Proving Trigonometric Identities, 2

Category: Trigonometry

"Published in Newark, California, USA"

Prove that

Solution:

In proving the trigonometric identities, we have to simplify the hardest or more complicated part of the equation. In this case, we have to simplify the left side of the given equation. Let's consider the following procedure.

but

The above equation becomes

but

The above equation becomes

Therefore,

Area Bounded - Two Curves

Category: Integral Calculus, Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the area bounded by the given curves

Solution: 

To illustrate the problem, it is better to draw or sketch the graph of two given equations above by using the principles of Analytic Geometry as follows

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Since the limits or their points of intersection are not given, then we have to solve their points of intersection using the two equations, two unknowns as follows

Subtract the second equation from the first equation, we have

Substitute the value of y to either of the two given equations in order to get the value of x as follows

Equate each factor to zero and solve for the value of x. The values of x are 3 and -1.

Their points of intersection are (-1, -4) and (3, -4).

Rewrite the two given equations as y = f(x), label further the figure, and use the vertical strip as follows

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The area bounded by the two curves is given by the formula

Substitute the values of the limits and the two functions to the above equation, we have

Area - Polar Functions, Curves

Category: Integral Calculus, Algebra, Trigonometry

"Published in Newark, California, USA"

Find the area of a four leaf rose

Solution:

To illustrate the problem, it is better to draw or sketch the figure as follows

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In this problem, we will get the area of one leaf and then multiply it by 4 later because the four leaves are symmetrical or identical to each other. Since the limits for one leaf are not given, then we have to solve these first as follows

set r = 0, we have

Since we have the limits for one leaf of a four leaf rose, we can further label the figure as follows

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The area of the sector is given by the formula

Integrate on both sides of the equation to get the area of one leaf, we have

Since one leaf is also symmetrical, then we can rewrite the above equation as follows

Therefore, the area of a four leaf rose is