MATHEMATICS HIGH SCHOOL

PLEASE HELP Use the squared identities to simplify 2cos2x cos2x.

Answers

Answer 1

Answer:

The correct option is option D as 2cos²(x)cos²(x) simplifies as follows:

2cos²(x)cos²(x) = {3 + 4cos(2x) + cos(4x)} / 4

Simplification:

The given expression is : 2cos²(x)cos²(x)

The square identity for cosine is given by:

2cos²(x) -1 = cos(2x)

Thus,

2cos²(x) = {cos(2x) +1}

simplifying again,

cos²(x) = {cos(2x) +1}/2

Simplifying the above using squared identities:

2cos²(x)cos²(x) = {cos(2x) +1}cos²x

= {cos(2x) +1} {{cos(2x) +1}/2}

$= \frac{\{cos(2x) +1\}^2}{2}\n\n=(cos^2(2x)+2cos(2x)+1)/(2)\n\n=((cos(4x)+1)/(2)+2cos(2x)+1)/(2)\n\n=(3+4cos(2x)+cos(4x))/(4)$

so,

2cos²(x)cos²(x) = {3 + 4cos(2x) + cos(4x)} / 4

Hence option D is correct.

Learn more about squared identities:

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Answer 2

Answer:

Answer:

Step-by-step explanation:

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Twenty percent of 150 is

Answers

20% of 150 is 30

Change the percentage into a decimal by dividing it over 100:

20 / 100 = 0.2

Multiply:

0.2 × 150 = 30

Answer: 30

Step-by-step explanation:

What does the 80:20 rule say?

Answers

Answer: The 80:20 rule says that asserts that 80% of outcomes result from 20% of all causes for any shown event.

Step-by-step explanation: I already learned about that.

Hope it helped :D

Please mark me brainliest.

Can someone help me solve for x

Answers

Answer:

3x x 2x = 6x

15+20 =35

x=30 x

Ans:

x=29

Step-by-step explanation:

it's a straight line so it's 180°

One less than the product of a number and 3 is 5 more than the number it self

Answers

Answer: 3n - 1 = n + 5

Step-by-step explanation:

If needed to solve 3n-1=n+5

3n-n=5+1

2n=6

n=3

This is extra information. I hope this helped you

Order from least to greatest 1/2,0.55,5/7

Answers

Believe it or not, it is already in that order because 1/2 is the same as 0.50 which is less than 0.55, and 5/7 is the same as 0.71 which is greater than all the above.

Round 9.0072 to 2 decimal places

Answers

You can round it to thousandths. 9.007
Or you can round it to hundredths 9.01

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