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vodomira [7]
4 years ago
5

Prove cos^4 X +sin^4 x= 2cos^2 x for any acute angle​

Mathematics
2 answers:
Nikolay [14]4 years ago
8 0

Answer:

Proof provided in the explanation.

Step-by-step explanation:

\cos^4(\theta)-\sin^4(\theta)+1

I'm going to rewrite that middle term so I can apply Pythagorean Identity, \sin^2(\theta)=1-\cos^2(\theta):

\cos^4(\theta)-(\sin^2(\theta))^2+1

\cos^4(\theta)-(1-\cos^2(\theta))^2+1

Use (a+b)^2=a^2+2ab+b^2:

\cos^4(\theta)-[1-2\cos^2(\theta)+\cos^4(\theta)]+1

Distribute:

\cos^4(\theta)-1+2\cos^2(\theta)-\cos^4(\theta)+1

Combine like terms:

\cos^4(\theta)-\cos^4(\theta)-1+1+2\cos^2(\theta)

Applying inverse property of addition:

0+0+2\cos^2(\theta)

Applying identity property of addition:

2\cos^2(\theta)

Nonamiya [84]4 years ago
8 0

To prove cos^4x-sin^4x +1= 2 cos^2 x

cos^4x -sin^4 x+1

= ( cos^2x +sin^2x)(cos^2x - sin^2x) +1

You need to remember 3 basic results

sin^2x + cos^2x= 1

cos2x = cos^2x - sin^2x

1+ cos 2x = 2 cos^2x

Now Let's revert back to question

= ( cos^2x + sin^2x )( cos^2x - sin^2x) +1

= 1( cos2x) +1

= cos2x +1

= 2 cos^2x

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Alchen [17]

Answer:

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3 years ago
Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?
photoshop1234 [79]

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326

What is the probability that they are both queens?

P = \frac{D}{T} = \frac{6}{1326} = 0.0045

0.45% probability that they are both queens.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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