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How do you simplify


( \sin(90 - \alpha ) * \tan( - \alpha ) * \cos( \alpha - 180) )/( \ \cos(100) \sin(135?) )


User Nir Levy
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1 Answer

7 votes

Answer:


(sin(90^(\circ) - \alpha) * tan (-\alpha) * cos(\alpha - 180^(\circ)))/(cos(100^(\circ)) * sin(135 ^(\circ))) = (√(2) * sin(2 * \alpha) )/(2 * cos(100^(\circ)) )

Step-by-step explanation:

The given expression is presented as follows;


(sin(90^(\circ) - \alpha) * tan (-\alpha) * cos(\alpha - 180^(\circ)))/(cos(100^(\circ)) * sin(135 ^(\circ)))

The sine, cosine, and tangent of the common angles are given as follows;

sin(A - B) = sin(A)·cos(B) - cos(A)·sin(B)

∴ sin(90° - α) = sin(90°)·cos(α) - cos(90°)·sin(α) = cos(α)

tan(-α) = -tan(α)

cos(A - B) = cos(A)·cos(B) + sin(A)·sin(B)

∴ cos(α - 180°) = cos(α)·cos(180°) + sin(α)·sin(180°) = -cos(α)

sin(135°) = sin(90° + 45°) = sin(90°)·cos(45°) + sin(45°)·cos(90°) = cos(45°) = √2/2

Therefore, we get;


(sin(90^(\circ) - \alpha) * tan (-\alpha) * -cos( 180^(\circ)))/(cos(100^(\circ)) * sin(135 ^(\circ))) = (cos( \alpha) * -tan (\alpha) * -cos(\alpha ))/(cos(100^(\circ)) * (√(2) )/(2))


(cos( \alpha) * -(sin(\alpha))/(cos(\alpha)) * -cos(\alpha ))/(cos(100^(\circ)) * (√(2) )/(2)) = (sin( \alpha) * cos(\alpha ))/(cos(100^(\circ)) * (√(2) )/(2)) = (2)/(√(2) ) * (sin( \alpha) * cos(\alpha ))/(cos(100^(\circ)) )


(2)/(√(2) ) * (sin( \alpha) * cos(\alpha ))/(cos(100^(\circ)) ) = (sin(2 * \alpha) )/(√(2) * cos(100^(\circ)) )

Therefore;


(sin(90^(\circ) - \alpha) * tan (-\alpha) * cos(\alpha - 180^(\circ)))/(cos(100^(\circ)) * sin(135 ^(\circ))) = (√(2) * sin(2 * \alpha) )/(2 * cos(100^(\circ)) )

User Apogee
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