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3 years ago

Find the values of the trigonometric functions of θ from the information given.

Mathematics

1 answer:

OLga [1]3 years ago

3 0

Answer:

sin(θ) = -(4√15)/17
cos(θ) = 7/17 . . . . . . . given
tan(θ) = -(4√15)/7
csc(θ) = -(17√15)/60
sec(θ) = 17/7
cot(θ) = -(7√15)/60

Step-by-step explanation:

The relationship between sine and cosine is ...

sin² + cos² = 1

Solving for sine gives ...

sin = ±√(1 -cos²)

In this problem, we want the negative root.

sin(θ) = -√(1 -(7/17)²) = -√(240/289) = -(4√15)/17

tan(θ) = sin(θ)/cos(θ) = ((-4√15)/17)/(7/17) = -(4√15)/7

___

And the inverse functions are ...

sec(θ) = 1/cos(θ) = 17/7

csc(θ) = 1/sin(θ) = -17/(4√15) = -(17√15)/60

cot(θ) = 1/tan(θ) = -(7√15)/60

_____

Of course, you're aware that 1/√15 = (√15)/15.

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I need help with substitution

Neporo4naja [7]

Answer:

(-7,-15)

Step-by-step explanation:

We can substitute first y in the second y. Both equations are also functions. We can merge the equations.

$2x - 1 = x - 8 \\ 2x - x = - 8 + 1 \\ x = - 7$

Substitute x = -7 in any given equations. I will choose the second equation to substitute in.

$y = x - 8 \\ y = - 7 - 8 \\ y = - 15$

Answer Check

Substitute both x and y in both equations.

$- 15 = 2( - 7) - 1 \\ - 15 = - 14 - 1 \\ - 15 = - 15$

The equation is true for (-7,-15).

$- 15 = - 7 - 8 \\ - 15 = - 15$

The second equation is true for (-7,-15).

Therefore our answer is (-7,-15)

6 0

3 years ago

Read 2 more answers

Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .