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Brums [2.3K]
3 years ago
8

How would you solve the following trig equation sin(x)=-0.5 by isolating the x?

Mathematics
2 answers:
IrinaVladis [17]3 years ago
5 0

To get x by itself in a trig equation like this, you use the inverse sine. It's written on your calculator as sin⁻¹ and often in books as arcsin to differentiate it from the negative exponent. Taking the sine and inverse sin of any angle brings you back to the original angle.

sin (x) = -0.5

arcsin (sin(x)) = arcsin (-0.5)

x = arcsin (-0.5)

Since you are asking "what angle has its sine as -1/2?" this problem can be answered with a 30- 60 - 90 triangle or with a calculator. In degree mode, you would this be -30 degrees, or 330 degrees and 210 degrees, as the sine function is negative in quadrants III and IV.

Vitek1552 [10]3 years ago
3 0

Make use of the inverse sine function. Take the inverse sine of both sides of the equation. Of course, within the appropriate limits, the inverse sine of the sine function is the original argument, as is the case with any inverse function: f⁻¹(f(x)) = x.

... sin⁻¹(sin(x)) = sin⁻¹(-0.5)

... x = sin⁻¹(-0.5)

... x = -30°

_____

You need to be careful with inverses of trig functions, because they are only defined over a limited domain and range. The range of the inverse sine function is -90° to 90°, so, for example, sin⁻¹(sin(150°)) = sin⁻¹(0.5) = 30°.

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Which of the following is the solution to |x+8|=1?
KonstantinChe [14]

Answer:

-7 OR -9

Step-by-step explanation:

8-7 in the inside

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Read 2 more answers
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
2 years ago
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dezoksy [38]

Answer:

Given:

v = 150d

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The rate of change of the function representing the number of vehicles manufactured for the coming year is CONSTANT (150) , and its graph is a STRAIGHT LINE . So, the function is a LINEAR function.

Step-by-step explanation:

7 0
2 years ago
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