sin
(
2
θ
)
=
2
(
3
5
)
(
−
4
5
)
=
−
24
25
Write the double-angle formula for cosine.
\displaystyle \cos \left(2\theta \right)={\cos }^{2}\theta -{\sin }^{2}\thetacos(2θ)=cos
2
θ−sin
2
θ
Again, substitute the values of the sine and cosine into the equation, and simplify.
)
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where \displaystyle \alpha =\betaα=β. Deriving the double-angle formula for sine begins with the sum formula,
\displaystyle \sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \betasin(α+β)=sinαcosβ+cosαsinβ
If we let \displaystyle \alpha =\beta =\thetaα=β=θ, then we have
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, \displaystyle \cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \betacos(α+β)=cosαcosβ−sinαsinβ, and letting \displaystyle \alpha =\beta =\thetaα=β=θ, we have
Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:
Similarly, to derive the double-angle formula for tangent, replacing \displaystyle \alpha =\beta =\thetaα=β=θ in the sum formula gives
\displaystyle \begin{array}{c}\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \tan \left(\theta +\theta \right)=\frac{\tan \theta +\tan \theta }{1-\tan \theta \tan \theta }\\ \tan \left(2\theta \right)=\frac{2\tan \theta }{1-{\tan }^{2}\theta }\end{array}
tan(α+β)=
1−tanαtanβ
tanα+tanβ
tan(θ+θ)=
1−tanθtanθ
tanθ+tanθ
tan(2θ)=
1−tan
2
θ
2tanθ
A GENERAL NOTE: DOUBLE-ANGLE FORMULAS
The double-angle formulas are summarized as follows:
\displaystyle \sin \left(2\theta \right)=2\sin \theta \cos \thetasin(2θ)=2sinθcosθ
\displaystyle \tan \left(2\theta \right)=\frac{2\tan \theta }{1-{\tan }^{2}\theta }tan(2θ)=
1−tan
2
θ
2tanθ
HOW TO: GIVEN THE TANGENT OF AN ANGLE AND THE QUADRANT IN WHICH IT IS LOCATED, USE THE DOUBLE-ANGLE FORMULAS TO FIND THE EXACT VALUE.
Draw a triangle to reflect the given information.
Determine the correct double-angle formula.
Substitute values into the formula based on the triangle.
Simplify.
EXAMPLE 1: USING A DOUBLE-ANGLE FORMULA TO FIND THE EXACT VALUE INVOLVING TANGENT
Given that
and \displaystyle \thetaθ is in quadrant II, find the following:
\displaystyle \sin \left(2\theta \right)sin(2θ)
\displaystyle \cos \left(2\theta \right)cos(2θ)
\displaystyle \tan \left(2\theta \right)tan(2θ)
SOLUTION
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given \displaystyle \tan \theta =-\frac{3}{4}tanθ=−
4
3
, such that \displaystyle \thetaθ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because \displaystyle \thetaθ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
Now we can draw a triangle similar to the one shown in Figure 2.
Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.
Figure 2
Let’s begin by writing the double-angle formula for sine.
\displaystyle \sin \left(2\theta \right)=2\sin \theta \cos \thetasin(2θ)=2sinθcosθ
We see that we to need to find \displaystyle \sin \thetasinθ and \displaystyle \cos \thetacosθ. Based on Figure 2, we see that the hypotenuse equals 5, so \displaystyle \sin \theta =\frac{3}{5}sinθ=
. Substitute these values into the equation, and simplify.
Thus,
Solution
EXAMPLE 2: USING THE DOUBLE-ANGLE FORMULA FOR COSINE WITHOUT EXACT VALUES
Use the double-angle formula for cosine to write \displaystyle \cos \left(6x\right)cos(6x) in terms of \displaystyle \cos \left(3x\right)cos(3x).
SOLUTION
c
Analysis of the Solution
This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.
Using Double-Angle Formulas to Verify Identities
Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.
EXAMPLE 3: USING THE DOUBLE-ANGLE FORMULAS TO ESTABLISH AN IDENTITY
Establish the following identity using double-angle formulas:
\displaystyle 1+\sin \left(2\theta \right)={\left(\sin \theta +\cos \theta \right)}^{2}1+sin(2θ)=(sinθ+cosθ)
2
θ
)
Step-by-step explanation:
