All categories

3 years ago

If sin(x) = squareroot 2 over 2 what is cos(x) and tan(x)

Mathematics

1 answer:

Bumek [7]3 years ago

7 0

Answer:

cos(x) = square root 2 over 2; tan(x) = 1

Step-by-step explanation:

$\frac{\sqrt{2} }{2}$

was, before it was rationalized,

$\frac{1}{\sqrt{2} }$

Therefore,

$sin(x)=\frac{1}{\sqrt{2} }$

The side opposite the reference angle measures 1, the hypotenuse measures square root 2. That makes the reference angle a 45 degree angle. From there we can determine that the side adjacent to the reference angle also has a measure of 1. Therefore,

$cos(x)=\frac{1}{\sqrt{2} }=\frac{\sqrt{2} }{2}$ and

since tangent is side opposite (1) over side adjacent (1),

tan(x) = 1

You might be interested in

a woman 1.65. tall stood 50m away from the foot of a tower,and observe that the angle of elevation of the top of the tower to be

ElenaW [278]

$\sf\huge\underline{\star Solution:-}$

$\rightarrow$ Let the women's height be AE and distance between the women and tower be AC.

Also let the height of tower be BC.

$\rightarrow$ Now, clearly it is forming a triangle.

So, in triangle ABC,

$\rightarrow$ $\sf{Tan50° \:= \: \frac{BC}{AC}}$

$\rightarrow$ $\sf{1.19 \:= \: \frac{BC}{50}}$

$\rightarrow$ $\sf{1.19 ×50\:= \: BC}$

$\rightarrow$ $\sf{59.5m\:= \: BC}$

$\rightarrow$ Hence, BC = 59.5m

So, BD(total height of the tower) $\sf{ = \:BC+CD}$

$\sf{= \:59.5+1.65}$

$\sf{=\: 61.15m}$

Therefore, total height of the tower = $\sf\purple{ 61.15m.}$

________________________________

Hope it helps you:)

5 0

2 years ago

Read 2 more answers

Ya'll FR get 50 points if you answer this question.

NeX [460]

its 0.37

the way to find out is to multiply each of the options to the x and if you don't get the y its wrong

pls give me br ai nli es t im try na rank up

7 0

2 years ago

Read 2 more answers

How do you work out <br> 5x71

Gwar [14]

Answer:

355

Step-by-step explanation:

8 0

3 years ago

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

Zina [86]

Answer:

Step-by-step explanation:

The given differential equation is:

$x^3y'' + 2x^2y' + 4y$

the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.

So, for a regular singular point ; $x=x_o$ is located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.

Let first convert it to standard form by dividing through with x³

$y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0$

$y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0$

The standard form of the differential equation is :

$\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0$

Thus;

$P(x) = \dfrac{2}{x}$

$Q(x) = \dfrac{4}{x^3}$

The zeros of $x,x^3$ is 0

Therefore , the singular points of above given differential equation is 0

Classify each singular point as regular or irregular.

Let p(x) = xP(x) and q(x) = x²Q(x)

p(x) = xP(x)

p(x) = $x*\dfrac{2}{x}$

p(x) = 2

q(x) = x²Q(x)

q(x) = $x^2 * \dfrac{4}{x^3}$

q(x) = $\dfrac{4}{x}$

The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.

Thus ; from above; we can say that q(x) is not analytic at x = 0

$Q(x) = \dfrac{4}{x^3}$ do not satisfy the condition,at most to the second power in the denominator of Q(x).

Thus, the point x =0 is an irregular singular point

6 0

3 years ago

A building is 4,500 feet tall while the scale model of the mountain is 15 inches tall. What is the scale factor?

Natasha_Volkova [10]

Answer:

I think it is 300.

Step-by-step explanation:

5 0

3 years ago