If cos(x)cos(π/7)+sin(x)sin(π/7)= - (√2)/2, then x can equal:

1 answer:

8 0

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta) \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] ~\dotfill\\\\$

$\bf cos(x)cos\left( \cfrac{\pi }{7} \right)+sin(x)sin\left( \cfrac{\pi }{7} \right)=-\cfrac{\sqrt{2}}{2}\implies cos\left( x-\cfrac{\pi }{7} \right)=-\cfrac{\sqrt{2}}{2} \\\\\\ x-\cfrac{\pi }{7}=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right)\implies x-\cfrac{\pi }{7}= \begin{cases} \frac{3\pi }{4}\\\\ \frac{5\pi }{4} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf x-\cfrac{\pi }{7}=\cfrac{3\pi }{4}\implies x=\cfrac{3\pi }{4}+\cfrac{\pi }{7}~\hfill \stackrel{n ~\in~ \mathbb{Z}}{x=\cfrac{3\pi }{4}+\cfrac{\pi }{7}~~~~+2\pi n} \\\\[-0.35em] ~\dotfill\\\\ x-\cfrac{\pi }{7}=\cfrac{5\pi }{4}\implies x=\cfrac{5\pi }{4}+\cfrac{\pi }{7}~\hfill \stackrel{n ~\in~ \mathbb{Z}}{x=\cfrac{5\pi }{4}+\cfrac{\pi }{7}~~~~+2\pi n}$

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Step-by-step explanation:

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1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

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3 years ago

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A, B and E.

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4 years ago

A store sells pool floats.

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Step-by-step explanation:

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