Rewrite with only sin x and cos x. cos 2x + sin x

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

8 0

we can pretty much use any of the 3 double-angle identities for cosine, so

$\bf \textit{Double Angle Identities} \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ 2cos^2(\theta)-1&\leftarrow \textit{let's use this one} \end{cases}\\\\ -------------------------------\\\\ cos(2x)+sin(x)\implies 2cos^2(x)-1+sin(x)$

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The vertices of a backyard are W(10,30),X(10,100),Y(110,100),Z(50,30). The coordinates are measured in feet. The line segment XZ

MariettaO [177]

Given vertices W(10,30), X(10,100), Y(110,100), Z(50,30) of a backyard, find distances WX, XY, YZ, ZW and XZ:

1. $WX=\sqrt{(10-10)^2+(30-100)^2}=\sqrt{70^2}=70\ ft;$

2. $XY=\sqrt{(110-10)^2+(100-100)^2}=\sqrt{100^2}=100\ ft;$

3. $YZ=\sqrt{(110-50)^2+(100-30)^2}=\sqrt{60^2+70^2}=10\sqrt{85}\ ft;$

4. $ZW=\sqrt{(50-10)^2+(30-30)^2}=\sqrt{40^2}=40\ ft;$

5. $XZ=\sqrt{(10-50)^2+(100-30)^2}=\sqrt{40^2+70^2}=10\sqrt{65}\ ft.$

Then:

1. the area

$A_{WXZ}=\sqrt{\frac{70+40+10\sqrt{65}}{2}\\\cdot (\frac{70+40+10\sqrt{65}}{2}-70)\cdot (\frac{70+40+10\sqrt{65}}{2}-40)\cdot (\frac{70+40+10\sqrt{65}}{2}-10\sqrt{65})}=\\ \\=1400\ ft^2.$

2. the area

$A_{XYZ}=\sqrt{\frac{100+10\sqrt{85}+10\sqrt{65}}{2}\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-100)}\cdot\\ \\\cdot\sqrt{(\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{85})\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{65})}=3500\ ft^2.$