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bixtya [17]
3 years ago
8

Prove that cos 3A = cos (2A+A)

Mathematics
2 answers:
Blababa [14]3 years ago
6 0

Not sure what the question is, but I guess it's to prove that

\cos3A=4\cos^3A-3\cos A

Expand the left side as

\cos3A=\cos(2A+A)=\cos2A\cos A-\sin2A\sin A

and use the double angle identities to write

\cos3A=(\cos^2A-\sin^2A)\cos A-2\sin^2A\cos A

\cos3A=\cos^3A-3\sin^2A\cos A

Recall the Pythagorean identity:

\cos3A=\cos^3A-3(1-\cos^2A)\cos A

\cos3A=\cos^3A-3\cos A+3\cos^3A

\implies\cos3A=4\cos^3A-3\cos A

as required.

olga55 [171]3 years ago
4 0

Answer:

you useless coomstain

Step-by-step explanation:

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4 0
3 years ago
2(x-1)=3(x-4) how many solutions
34kurt
Distribute
2(x-1) =3(x-4)
2x -2 =3 (x-4)
Distribute
2x-2 =3 (x-4)
2x-2=3x-12
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2x -3x =3x -10 -3x
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6 0
3 years ago
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what is the order from least to greatest - 4 with the exponent of 3, square root of 40, square root of 13, square root of 7, , 1
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Answer: -4^{3} , 14/9 , \sqrt{7} , \sqrt{13} , \sqrt{40}

Step-by-step explanation:

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Arranging from least to greatest , we have :

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Answer:

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