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[Risolto] Seno e coseno

  

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2
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$cos(β) = cos\big[180-2·cos^{-1}\big(\frac{1}{4}\big)\big] = \dfrac{7}{8}$;

$sen(β) = sen\big[180-2·cos^{-1}\big(\frac{1}{4}\big)\big] = 0,484123$.



2

COS(β) = COS(pi - 2·α) = - COS(2·α)

COS(β) = - (COS(α)^2 - SIN(α)^2)

COS(α) = 1/4

SIN(α) = √(1 - (1/4)^2) = √15/4

Quindi:

COS(β) = - ((1/4)^2 - (√15/4)^2) = 7/8

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SIN(β) = √(1 - (7/8)^2) = √15/8

 



2

* cos(α) = 1/4
* cos^2(α) = 1/16
* sin^2(α) = 15/16
* sin(α) = ± √15/4
* β = π - 2*α
* sin(β) = sin(π - 2*α) = sin(2*α) = 2*sin(α)*cos(α)
* cos(β) = cos(π - 2*α) = - cos(2*α) = sin^2(α) - cos^2(α)



2
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sin β/2 = cos α = 0,25

cos β/2 = √1-1/16 = (√15)/4

sin β = 2*sin β/2*cos β/2 = (√15)/8 

cos β = √1-sin^2 β = (√1-15/64 = 7/8 

 



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