All categories

2 years ago

Write the product as a sum: 14cos(39x)sin(19x)

Mathematics

1 answer:

charle [14.2K]2 years ago

4 0

Recall that

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

sin(a - b) = sin(a) cos(b) - cos(a) sin(b)

Adding these together gives

sin(a + b) + sin(a - b) = 2 sin(a) cos(b)

To get 14 cos(39x) sin(19x) on the right side, multiply both sides by 7 and replace a = 19x and b = 39x :

7 (sin(19x + 39x) + sin(19x - 39x)) = 14 cos(39x) sin(19x)

7 (sin(58x) + sin(-20x)) = 14 cos(39x) sin(19x)

7 (sin(58x) - sin(20x)) = 14 cos(39x) sin(19x)

You might be interested in

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

$(\sqrt[4]{13},0,0)$

$(-\sqrt[4]{13},0,0)$

$(0,\sqrt[4]{13},0)$

$(0,-\sqrt[4]{13},0)$

$(0,0,\sqrt[4]{13})$

$(0,0,-\sqrt[4]{13})$

$\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

5 0

3 years ago

A bike store sells scooters at a 54% markup. If the store bought each scooter for $29.25, what is the resale price to the neares

7nadin3 [17]

The purchasing price of the store = $29.25.

They want to resale it at 54% markup of the purchasing price.

54% of 29.25 = 0.54 × 29.25.

On multiplying 0.54 and 29.25, we get 15.795.

Resale price = The purchasing price + Markup

= 29.25 + 15.795.

On adding 29.25 and 15.795, we get 45.045.

We can round it to 45 to the nearest dollar.

<h3>Therefore, the resale price to the nearest dollar is $45.</h3>

5 0

3 years ago

Can someone help me with this :(

julia-pushkina [17]

Answer:

Step-by-step explanation:

6+x, when x = 3

6+(3)

6+3=9

5 0

2 years ago

Read 2 more answers

Saige's spaceship traveled 588 588588 kilometers ( km ) (km)left parenthesis, start text, k, m, end text, right parenthesis in 6

pshichka [43]

Answer:

The first ship has the same speed as Saige's ship while the second and third do not have the same speed.

Step-by-step explanation:

d = Distance traveled by Saige's spaceship = 588 km

t = Time taken = 60 seconds

Speed is given by

$s=\dfrac{d}{t}\\\Rightarrow s=\dfrac{588}{60}\\\Rightarrow s=9.8\ \text{km/s}$

The other ships

$s=\dfrac{441}{45}=9.8\ \text{km/s}$

$s=\dfrac{215}{25}=8.6\ \text{km/s}$

$s=\dfrac{649}{110}=5.9\ \text{km/s}$

So, the first ship has the same speed as Saige's ship while the second and third do not have the same speed.

8 0

2 years ago

Can someone pls help me with this problem

VLD [36.1K]

Answer:

Rule= Y × 10

X ÷ 10

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers