/3 <=> (sinx)/(cosx)= 1/3 `

`<=> 3sinx = cosx `

`<=> (3sinx)^2 = (cosx)^2 ` car ` 3sinx` et `cosx` ont meme signe

`<=> 9sin^2x = cos^2x `

`<=> 10sin^2x = cos^2x + sin^2x = 1 `

`<=> sin^2x = 1/(10) `

`<=> sinx = -sqrt( 1/10) = -(sqrt(10))/(10) ` car ` sinx < 0 `

` sinx = -(sqrt(10))/(10) `


On a ` tanx = (sinx)/(cosx) `

`=> cosx = (sinx)/(tanx) = (-(sqrt(10))/(10))/( 1/3) `

`=> cosx = (-3sqrt(10))/(10) `

`=> cosx = (-3sqrt(10))/(10) `