For example, consider a point \(P\) on the unit circle, with coordinates \((x,y)\text{.}\) If we draw a right triangle (as shown in the figure below), the Pythagorean Theorem says that \(x^2+y^2=1\text{.}\)
But, remember, that the \(x\)-coordinate of the point corresponds to \(\cos\theta\) and the \(y\)-coordinate corresponds to \(\sin\theta\text{.}\) Thus, we get:
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 next two activities will lead us to find the other two Pythagorean Identities.
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.
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.
Students will probably make the conjecture that the value of \(\sin\left( \pi + \dfrac{\pi}{2}\right)\) is just \(\sin \pi + \sin \left( \dfrac{\pi}{2} \right)\text{,}\) which equals \(1\text{.}\)
So from this example, we can see that it is NOT safe to assume that \(\sin\left( \pi + \dfrac{\pi}{2}\right) = \sin \pi + \sin \left( \dfrac{\pi}{2} \right)\text{.}\)
Notice that in Activity 8.1.9, we saw that \(\sin\left( \pi + \dfrac{\pi}{2}\right) \neq \sin \pi + \sin \dfrac{\pi}{2}\text{.}\) 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, \(\cos\left( \pi + \dfrac{\pi}{2}\right) \neq \cos \pi + \cos \dfrac{\pi}{2}\) and \(\tan\left( \pi + \dfrac{\pi}{2}\right) \neq \tan \pi + \tan \dfrac{\pi}{2}\text{.}\) The same is true for finding the difference of two angles.
Recall that the coordinates of points on the unit circle are given by \((\cos\theta, \sin\theta)\text{.}\) In the first 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 \((\cos\beta, \sin\beta)\text{.}\) In the second figure, the triangle is rotated so that point \(B\) has coordinates \((1,0)\text{.}\)
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 \(B\text{.}\)
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 \(Q\text{?}\)
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?
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?
In Activity 8.1.11, 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 \(\tan(\alpha+\beta)\text{.}\)
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 \(180^\circ\text{.}\)
Split \(75^\circ\) into the sum of two angles which can be found on the unit circle (use unique angles such as \(0^\circ\text{,}\)\(30^\circ\text{,}\)\(45^\circ\text{,}\)\(60^\circ\text{,}\)\(90^\circ\text{,}\) and \(180^\circ\text{,}\) etc.).
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 \(\alpha\text{.}\)
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 \(\alpha\text{.}\)
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 \(\alpha\text{.}\)
Double angle formulas are used to express the trigonometric ratios of double angles \((2\theta)\) in terms of trigonometric ratios of single angle \((\theta)\text{.}\)
For example, suppose we start with \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\text{.}\) We can substitute the first \(\cos^2\theta\) with \(1-\sin^2\theta\) (think about how you can rewrite the first Pythagorean Identity, \(\cos^2\theta+\sin^2\theta=1\)). We will then have \(1-\sin^2\theta-\sin^2\theta\text{,}\) which simplifies to \(1-2\sin^2\theta\text{.}\) Thus, \(\cos(2\theta)=1-2\sin^2\theta\text{.}\) Using the same method, you can get \(2\cos^2\theta-1\text{.}\)
Use \(\sin^2\theta+\cos^2\theta=1\) to find the value of \(\cos\alpha\text{,}\) which is equal to \(\dfrac{\sqrt5}{3}\text{.}\) Then, use the double angle formula to find that \(\sin2\alpha=\dfrac{4\sqrt5}{9}\text{.}\)
From part (a), we know that \(\cos\alpha = \dfrac{\sqrt5}{3}\) and \(\sin\alpha = \dfrac{2}{3}\text{.}\) So, by using the double angle formula \(\cos^2\alpha - \sin^2\alpha\text{,}\) we know that \(\cos2\alpha = \dfrac{5}{9} - \dfrac{4}{9}\text{,}\) which equals \(\dfrac{1}{9}\text{.}\)
From part (a), we know that \(\cos\alpha = \dfrac{\sqrt5}{3}\) and \(\sin\alpha = \dfrac{2}{3}\text{.}\) So, \(\tan\alpha = \dfrac{\dfrac{2}{3}}{\dfrac{\sqrt5}{3}}\text{,}\) which is equal to \(\dfrac{2}{\sqrt5}\text{.}\) By using the double angle formula \(\dfrac{2\tan\alpha}{1-\tan^2\alpha}\text{,}\) we know that \(\tan2\alpha = \dfrac{\dfrac{4}{\sqrt5}}{1-\dfrac{4}{5}}\text{,}\) which equals \(4\sqrt5\text{.}\)
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.
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 \(\sin^2\theta\text{.}\)
To generate the power reduction formula for tangent, let’s begin with its definition: \(\tan^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta}\text{.}\) Use the formulas you created in parts (a) and (b) to rewrite \(\tan^2\theta\text{.}\)
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 Theorem 8.1.24 and then simplify
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 \(\sin\dfrac{\theta}{2}\text{.}\)
The derivation for \(\cos\left(\dfrac{\theta}{2}\right)\) and \(\tan\left(\dfrac{\theta}{2}\right)\) is similar to that in Activity 8.1.26 when starting with the power-reduction formulas for each trigonometric function.
\(\sin\dfrac{\theta}{2}=\sqrt{\dfrac{1-\cos\theta}{2}}\) or \(\sin\dfrac{\theta}{2}=-\sqrt{\dfrac{1-\cos\theta}{2}}\) depending on which quadrant \(\theta\) is in.
\(\cos\dfrac{\theta}{2}=\sqrt{\dfrac{1+\cos\theta}{2}}\) or \(\cos\dfrac{\theta}{2}=-\sqrt{\dfrac{1+\cos\theta}{2}}\) depending on which quadrant \(\theta\) is in.
\(\tan\dfrac{\theta}{2}=\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{1-\cos\theta}{\sin\theta}\) or \(\tan\dfrac{\theta}{2}=-\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{1-\cos\theta}{\sin\theta}\) depending on which quadrant \(\theta\) is in.
