Proving Trigonometric Identities, 3

Category: Trigonometry, Algebra

"Published in Suisun City, California, USA"

The lower right-hand corner of a long piece of paper 6 in. wide is folded over to the left-hand edge as shown. The length L of the fold depends on the angle θ. Show that 

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Solution:

Consider the given figure above and label further using the principles of Plane Geometry, we have

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By Pythagorean Theorem, 

We need to use another equations in order to eliminate x and y. By using trigonometric functions of right triangles, we have

Using a small right triangle

Using a large right triangle

Consider again the equation,

Substitute the values of x and y, we have

Take the square root on both sides of the equation, we have

Multiply both the numerator and the denominator by 1 - sin² θ, we have

Therefore,

Logarithmic Expression

Category: Algebra

"Published in Suisun City, California, USA"

Rewrite the expression for

Solution:

Consider the given equation above

Expand the logarithmic expression for the main fraction, as follows

Expand the logarithmic expression for the second term, as follows

Rewrite the logarithmic expression for the square root and cube root functions and therefore,

Graphical Sketch - Rational Function

Category: Analytic Geometry, Algebra

"Published in Suisun City, California, USA"

Sketch the graph for

Solution:

Consider the given equation above

Factor the numerator and denominator, if possible, we have

To get the x-intercept, we have to set the numerator to zero, as follows

The coordinates of the x-intercept is (2, 0).

To get the y-intercept, substitute x = 0 to the given equation as follows

The coordinates of the y-intercept is (0, 2).

To get the vertical asymptotes, we have to set the factors of the denominator to zero, as follows

If

If 

The vertical asymptotes are x = 1 and x = -1.

To get the horizontal asymptote, consider the given equation above, as follows

Divide both the numerator and the denominator by the variable with the highest degree, we have

Take the limit of the given equation above as x approaches to infinity, we have

The horizontal asymptote is y = 0.

Next, we need to draw the dotted lines for vertical asymptotes and horizontal asymptote, we have

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As you noticed that the horizontal asymptote contains the x-intercept. Usually, horizontal asymptotes never passes the curve or point but there are times that the horizontal asymptote will pass the curve.

The vertical asymptotes never passes the curve or point.

To start in sketching the curve, let's consider the given equation again

If x < -1, then

Draw the curve at the lower left side between the vertical and horizontal asymptotes.

If -1 < x < 0, then

From y-intercept, draw the curve upward approaching to vertical asymptote. 

If 0 < x < 1, then

From y-intercept, draw the curve upward approaching to vertical asymptote.

If 1 < x < 2, then

From x-intercept, draw the curve downward approaching to vertical asymptote.

If x > 2, then

From x-intercept, draw the curve going to the right approaching to horizontal asymptote.

The final sketch of the graph for the given equation should be like this.

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