All categories

3 years ago

use the power-reducing formulas to rewrite the expression in terms of the first power of the cosine . . sin^6 x

1 answer:

andrey2020 [161]3 years ago

4 0

1/128 (3 - 4 Cos[4 x] + Cos[8 x])( sin^2(x) )^2 cos^4(x)

( (1 - cos^2(x) )^2 cos^4(x) 

(1 - 2cos^2(x) + cos^4(x) ) cos^4(x) 

cos^4(x) - 2cos^6(x) + cos^8(x) 

( cos^2(x) )^2 - 2( cos^2(x) )^3 + ( cos^2(x) )^4 <==> knowing the identity; cos^2(x) = (1/2) * (1 + cos(2x)) 

( (1/2) * (1 + cos(2x)) )^2 - 2( (1/2) * (1 + cos(2x)) )^3 + ( (1/2) * (1 + cos(2x)) )^4 

( (1/4) * (1 + 2cos(2x) + cos^2(2x) )) - 2( (1/8) * (cos^3(2x) + 3cos^2(2x) + 3cos(2x) + 1) + ( (1/16) * cos^4(2x) + 4cos^3(2x) + 6cos^2(2x) + 4cos(2x) + 1)

You might be interested in

Stacey wishes to make a hamburger which is lower in fat without having to give up the great taste. She purchases 5 pounds of gro

STALIN [3.7K]

Answer:

%P = 8%

the percentage of fat in the mixture is 8%

Step-by-step explanation:

Given;

The percentage of fat in the mixture is;

%P = total mass of fat in mixture/total mass of mixture × 100%

%P = F/T × 100% ….....1

Mass of fat in mixture is;

She purchases 5 pounds of ground beef which is 15% fat

F1 = 15% of 5 pounds = 0.15×5 pounds

F1 = 0.75 pound

and combines it with 7 pounds of ground turkey which is only 3% fat

F2 = 3% of 7 pounds = 0.03 × 7

F2 = 0.21 pound

Total mass F = F1 + F2

F = 0.75+0.21

F = 0.96 pound

Total mass of mixture T = 5 + 7 pounds

T = 12 pounds.

From equation 1,

%P = F/T × 100%

%P = 0.96/12 × 100%

%P = 0.08×100%

%P = 8%

the percentage of fat in the mixture is 8%

5 0

3 years ago

³√73-(-3)², calculate the value

Vitek1552 [10]

The value of this equation is: 4
Answer: 4

6 0

3 years ago

Find the value of n.

Yuki888 [10]

The answer is 30. You are welcome

5 0

3 years ago

Read 2 more answers

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago

Solve for a 5 + 7a = 19

Bess [88]

A = 2 all you gotta do is 7 x 2 which is 14 and add 5 to it which equals 19

8 0

2 years ago

Read 2 more answers