how do you simplify
[tex] \frac{ \sin(90 - \alpha ) \times \tan( - \alpha ) \times \cos( \alpha - 180) }{ \ \cos(100) \sin(135?) } [/tex]

Answers

Answer 1

Answer:

[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times cos(\alpha - 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})} = \dfrac{\sqrt{2} \times sin(2 \times \alpha) }{2 \times cos(100^{\circ}) }[/tex]

Explanation:

The given expression is presented as follows;

[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times cos(\alpha - 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})}[/tex]

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;

[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times -cos( 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})} = \dfrac{cos( \alpha) \times -tan (\alpha) \times -cos(\alpha )}{cos(100^{\circ}) \times \dfrac{\sqrt{2} }{2}}[/tex]

[tex]\dfrac{cos( \alpha) \times -\dfrac{sin(\alpha)}{cos(\alpha)} \times -cos(\alpha )}{cos(100^{\circ}) \times \dfrac{\sqrt{2} }{2}} = \dfrac{sin( \alpha) \times cos(\alpha )}{cos(100^{\circ}) \times \dfrac{\sqrt{2} }{2}} = \dfrac{2}{\sqrt{2} } \times \dfrac{sin( \alpha) \times cos(\alpha )}{cos(100^{\circ}) }[/tex]

[tex]\dfrac{2}{\sqrt{2} } \times \dfrac{sin( \alpha) \times cos(\alpha )}{cos(100^{\circ}) } = \dfrac{sin(2 \times \alpha) }{\sqrt{2} \times cos(100^{\circ}) }[/tex]

Therefore;

[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times cos(\alpha - 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})} = \dfrac{\sqrt{2} \times sin(2 \times \alpha) }{2 \times cos(100^{\circ}) }[/tex]


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Answers

Answer:

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