Application of the Cosine Rule (Law Of Cosines)

In this video, I demonstrate how to use the Law of Cosines to find an unknown long side of a scalene triangle, given an angle and adjacent 2 shorter sides.

The triangle has been constructed from 2 shorter sides of lengths 2.0m and 3.0m, subtended by an angle of 135°. We are required to find the length of the longest side - and we can do that by using the Cosine Rule (aka Law of Cosines) which states...

c^2 = a^2 + b^2 - 2ab\cos \theta

where c is the longest side and θ is the angle between the 2 shorter sides a and b.

Equation for ellipses not centred at Origin

The equation of an ellipse that is offset from (0,0) is very similar to the standard form, except that the centre coordinates (p,q) are subtracted from the variables x and y respectively. That is...

\frac{\left ( x-p \right )^2}{a^2}-\frac{\left ( y-q \right )^2}{b^2}=1

Here, a and b still are the lengths of the semi-major and semi-minor axes respectively.

For an ellipse with a given semi-major and semi-minor axis, the shape and every other feature remains the same as one located at (0,0), except all features have been offset by the centre point coordinates (p, q). That is...

The centre is located at: (p,q) rather than (0,0)
The foci are located at: (±ae+p , q) rather than (±ae , 0)
The horizontal vertices are located at: (±a+p , q) rather than (±a , 0)
The vertical vertices are located at: (±p, b+q) rather than (0 , ±b)
The directrices have the equations: x = ±(a/e)+p rather than x = ±a/e

In this video, we also sketch the ellipse formed by the equation:

9x^2 + 36x + 16y^2 + 96y = -36

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Ellipses - Example 2: Finding the Equation

In this lesson, we find the equation of an ellipse given it has vertices or x-intercepts of (8,0) and (-8,0) and focal points (5,0) and (-5,0).

The standard form of the ellipse is:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

And the x-intercepts are given by:

(\pm a,0)

which means that a = 8.

For the term b², it can be expressed as...

b^2 = a^2 - (ae)^2

where a is the length of the semi-major axis, e is the eccentricity, ae is the distance from the centre of the ellipse to a focal point. So in this case, we have...

ae = 5

Thus

b^2 = 8^2 - 5^2 = 64 - 25 = 39

So the equation of the ellipse is...

\frac{x^2}{64} + \frac{y^2}{39} = 1

or...

39x^2 + 64y^2 = 2496

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Ellipses - Example 1: Sketching 4x^2 + 9y^2 = 36

4x² + 9y² = 36 is the equation of an ellipse centred at the origin (0,0). Before we can sketch the ellipse, we need to find the vertices (i.e. the x and y intercepts) by transforming the equation to the standard form, which is:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

where...

a is the semi-major axis
b is the semi-minor axis

So, dividing the equation by 36, we get:

\frac{x^2}{9} + \frac{y^2}{4} = 1

Or...

\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1

Thus a = 3 and b = 2, and hence the vertices are:

A = (3,0)
A' = (-3,0)
B = (0,2)
B' = (0,-2)

To fully define the ellipse, we should also find the focal points and the directrices. Thus we to find the eccentricity. We can do this through the relationship:

b^2 = a^2 - (ae)^2

The focal points (foci) are given by:

F = (ae,0)
F' = (-ae,0)

And the equations are of the directrices are:

x = \frac{a}{e}

x = -\frac{a}{e}

Integral of ∫dx/sqrt(a^2 + x^2)

The integral ∫dx/√(a² + x²) may be approached in 2 ways:

1. Trigonometric substitution, or
2. Hyperbolic substitution

The goal of both methods is to simplify the denominator expression: √(a² + x²).

With trigonometric substitution, we can let x = atanθ, and thus dx = asec²θdθ, and √(a² + x²) simplifies to √[a²(1 + tan²θ)] = asecθ. And the integral becomes ∫secθdθ.

With hyperbolic substitution, we can let x = asinhu, and thus dx = acoshudu, and √(a² + x²) simplifies to √[a²(1 + sinh²u)] = acoshu, so the integral becomes ∫du.

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Conic Sections: The Ellipse - Part 2

With the standard form of the ellipse being x²/a² + y²/b² = 1 and given the eccentricity, e, we can define and determine all of the features of the ellipse.

For instance, the:

• x-intercepts are located at coordinates (a,0) and (-a,0)
• y-intercepts are located at coordinates (0,b) and (0,-b)
• semi-major axis has a length of a
• semi-minor axis has a length of b
• foci are located at (ea,0) and (-ea,0)
• directrices have equations x = a/e and x = -a/e

Suggested tutorials:

• Conic Sections: The Ellipse - Part 1

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How to Integrate ∫sin^3(x)cos^3(x)dx - odd powers

Integrals of the product of the powers of sine and cosine come in 4 permutations:

1. The powers m and n are both even
2. The powers m and n are even and odd respectively
3. The powers m and n are odd and even respectively
4. The powers m and n are both odd

In this video, we explore case 4 where both powers are odd. The example we use to demonstrate the methodology in this case is...

∫sin³(x)cos³(x)dx

Remember that cos(x) is the first derivative of sin(x), so we reserve one and write the integrand as...

sin³(x)cos³(x) = sin³(x)cos²(x)cos(x)

Then if we write the cos²(x) term as...

cos²(x) = 1 - sin²(x)

we have...

sin³(x)[1 - sin²(x)]cos(x) = [sin³(x) - sin⁵(x)]cos(x)

Then using the substitution u = sin(x), we can easily evaluate the integral.

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Integrals of powers of cos(x) - ∫cos^6(x)dx

Integrals of powers of trigonometric functions get increasingly difficult and tedious as the power increases. The idea is to reduce the power to the first order, and this can be done with power reduction formulas such as the half-angle formula. But as the power increases, the number of iterations of the half-angle formula also increases accordingly.

But there is a neat alternative for power reduction in complex numbers. Consider De Moivre's Theorem. It is probably the most efficient method of power reduction. For instance, in polar form...

z = cos(x) + isin(x)

If we raise z to the n^th power, then by De Moivre's Theorem...

zⁿ = [cos(x) + isin(x)]ⁿ = cos(nx) + isin(nx)

So we've reduced the power from n to 1 in just one step! Similarly...

1/z = z^-1 = [cos(x) + isin(x)]^-1 = cos(-x) + isin(-x) = cos(x) - isin(x)

1/zⁿ = z^-n = cos(-nx) + isin(-nx) = cos(nx) - isin(nx)

Now adding:

z + 1/z = 2cos(x)

zⁿ + 1/zⁿ = 2cos(nx)

These results are the key to performing integrals of larger powers of cosine.

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