All categories

3 years ago

If sin theta = cos theta then find the value of theta

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

4 0

Step-by-step explanation:

$\because \sin \theta = \cos \theta \\ \\ \therefore \: \sin \theta = \sin(90 \degree - \theta) \\ \\ \therefore \: \theta = 90 \degree - \theta \\ \\ \therefore \: \theta + \theta= 90 \degree \\ \\ \therefore \: 2\theta = 90 \degree \\ \\\therefore \: \theta = \frac{90 \degree }{2} \\ \\ \huge \orange{\boxed{\therefore \: \theta = 45 \degree}}\\ \\$

You might be interested in

Identify an equation in point-slope form for the line parallel to y=-2/3x+8 that

UNO [17]

Answer:

$\large\boxed{y+5=\dfrac{2}{3}(x-4)}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

Parallel lines have the same slope.

We have the equation in the slope-intercept form (y = mx + b)

$y=-\dfrac{2}{3}x+8\to m=\dfrac{2}{3}$

Put to the point-slope equation value of the slope and the coordinates of the point (4, -5):

$y-(-5)=\dfrac{2}{3}(x-4)\\\\y+5=\dfrac{2}{3}(x-4)$

7 0

3 years ago

What are two ways that you can find the surface area of a prism

3241004551 [841]

Answer:

Step-by-step explanation:

4 0

3 years ago

Consider the sequence 8, 14, 20, 26, . . .. (a) What is the next term in the sequence? 2.2 Exercises 98 (b) Find a formula for t

omeli [17]

Answer:

Step-by-step explanation:

Each term of the given arithmetic sequence is 6 more than the previous one. Thus, the next term is 26+6, or 32.

a(n) = 8 + 6(n-1)

3 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is