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To delve into the concept of triple angle formulae, we need to first refresh our understanding of what the double angle formulae help us achieve along with the area of a right triangle.
Double angle formulae are an excellent choice for calculating the value of trigonometric functions in specific situations. However, double angle formulae are limited to instances where we need up to two times the value of the desired angle. For starters, let’s consider the following example that asks us to derive a double angle formula to solve an equation.
y=4cos²(x)-2 x∈[0,2π] |
Here, we already know the interval oscillating between 0 and 2π.
Let’s try to analyse it using a double angle formula. In this case, we can convert the equation to a double angle formula as follows.
y=y=4cos²(x)-2 =2(2cos²(x)-1) =2 cos 2x. |
We read this interval because the amplitude of this function is 2 with a period of π.
Computing triple angle formulae
We now know how to convert a function into a double angle formula to find the value of an angle twice the number assumed.
Now, what if we require an argument that employs three times the value of an angle in a trigonometric function? We will try to decode this enigma by using a simple method: By combining the sum formula and the double angle formula for any given angle. And what we will have is a triple angle formula. However, the approach here would be to consider an equation that uses a linear combination of sine and cosine and then converts it into a simple cosine function.
Let’s try to expand this idea with the help of an example.
Equation: Acos(x)+Bsin(x)=Ccos(x-D) Where C=√A²+B² ), Cos D =A/C And sin D= B/C. |
Example:
Let’s take a simple example here, so that we understand the concept of a triple angle formula. A generic template for this should look something like this:
sin3θ=sin(2θ+θ) =sin2θ.cosθ+cos2θsinθ =(2sinθcosθ)cosθ+(1-2sin² θ)sinθ =2sinθcos2 θ+sinθ -2sin3 θ =2sinθ(1-sin² θ)+sinθ-2sin³ θ =2sinθ-2sin³ θ+sinθ-2sin³ θ=3sinθ-4sin³ θ |
Using this template, let’s compute the triple angle formula for sin3x.
We do this by breaking this down into the sum formula and the double angle formula for sin3x.
sin3x=sin(2x+x) =sin(2x)cosx+co s(2x) sinx =(2sinxcosx)cosx+(cos² x-sin² x)sinx =2 sinx cos2 x+cos2 xsinx-sin3 x =3sinxcos2x-sin3 x =3 sinx cos²x=sin³ x =3 sinx(1-sin² x)-sin³ x =3 sinx-4sin³ x |
We choose to go with the sine function here to explain how a triple angle formula is derived. Cos functions are derived the same way. The template holds true by replacing sin with cos.
COSINE FUNCTION for deriving a triple angle formula
The cosine function for a triple angle formula follows a similar pattern. The base concept remains the same. Combine the sum formula and the double angle formula, similar to what we did for the sine function.
cos(3θ)=cos³ θ-3sin² θcosθ =4cos³ θ-3cosθ |
Let’s check if this statement is true with the help of a simple cosine triple angle problem. Let’s try solving the cosine function for cos3x.
Cos3x=cos(2x+x) =cos2xcosx-sin2xsinx =(2cos² x-1)cosx-(2sinxcosx)sinx =2cos³ x-cosx-2sin² cosx =2cos³ x-cosx-2cosx(1-cos² x) =2cos³ x-cosx-2cosx+2cos³ x =4cos³ x-3cosx |
Any value substituted in this derived formula will give you the triple angle identity of any equation that uses a cosine function.
Tangent function for a triple angle formula
The tangent formula allows the same methodology of breaking the formula into a total and a double angle component following the exact solution for tanθ.
For a triple angle, the base equation always carries 3.
Hence,
tan(3θ)=(3tanθ-tan³ θ)/(1-3tan² θ) |
Let’s try solving a small example using this as our result. Let’s try solving for tan3x.
Tan3x=tan(2x+x) =(tan2x+tanx)/(1-tan2xtanx) = ( 2tanx/(1-tan2 x)+tanx)⁄(1-(2tanx-tanx)/(1-tan² x)) =((2tanx+tanx-tan³x)/(1-tan²x-2tan² x)+tanx)⁄ =(3tanx-tan³ x)/(1-3tan² x) |
Basic methodology involved
Triple angle identities are useful in situations where trigonometric functions need to be simplified due to expressions that do not necessarily translate to an actual value.
We have successfully identified the HOW to do this aspect of solving for a triple angle identity. One of the typical techniques generally used – which we did here – is the substitution rule with a trigonometric function and then simplifying the result with a simple breakdown.
Because triple angle formulas for both sine and cosine functions involve powers of a single function to assemble from the breakdowns of two subparts, the end result is almost always predictable once you know what function the angle uses in its equation.
Trigonometric triple angle functions establish a relationship between basic trigonometric functions factored in three times a given angle in the angle’s trigonometric context. What this means, in simplified terms, is that we use basic trigonometric functions to derive a value for an angle that is three times above the well-known value. Only here, we do this in the angle’s trigonometric context and not its real value. Here, we can use the right triangle formula to compute values.
This method also gives us a firm hold over the concept of functions in angles. The general idea here is to identify which function plays what role in the angle provided, simplifying the equation and then deriving a working triple angle formula to supplement our understanding of the concept’s core.
So we can see that there are multiple ways to solve the triple angle using the various formulas. Initially, it might be a bit difficult to figure out which formula or method to use, but one will be able to master this concept with time and practice.
Conclusion
The reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine, can be derived using the double-angle formulas. They enable us to rewrite even sine or cosine powers in terms of the first power of cosine. Higher-level maths courses, particularly calculus, rely heavily on these formulas. Three identities are included in the power-reducing formulas, which are easily derived from the double angle formulas.
