All categories

2 years ago

SinA/sinB=2/√3 and cosA/cosB=1/√2 ,tan^2B=?

Mathematics

1 answer:

Talja [164]2 years ago

6 0

Answer:

3/8 tan²A

Step-by-step explanation:

sinB = √3/2 sinA and cos B= √2 cosA (from given)

You might be interested in

Find all x-intercepts and y-intercepts of the graph of the function.<br> F(x)=x³-4x² - 4x + 16

Lady bird [3.3K]

Answer: (0;16) (4;0) (2;0) (-2;0).

Step-by-step explanation:

Y-intercepts of the graph of the function F(x)=x³-4x² - 4x + 16:

$x=0\\F(x)=0^3-4*0^2-4*0+16\\F(x)=16.\\Hence,\ (0;16).$

X-intercepts of the graph of the function F(x)=x³-4x² - 4x + 16:

$F(x)=0\\0=x^3-4x^2-4x+16\\0=(x^3-4x^2)-(4x+16)\\0=x^2*(x-4)-4*(x-4)\\0=(x-4)*(x^2-4)\\0=(x-4)*(x^2-2^2)\\0=(x-4)*(x-2)*(x+2)\\x-4=0\\x_1=4.\\Hence,\ (4;0)\\x-2=0\\x_2=2.\\Hence,\ (2;0)\\x+2=0\\x_3=-2.\\Hence,\ (-2;0).$

8 0

1 year ago

(2x+5)/3=-7 <br> show all work

Phantasy [73]

→(2X+5)=-21→2X=-26→X=-13

3 0

3 years ago

Raise 1.00203333 to the negative 360 power

Dmitry [639]

Answer:0.4813042802

Step-by-step explanation:

8 0

3 years ago

Find the area of a rectangle with side lengths 3/5 inch and 5/6 inch

Helga [31]

Answer:

15/30 0r 1/2

Step-by-step explanation:

You multiply the fractions

8 0

2 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago

SinA/sinB=2/√3 and cosA/cosB=1/√2 ,tan^2B=?​

SinA/sinB=2/√3 and cosA/cosB=1/√2 ,tan^2B=?