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Precalculus for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

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Section 8.1 Trigonometric Identities (TE1)

Subsection 8.1.1 Activities

Remark 8.1.1.

Pythagorean Identities, as we have seen in a previous section, are derived from the Pythagorean Theorem \(\left( a^2+b^2=c^2 \right) \text{.}\) Pythagorean Identities are used in solving many trigonometric problems where one trigonometric ratio is given and we are expected to find the other trigonometric ratios. The most commonly used Pythagorean Identity is:
\begin{equation*} \sin^2\theta + \cos^2\theta = 1\text{.} \end{equation*}
The next two activities will lead us to find the other two Pythagorean Identities.

Activity 8.1.2.

Consider the triangle \(ABC\text{,}\) where \(C\) is the right angle and angle \(B\) is \(\theta\text{.}\)
(a)
Draw a diagram to represent the triangle. Be sure to label the sides with \(a\text{,}\) \(b\text{,}\) and \(c\) and angle \(B\) as \(\theta\text{.}\)
Answer.
Students should draw a right triangle so that \(C\) is the right angle, \(c\) is the hypotenuse, and \(B\) is \(\theta\text{.}\)
(b)
Let’s begin with the Pythagorean Theorem, \(a^2+b^2=c^2\text{.}\) If we divide each term by \(a^2\text{,}\) what would the resulting equation be?
Answer.
\begin{equation*} \frac{a^2}{a^2}+\frac{b^2}{a^2}=\frac{c^2}{a^2}\text{,} \end{equation*}
which simplifies to
\begin{equation*} 1+\frac{b^2}{a^2}=\frac{c^2}{a^2} \end{equation*}
(c)
How could we rewrite the equation we got in part (b) so that each term with a fraction can be represented as a fraction squared?
Answer.
\(1+ \left( \frac{b}{a} \right) ^2 = \left( \frac{c}{a} \right) ^2\)
(d)
Let’s now try and think of the relationships of the angle and sides (i.e., our trigonometric ratios). Label the sides of the triangle in terms of its relationship with \(\theta\) (i.e., angle \(B\)).
Answer.
Students should label the sides so that the following sides should be labeled as:
  • \(a\) is the adjacent side of \(\theta\)
  • \(b\) is the opposite side of \(\theta\)
  • \(c\) is the hypotenuse
(e)
Replace \(a\text{,}\) \(b\text{,}\) and \(c\) in the equation we got in part (c) with how you labeled the sides in part (d).
Answer.
Replacing the \(a\text{,}\) \(b\text{,}\) and \(c\) in the equation
\begin{equation*} 1+ \left( \frac{b}{a} \right)^2= \left( \frac{c}{a} \right)^2 \end{equation*}
with the labels from part (d) gives
\begin{equation*} 1+ \left( \frac{\text{opposite}}{\text{adjacent}} \right)^2=\left( \frac{\text{hypotenuse}}{\text{adjacent}} \right)^2 \end{equation*}
(f)
Recall the following trigonometric functions:
  • \(\displaystyle \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)
  • \(\displaystyle \csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}\)
  • \(\displaystyle \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)
  • \(\displaystyle \sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}\)
  • \(\displaystyle \tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)
  • \(\displaystyle \cot\theta=\frac{\text{adjacent}}{\text{opposite}}\)
Using these trigonometric ratios, how can we rewrite the equation in part (e)?
Answer.
\(1 + \tan^2\theta = \sec^2\theta\)

Remark 8.1.3.

In Activity 8.1.2, we derived the second Pythagorean Identity: \(\sec^2\theta-\tan^2\theta=1\text{.}\) Note that we can rewrite what we got in the previous activity \(\left( 1 + \tan^2\theta = \sec^2\theta \right) \) to get to this form.

Activity 8.1.4.

Consider the same triangle \(ABC\text{,}\) where \(C\) is the right angle and angle \(B\) is \(\theta\) like in Activity 8.1.2.
(a)
Beginning with the Pythagorean Theorem again \(\left( a^2+b^2=c^2 \right) \text{,}\) what would the resulting equation be if we divide each term by \(b^2\text{?}\)
Answer.
\begin{equation*} \frac{a^2}{b^2}+\frac{b^2}{b^2}=\frac{c^2}{b^2}\text{,} \end{equation*}
which simplifies to
\begin{equation*} \frac{a^2}{b^2}+1=\frac{c^2}{b^2} \end{equation*}
(b)
How could we rewrite the equation we got in part (a) so that each term with a fraction can be represented as a fraction squared?
Answer.
\(\left( \frac{a}{b} \right) ^2 + 1 =\left( \frac{c}{b} \right) ^2\)
(c)
Let’s now try and think of the relationships of the angle and sides (i.e., our trigonometric ratios). Replace \(a\text{,}\) \(b\text{,}\) and \(c\) in the equation we got in part (b) with the relationship each side has with \(\theta\) (i.e., opposite, adjacent, hypotenuse).
Answer.
Replace the \(a\text{,}\) \(b\text{,}\) and \(c\) in the equation
\begin{equation*} \left( \frac{a}{b} \right) ^2 + 1 = \left( \frac{c}{b} \right) ^2 \end{equation*}
will yield
\begin{equation*} \left( \frac{\text{adjacent}}{\text{opposite}} \right) ^2 + 1 = \left( \frac{\text{hypotenuse}}{\text{opposite}} \right) ^2 \end{equation*}
(d)
Recall the following trigonometric functions:
  • \(\displaystyle \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)
  • \(\displaystyle \csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}\)
  • \(\displaystyle \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)
  • \(\displaystyle \sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}\)
  • \(\displaystyle \tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)
  • \(\displaystyle \cot\theta=\frac{\text{adjacent}}{\text{opposite}}\)
Using these trigonometric ratios, how can we rewrite the equation in part (c)?
Answer.
\(\cot^2\theta + 1 = \csc^2\theta\)

Remark 8.1.5.

In Activity 8.1.4, we derived the third Pythagorean Identity: \(\csc^2\theta-\cot^2\theta=1\text{.}\) Note that we can rewrite what we got in the previous activity \(\left( \cot^2\theta + 1 = \csc^2\theta \right)\) to get to this form.

Definition 8.1.6.

The three Pythagorean Identities are:
  • \(\displaystyle \sin^2\theta + \cos^2\theta = 1\)
  • \(\displaystyle \sec^2\theta - \tan^2\theta = 1\)
  • \(\displaystyle \csc^2\theta - \cot^2\theta = 1\)
Note that each of these identities can be written in different forms, which are all equivalent to one another. For example, \(\sin^2\theta + \cos^2\theta = 1\) can be rewritten as \(1 - \sin^2\theta = \cos^2\theta\) or \(1 - \cos^2\theta = \sin^2\theta\text{.}\) You will want to become familiar with all these forms.

Remark 8.1.7.

Other identities and formulas (in addition to the Pythagorean Identities) can be used to solve various mathematical problems. The next few activities will lead us through an exploration of other types of identities and formulas.

Activity 8.1.8.

Refer back to the unit circle to determine the exact value of \(\sin\left( \pi + \frac{\pi}{2}\right)\text{.}\)
(a)
What is the value of \(\sin \pi\text{?}\)
  1. \(\displaystyle -1\)
  2. \(\displaystyle 0\)
  3. \(\displaystyle 1\)
Answer.
B
(b)
What is the value of \(\sin \frac{\pi}{2}\text{?}\)
  1. \(\displaystyle -1\)
  2. \(\displaystyle 0\)
  3. \(\displaystyle 1\)
Answer.
C
(c)
Based on your answers from parts (a) and (b), what do you think the value of \(\sin\left( \pi + \frac{\pi}{2}\right)\) is?
Answer.
Students will probably make the conjecture that the value of \(\sin\left( \pi + \frac{\pi}{2}\right)\) is just \(\sin \pi + \sin \left( \frac{\pi}{2} \right)\text{,}\) which equals \(1\text{.}\)
(d)
Let’s test your conjecture from part (c). What is the value of \(\pi + \frac{\pi}{2}\text{?}\)
  1. \(\displaystyle \frac{2\pi}{2}\)
  2. \(\displaystyle \pi\)
  3. \(\displaystyle \frac{3\pi}{2}\)
  4. \(\displaystyle \frac{2\pi}{3}\)
Answer.
C
(e)
What is the value of \(\sin \left( \frac{3\pi}{2} \right)\text{?}\)
  1. \(\displaystyle -1\)
  2. \(\displaystyle 0\)
  3. \(\displaystyle 1\)
Answer.
A
(f)
Is it safe to assume that \(\sin\left( \pi + \frac{\pi}{2}\right) = \sin \pi + \sin \left( \frac{\pi}{2} \right)\text{?}\)
Answer.
From the previous parts, students should see that
\begin{equation*} \sin \pi + \sin \frac{\pi}{2} = 1 \end{equation*}
and
\begin{equation*} \sin\left( \pi + \frac{\pi}{2}\right) = -1\text{.} \end{equation*}
So from this example, we can see that it is NOT safe to assume that \(\sin\left( \pi + \frac{\pi}{2}\right) = \sin \pi + \sin \left( \frac{\pi}{2} \right)\text{.}\)

Remark 8.1.9.

Notice that in Activity 8.1.8, we saw that \(\sin\left( \pi + \frac{\pi}{2}\right) \neq \sin \pi + \sin \frac{\pi}{2}\text{.}\) This is also true for both cosine and tangent as well - that is, if we want to find the cosine or tangent of the sum of two angles, we cannot assume that it is equal to the sum of the two trigonometric functions of each angle. For example, \(\cos\left( \pi + \frac{\pi}{2}\right) \neq \cos \pi + \cos \frac{\pi}{2}\) and \(\tan\left( \pi + \frac{\pi}{2}\right) \neq \tan \pi + \tan \frac{\pi}{2}\text{.}\) The same is true for finding the difference of two angles.

Activity 8.1.10.

Recall that the coordinates of points on the unit circle are given by \((\cos\theta, \sin\theta)\text{.}\) In the unit circle shown below, point \(P\) makes an angle \(\alpha\) with the positive \(x\)-axis and has coordinates \((\cos\alpha, \sin\alpha)\) and point \(Q\) makes an angle \(\beta\) with the positive \(x\)-axis and has coordinates \((\cos\beta, \sin\beta)\text{.}\) Point \(B\) has coordinates \((1,0)\text{.}\)
Figure 8.1.11. Image of the unit circle used to derive the difference formula.
(a)
Note that angle \(AOB\) is equal to \(\left(\alpha-\beta\right)\text{.}\) What are the coordinates of point \(A\text{?}\)
Answer.
Point \(A\) has coordinates \((\cos (\alpha - \beta), \sin (\alpha - \beta))\text{.}\)
(b)
Triangles \(POQ\) and \(AOB\) are rotations of one another. What can we say about the lengths of \(PQ\) and \(AB\text{?}\)
Answer.
Because triangles \(POQ\) and \(AOB\) are rotations of one another, we know that the lengths of \(PQ\) and \(AB\) are the same. That is, the distance from \(P\) to \(Q\) is equal to the distance from \(A\) to \(B\text{.}\)
(c)
Let’s use the distance formula, \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\text{,}\) to find the length of \(PQ\text{.}\) What do you get when you plug in the coordinates of point \(P\) and point \(Q\text{?}\)
  1. \(\displaystyle d=\sqrt{(\cos\beta-\cos\alpha)^2+(\sin\beta-\sin\alpha)^2}\)
  2. \(\displaystyle d=\sqrt{(\cos\alpha-\sin\beta)^2+(\sin\alpha-\cos\beta)^2}\)
  3. \(\displaystyle d=\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}\)
  4. \(\displaystyle d=\sqrt{(\sin\alpha-\cos\beta)^2+(\cos\alpha-\sin\beta)^2}\)
Hint.
Remember that point \(P\) has coordinates \((\cos\alpha, \sin\alpha)\) and point \(Q\) has coordinates \((\cos\beta, \sin\beta)\text{.}\)
Answer.
A or C, depending on which point students started with first
(d)
Begin simplifying your answer from part (c) by applying the algebraic identity \(\left(a-b\right)^2=a^2-2ab+b^2\text{.}\) What do you get when squaring the two binomials under the radical?
  1. \(\displaystyle d=\sqrt{\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta+\cos^2\alpha-2\cos\alpha\cos\beta+cos^2\beta}\)
  2. \(\displaystyle d=\sqrt{\cos^2\alpha+2\cos\alpha\cos\beta+cos^2\beta+\sin^2\alpha+2\sin\alpha\sin\beta+\sin^2\beta}\)
  3. \(\displaystyle d=\sqrt{\cos^2\alpha-\cos\alpha\cos\beta+cos^2\beta+\sin^2\alpha-\sin\alpha\sin\beta+\sin^2\beta}\)
  4. \(\displaystyle d=\sqrt{\cos^2\alpha-2\cos\alpha\cos\beta+cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta}\)
Answer.
A or D, depending on which point students started with first
(e)
Simplify your answer from part (d) even further by applying the Pythagorean Identity, \(\cos^2\theta+\sin^2\theta=1\text{.}\) What does your answer from part (d) simplify to?
  1. \(\displaystyle d=\sqrt{1-\sin\alpha\sin\beta-\cos\alpha\cos\beta}\)
  2. \(\displaystyle d=\sqrt{2-2\sin\alpha\sin\beta-2\cos\alpha\cos\beta}\)
  3. \(\displaystyle d=\sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}\)
  4. \(\displaystyle d=\sqrt{2-\sin\alpha\sin\beta-\cos\alpha\cos\beta}\)
Hint.
You may need to rearrange some of your terms before applying the Pythagorean Identity.
Answer.
B or C
(f)
Using the same steps as in parts (c-e), use the distance formula to find the distance between points \(A\) and \(B\text{.}\)
  1. \(\displaystyle d=\sqrt{1-2\cos\left(\alpha-\beta\right)}\)
  2. \(\displaystyle d=\sqrt{2-2\cos\left(\alpha-\beta\right)}\)
  3. \(\displaystyle d=\sqrt{1-\cos\left(\alpha-\beta\right)}\)
  4. \(\displaystyle d=\sqrt{-2\cos\left(\alpha-\beta\right)+2}\)
Hint.
Point \(A\) has coordinates \((\cos (\alpha - \beta), \sin (\alpha - \beta))\) and point \(B\) has coordinates \((1,0)\text{.}\)
Answer.
B or D
(g)
From part (b), we know that the length of \(PQ\) is equal to the length of \(AB\text{.}\) Therefore,
\begin{equation*} \sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}=\sqrt{2-2\cos\left(\alpha-\beta\right)}\text{.} \end{equation*}
Square both sides of this equation and solve for \(\cos\left(\alpha-\beta\right)\text{.}\)
Answer.
\(\cos\left(\alpha-\beta\right)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\)

Remark 8.1.12.

In Activity 8.1.10, we derived the difference formula for cosine. The development of the sum/difference formulas are similar for sine and cosine and same for \(\tan(\alpha-\beta)\) compared to \(\tan(\alpha+\beta)\text{.}\)

Definition 8.1.13.

The Sum and Difference Formulas in trigonometry are used to find the value of the trigonometric functions at specific angles where it is easier to express the angle as the sum or difference of unique angles such as \(0^\circ \text{,}\) \(30^\circ\text{,}\) \(45^\circ\text{,}\) \(60^\circ\text{,}\) \(90^\circ\text{,}\) and \(180^\circ\text{.}\)
The six main trigonometric sum and difference formulas are:
  • \(\displaystyle \sin \left(\alpha+\beta\right) = \sin\alpha\cos\beta+\cos\alpha\sin\beta\)
  • \(\displaystyle \sin \left(\alpha-\beta\right) = \sin\alpha\cos\beta-\cos\alpha\sin\beta\)
  • \(\displaystyle \cos \left(\alpha+\beta\right) = \cos\alpha\cos\beta-\sin\alpha\sin\beta\)
  • \(\displaystyle \cos \left(\alpha-\beta\right) = \cos\alpha\cos\beta+\sin\alpha\sin\beta\)
  • \(\displaystyle \tan \left(\alpha+\beta\right) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)
  • \(\displaystyle \tan \left(\alpha-\beta\right) = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\)

Activity 8.1.14.

Use the unit circle and Definition 8.1.13 to find the exact value of \(\cos75\)°.
(a)
Split \(75\)° into the sum of two angles which can be found on the unit circle (use unique angles such as \(0\)°, \(30\)°, \(45\)°, \(60\)°, \(90\)°, and \(180\)°, etc.).
Answer.
\(30\)° and \(45\)°
(b)
Rewrite \(\cos75\)° as \(\cos\left(\alpha+\beta\right)\) where \(\alpha\) and \(\beta\) are the two angles you found in part (a).
Answer.
\(\cos75= \cos\left(30 + 40\right)\)
(c)
Apply Definition 8.1.13 to rewrite \(\cos75\)° with four trigonometric functions.
Answer.
\(\cos75=\cos30\cos45-\sin30\sin45\)
(d)
Use the unit circle to find the value of each trigonometric function in the formula you wrote in part (c).
Answer.
\(\cos75\)°\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right) - \left(\frac{1}{2}\right) \left(\frac{\sqrt2}{2}\right)\)
(e)
Simplify your answer from part (d).
Answer.
\(\cos75\)°\(=\frac{\sqrt6-\sqrt2}{4}\)

Activity 8.1.15.

Determine whether each statement is true or false by referring to Definition 8.1.13.
(a)
\(\sin(45°- 30°) = \sin45°- \sin30°\)
Answer.
False
(b)
\(\cos15° = \cos60°\cos45° + \sin60°\sin45°\)
Answer.
True
(c)
\(\tan 225° = \frac{\tan180° - \tan45°}{1+\tan180°\tan45°}\)
Answer.
False

Activity 8.1.16.

Apply Definition 8.1.13 to find the exact value for each of the following trigonometric functions.
(a)
\(\sin165\)°
Answer.
\(\frac{\sqrt6-\sqrt2}{4}\)
(b)
\(\tan195\)°
Answer.
\(2-\sqrt3\)
(c)
\(\cos\left(\frac{7\pi}{12}\right)\)
Answer.
\(\frac{\sqrt2-\sqrt6}{4}\)
(d)
\(\tan\left(\frac{11\pi}{12}\right)\)
Answer.
\(\sqrt3-2\)

Activity 8.1.17.

Apply Definition 8.1.13 to find the exact value for each of the following trigonometric functions.
(a)
\(\cos87\)°\(\cos33\)°\(-\sin87\)°\(\sin33\)°
Answer.
\(-\frac{1}{2}\)
(b)
\(\sin20\)°\(\cos80\)°\(-\cos20\)°\(\sin80\)°
Answer.
\(-\frac{\sqrt3}{2}\)
(c)
\(\sin\frac{\pi}{12}\cos\frac{5\pi}{12}+\sin\frac{5\pi}{12}\cos\frac{\pi}{12}\)
Answer.
\(1\)
(d)
\(\frac{\tan40°-\tan10°}{1+\tan40°\tan10°}\)
Answer.
\(\frac{\sqrt3}{3}\)

Activity 8.1.18.

Recall from Definition 8.1.13 the sum formulas for sine, cosine, and tangent.
(a)
Suppose \(\alpha=\beta\text{.}\) Rewrite the left-hand and right-hand sides of the sum formula for sine \(\left(\sin\left(\alpha+\beta\right)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\right)\) in terms of \(\alpha\text{.}\)
Answer.
\(\sin\left(\alpha+\alpha\right)=\sin\alpha\cos\alpha+\cos\alpha\sin\alpha\)
\(\sin(2\alpha)=2\sin\alpha\cos\alpha\)
(b)
Suppose \(\alpha=\beta\text{.}\) Rewrite the left-hand and right-hand sides of the sum formula for cosine \(\left(\cos\left(\alpha+\beta\right) =\cos\alpha\cos\beta-\sin\alpha\sin\beta\right)\) in terms of \(\alpha\text{.}\)
Answer.
\(\left(\cos\alpha+\alpha\right)=\cos\alpha\cos\alpha-\sin\alpha\sin\alpha\)
\(\cos(2\alpha)=\cos^2\alpha-sin^2\alpha\)
(c)
Suppose \(\alpha=\beta\text{.}\) Rewrite the left-hand and right-hand sides of the sum formula for tangent \(\left(\tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\right)\) in terms of \(\alpha\text{.}\)
Answer.
\(\left(\tan\alpha+\alpha\right)=\frac{\tan\alpha+\tan\alpha}{1-\tan\alpha\tan\alpha}\)
\(\tan(2\alpha)=\frac{2\tan\alpha}{1-\tan^2\alpha}\)

Definition 8.1.19.

Double angle formulas are used to express the trigonometric ratios of double angles \((2\theta)\) in terms of trigonometric ratios of single angle \((\theta)\text{.}\)
The Double Angle Formulas of sine, cosine, and tangent are:
  • \(\displaystyle \sin(2\theta)=2\sin\theta\cos\theta\)
  • \(\displaystyle \cos(2\theta)=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1\)
  • \(\displaystyle \tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}\)

Activity 8.1.20.

Suppose \(\sin\alpha=\frac{2}{3}\) and \(\alpha\) lies in Quadrant I.
(a)
Use the Pythagorean Identities and Definition 8.1.19 to find \(\sin2\alpha\text{.}\)
Answer.
Use \(\sin^2\theta+\cos^2\theta=1\) to find the value of \(\cos\alpha\text{,}\) which is equal to \(\frac{\sqrt5}{3}\text{.}\) Then, use the double angle formula to find that \(\sin2\alpha=\frac{4\sqrt5}{9}\text{.}\)
(b)
Use the Pythagorean Identities and Definition 8.1.19 to find \(\cos2\alpha\text{.}\)
Answer.
From part (a), we know that \(\cos\alpha = \frac{\sqrt5}{3}\) and \(\sin\alpha = \frac{2}{3}\text{.}\) So, by using the double angle formula \(\cos^2\alpha - \sin^2\alpha\text{,}\) we know that \(\cos2\alpha = \frac{5}{9} - \frac{4}{9}\text{,}\) which equals \(\frac{1}{9}\text{.}\)
(c)
Use the Pythagorean Identities and Definition 8.1.19 to find \(\tan2\alpha\text{.}\)
Answer.
From part (a), we know that \(\cos\alpha = \frac{\sqrt5}{3}\) and \(\sin\alpha = \frac{2}{3}\text{.}\) So, \(\tan\alpha = \frac{\frac{2}{3}}{\frac{\sqrt5}{3}}\text{,}\) which is equal to \(\frac{2}{\sqrt5}\text{.}\) By using the double angle formula \(\frac{2\tan\alpha}{1-\tan^2\alpha}\text{,}\) we know that \(\tan2\alpha = \frac{\frac{4}{\sqrt5}}{1-\frac{4}{5}}\text{,}\) which equals \(-4\sqrt5\text{.}\)

Activity 8.1.21.

The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine.
(a)
Let’s start with the double angle formula for cosine to find our first power-reduction formula: \(\cos2\theta=1-2\sin^2\theta\text{.}\) Use your algebra skills to solve for \(\sin^2\theta\text{.}\)
Answer.
\(\sin^2\theta=\frac{1-\cos2\theta}{2}\)
(b)
Given the second double angle formula,\(\cos2\theta=2\cos^2\theta-1\) \(\text{,}\) solve for \(\cos^2\theta\text{.}\)
Answer.
\(\cos^2\theta=\frac{1+\cos2\theta}{2}\)
(c)
To generate the power reduction formula for tangent, let’s begin with its definition: \(\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}\text{.}\) Use the formulas you created in parts (a) and (b) to rewrite \(\tan^2\theta\text{.}\)
Answer.
\(\tan^2\theta=\frac{1-\cos2\theta}{1+cos2\theta}\)

Definition 8.1.22.

The trigonometric power reduction formulas allow us to rewrite expressions involving trigonometric terms with trigonometric terms of smaller powers.
The Power Reduction Formulas are:
  • \(\displaystyle \sin^2\theta=\frac{1-\cos2\theta}{2}\)
  • \(\displaystyle \cos^2\theta=\frac{1+\cos2\theta}{2}\)
  • \(\displaystyle \tan^2\theta=\frac{1-\cos2\theta}{1+\cos\theta}\)

Activity 8.1.23.

Rewrite \(\sin^4\beta\) as an expression without powers greater than one using.
(a)
How can we rewrite \(\sin^4\beta\) as a function squared?
Answer.
\(\sin^4\beta=\left(\sin^2\beta\right)^2\)
(b)
Rewrite \(\sin^2\beta\) using Definition 8.1.22.
Answer.
\(\sin^4\beta=\left(\frac{1-\cos2\beta}{2}\right)^2\)
(c)
Square the right-hand side of the equation you found in part (b).
Answer.
\(\sin^4\beta=\frac{1-2\cos2\beta+\cos^22\beta}{4}\)
(d)
Notice that you still have a cosine function being squared in your equation in part (c). Substitute the \(\cos^2\beta\) using the power reduction formula in Definition 8.1.22 and then simplify
Answer.
\(\frac{2-4\cos\beta+1+\cos4\beta}{8}\) or \(\frac{3-4\cos2\beta+\cos4\beta}{8}\)

Activity 8.1.24.

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas. Let’s derive the half-angle formula for \(\sin\frac{\theta}{2}\text{.}\)
(a)
Recall the power reduction formula: \(\sin^2\theta=\frac{1-\cos2\theta}{2}\text{.}\) Replace \(\theta\) with \(\frac{\theta}{2}\text{.}\)
Answer.
\(\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos2\left(\frac{\theta}{2}\right)}{2}\)
(b)
Simplify the right-hand side of your equation in part (a).
Answer.
\(\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{2}\)
(c)
Now take the square root of both sides to solve for \(\sin\theta\text{.}\)
Answer.
\(\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}\)

Remark 8.1.25.

The derivation for \(\cos\left(\frac{\theta}{2}\right)\) and \(\tan\left(\frac{\theta}{2}\right)\) is similar to that in Activity 8.1.24 when starting with the power-reduction formulas for each trigonometric function.

Definition 8.1.26.

The Half Angle Formulas are:
  • \(\displaystyle \sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}\)
  • \(\displaystyle \cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}\)
  • \(\displaystyle \tan\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}=\frac{\sin\alpha}{1+\cos\alpha}=\frac{1-\cos\alpha}{\sin\alpha}\)

Activity 8.1.27.

Find \(\sin15°\) using a half-angle formula in Definition 8.1.26.
(a)
What angle is \(15\)° half of?
Answer.
\(30\)°
(b)
Substitute \(\theta\) with \(30\)° in the half-angle formula for sine.
Answer.
\(\sin\left(\frac{30}{2}\right)=\pm\sqrt{\frac{1-\cos30}{2}}\)
(c)
Use the unit circle to find the exact value of \(\cos30\)° and substitute that into the equation you got in part (b).
Answer.
\(\sin\left(\frac{30}{2}\right)=\pm\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}\)
(d)
Simplify your equation in part (c) to find the exact value of \(\sin\left(\frac{30}{2}\right)\text{.}\)
Answer.
\(\sin\left(\frac{30}{2}\right)=\pm\frac{\sqrt{2-\sqrt3}}{2}\)
(e)
Notice that there are two values for \(\sin15\)° when using the half-angle formula. Which value should we use and why?
Answer.
Because \(15\)° is in Quadrant I, we know that \(\sin15\)° should be positive. So, we should use \(\frac{\sqrt{2-\sqrt3}}{2}\text{.}\)

Exercises 8.1.2 Exercises

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