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Leona [35]
2 years ago
9

Show that each statement is true: Cos^4X-sin^4X=2cos^2X-1

Mathematics
1 answer:
Ivan2 years ago
4 0
cox^4x-sin^4x=2cos^2x-1\\\\
(cosx^2-sin^2x)(cosx^2x+sin^2x)=2cos^2x-1\\\\
cosx^2-sin^2x*1=2cos^2x-1\ \ \ | subtract\ 2cos^2x\\\\
-cosx^2-sin^2x=-1\ \ \ | multiply\ by\ -1\\\\
cosx^2+sin^2x=1\\\\
1=1
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Deffense [45]

Answer:

Part 1) MD=9\ units

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Step-by-step explanation:

Step 1

Find the length of MD

we know that

The incenter is the intersection of the angle bisectors of the three vertices of the triangle. Is the point forming the origin of a circle inscribed inside the triangle

so

In this problem

MD=ME=MF ------> is the radius of a circle inscribed inside the triangle

we have that

MF=9\ units

therefore

ME=9\ units

MD=9\ units

Step 2

Find the length of DC

we know that

In the right triangle MDC

Applying the Pythagoras theorem

MC^{2} =MD^{2}+DC^{2}

we have

MD=9\ units

MC=15\ units

substitute

15^{2} =9^{2}+DC^{2}

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