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2 years ago

Simplify the trigonometric expression sin(4x) +2 sin(2x) using Double-Angle Identities.

Mathematics

1 answer:

emmasim [6.3K]2 years ago

6 0

Answer:

8sin(x)cos³(x)

Step-by-step explanation:

sin(4x) +2 sin(2x) = 2sin(2x)*cos(2x) + 2sin(2x) = 2sin(2x)(cos2x + 1)=

= 2sin(2x)(cos²x - sin²x + cos²x + sin²x)=²2sin(2x)*(2cos²x)=

= 4*2sin(x)*cos(x)*cos²(x)= 8sin(x)cos³(x)

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Each month for 7 months, Eve reads 3 books. How many more books does she need to read before she has read 30 books?

enot [183]

How many books has she read so far? She's been doing this for 7 months, so so far she's read:

7*3=21

how many does she still need to read?

30-21=9

so she needs to read 9 more books. With 3 books a month, this means:

9/3=3

so she needs to read 9 more books, which will take her 3 months!

6 0

2 years ago

Hyun Woo is riding a ferris wheel. H t H(t) models his height (in m m) above the ground, t t seconds after the ride starts. Here

worty [1.4K]

Answer:

53 seconds.

Explanation:

The text and the model are garbled.

This is the question amended:

Hyun Woo is riding a ferris wheel. H(t) models his height (in m) above the ground, t seconds after the ride starts. Here, t is entered in radians.

H(t) = -10 cos(2π/150 t)+10

When does Hyun Woo first reach a height of 16 m?

<h2>Solution</h2>

When Hyun Woo reaches a height of 16 m the model states:

16 = -10 cos(2π/150 t)+10

Then you must find the lowest positive value of t that is a solution of the equation.

Solve the equation:

16 = -10 cos(2π/150 t)+10

10cos(2π/150 t) = - 6

cos(2π/150t) = -0.6

2π/150t = arccos(-0.6)

2π/150t = 2.214297

t = 52.86s ≈ 53 s ← answer

4 0

3 years ago

Help Me fast!!!!!!!!!!!!

Ugo [173]

Answer:

12.25 i think sorry if im wrong

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A diver went 25.65 feet below the surface of the ocean and then 16.5 feet further down he then rose 12.45 feet. Write and solve

trasher [3.6K]

X= 25.65 + 16.5 - 12.45
X=29.70

6 0

3 years ago

Read 2 more answers

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago