Analysis of the Solution

We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.

We will start on the left side, as it is the more complicated side:

[latex]\begin{align}\tan \theta \cos \theta &=\left(\frac{\sin \theta }{\cos \theta }\right)\cos \theta \\ &=\left(\frac{\sin \theta }{\cancel{\cos \theta }}\right)\cancel{\cos \theta } \\ &=\sin \theta \end{align}[/latex]

Analysis of the Solution

This identity was fairly simple to verify, as it only required writing [latex]\tan \theta[/latex] in terms of [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex].

Working on the left side of the equation, we have

[latex]\begin{align}\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]&=\left(1+\sin x\right)\left(1-\sin x\right)&& \text{Since sin(-}x\text{)=}-\sin x \\ &=1-{\sin }^{2}x&& \text{Difference of squares} \\ &={\cos }^{2}x&& {\text{cos}}^{2}x=1-{\sin }^{2}x\end{align}[/latex]

As the left side is more complicated, let’s begin there.

[latex]\begin{align}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }&=\frac{\left({\tan }^{2}\theta +1\right)-1}{{\sec }^{2}\theta }&& {\sec}^{2}\theta ={\tan }^{2}\theta +1 \\ &=\frac{{\tan }^{2}\theta }{{\sec }^{2}\theta } \\ &={\tan }^{2}\theta \left(\frac{1}{{\sec }^{2}\theta }\right) \\ &={\tan }^{2}\theta \left({\cos }^{2}\theta \right)&& {\cos }^{2}\theta =\frac{1}{{\sec }^{2}\theta } \\ &=\left(\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta }\right)\left({\cos }^{2}\theta \right)&& {\tan}^{2}\theta =\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } \\ &=\left(\frac{{\sin }^{2}\theta }{\cancel{{\cos }^{2}\theta}}\right)\left(\cancel{{\cos }^{2}\theta} \right) \\ &={\sin }^{2}\theta \end{align}[/latex]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

[latex]\begin{align}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }&=\frac{{\sec }^{2}\theta }{{\sec }^{2}\theta }-\frac{1}{{\sec }^{2}\theta } \\ &=1-{\cos }^{2}\theta \\ &={\sin }^{2}\theta \end{align}[/latex]

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

[latex]\begin{align}2\tan \theta \sec \theta &=2\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{1}{\cos \theta }\right) \\ &=\frac{2\sin \theta }{{\cos }^{2}\theta } \\ &=\frac{2\sin \theta }{1-{\sin }^{2}\theta }&& \text{Substitute }1-{\sin }^{2}\theta \text{ for }{\cos }^{2}\theta \end{align}[/latex]

Thus,

[latex]2\tan \theta \sec \theta =\frac{2\sin \theta }{1-{\sin }^{2}\theta }[/latex]

Let’s start with the left side and simplify:

[latex]\begin{align}\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}&=\frac{{\left[\sin \left(-\theta \right)\right]}^{2}-{\left[\cos \left(-\theta \right)\right]}^{2}}{\sin \left(-\theta \right)-\cos \left(-\theta \right)} \\ &=\frac{{\left(-\sin \theta \right)}^{2}-{\left(\cos \theta \right)}^{2}}{-\sin \theta -\cos \theta }&& \sin \left(-x\right)=-\sin x\text{ and }\cos \left(-x\right)=\cos x \\ &=\frac{{\left(\sin \theta \right)}^{2}-{\left(\cos \theta \right)}^{2}}{-\sin \theta -\cos \theta }&& \text{Difference of squares} \\ &=\frac{\left(\sin \theta -\cos \theta \right)\left(\sin \theta +\cos \theta \right)}{-\left(\sin \theta +\cos \theta \right)} \\ &=\frac{\left(\sin \theta -\cos \theta \right)\left(\cancel{\sin \theta +\cos \theta }\right)}{-\left(\cancel{\sin \theta +\cos \theta }\right)} \\ &=\cos \theta -\sin \theta\end{align}[/latex]

We will work on the left side of the equation.

[latex]\begin{align}\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)&=\left(1-{\cos }^{2}x\right)\left(1+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) \\ &=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) && \text{Find the common denominator}. \\ &=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\right) \\ &=\left({\sin }^{2}x\right)\left(\frac{1}{{\sin }^{2}x}\right) \\ &=1\end{align}[/latex]

Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c[/latex]. Letting [latex]\cos \theta =x[/latex], we can rewrite the expression as follows:

[latex]2{x}^{2}+x - 1[/latex]

This expression can be factored as [latex]\left(2x+1\right)\left(x - 1\right)[/latex]. If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for [latex]x[/latex]. At this point, we would replace [latex]x[/latex] with [latex]\cos \theta[/latex] and solve for [latex]\theta[/latex].

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

[latex]\begin{align}4{\cos }^{2}\theta -1&={\left(2\cos \theta \right)}^{2}-1 \\ &=\left(2\cos \theta -1\right)\left(2\cos \theta +1\right) \end{align}[/latex]

We can start with the Pythagorean identity.

[latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta[/latex]

Now we can simplify by substituting [latex]1+{\cot }^{2}\theta[/latex] for [latex]{\csc }^{2}\theta[/latex]. We have

[latex]\begin{align}{\csc }^{2}\theta -{\cot }^{2}\theta &=1+{\cot }^{2}\theta -{\cot }^{2}\theta \\ &=1\end{align}[/latex]

Use the formula for the cosine of the difference of two angles. We have

[latex]\begin{align}\cos \left(\alpha -\beta \right)&=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos \left(\frac{5\pi }{4}-\frac{\pi }{6}\right)&=\cos \left(\frac{5\pi }{4}\right)\cos \left(\frac{\pi }{6}\right)+\sin \left(\frac{5\pi }{4}\right)\sin \left(\frac{\pi }{6}\right) \\ &=\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \\ &=-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ &=\frac{-\sqrt{6}-\sqrt{2}}{4} \end{align}[/latex]

As [latex]{75}^{\circ }={45}^{\circ }+{30}^{\circ }[/latex], we can evaluate [latex]\cos \left({75}^{\circ }\right)[/latex] as [latex]\cos \left({45}^{\circ }+{30}^{\circ }\right)[/latex]. Thus,

[latex]\begin{align}\cos \left({45}^{\circ }+{30}^{\circ }\right)&=\cos \left({45}^{\circ }\right)\cos \left({30}^{\circ }\right)-\sin \left({45}^{\circ }\right)\sin \left({30}^{\circ }\right) \\ &=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}}{2}\right)-\frac{\sqrt{2}}{2}\left(\frac{1}{2}\right) \\ &=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4} \end{align}[/latex]

1. Let’s begin by writing the formula and substitute the given angles.
   [latex]\begin{align}\sin \left(\alpha -\beta \right)&=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \sin \left({45}^{\circ }-{30}^{\circ }\right)&=\sin \left({45}^{\circ }\right)\cos \left({30}^{\circ }\right)-\cos \left({45}^{\circ }\right)\sin \left({30}^{\circ }\right) \end{align}[/latex]

   Next, we need to find the values of the trigonometric expressions.

   [latex]\sin \left({45}^{\circ }\right)=\frac{\sqrt{2}}{2},\text{ }\cos \left({30}^{\circ }\right)=\frac{\sqrt{3}}{2},\text{ }\cos \left({45}^{\circ }\right)=\frac{\sqrt{2}}{2},\text{ }\sin \left({30}^{\circ }\right)=\frac{1}{2}[/latex]

   Now we can substitute these values into the equation and simplify.

   [latex]\begin{align} \sin \left({45}^{\circ }-{30}^{\circ }\right)&=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}}{2}\right)-\frac{\sqrt{2}}{2}\left(\frac{1}{2}\right) \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}\hfill \end{align}[/latex]
2. Again, we write the formula and substitute the given angles.
   [latex]\begin{align}\sin \left(\alpha -\beta \right)&=\sin \alpha \cos \beta -\cos \alpha \sin \beta\\ \sin \left({135}^{\circ }-{120}^{\circ }\right)&=\sin \left({135}^{\circ }\right)\cos \left({120}^{\circ }\right)-\cos \left({135}^{\circ }\right)\sin \left({120}^{\circ }\right)\end{align}[/latex]

   Next, we find the values of the trigonometric expressions.

   [latex]\sin \left({135}^{\circ }\right)=\frac{\sqrt{2}}{2},\cos \left({120}^{\circ }\right)=-\frac{1}{2},\cos \left({135}^{\circ }\right)=\frac{\sqrt{2}}{2},\sin \left({120}^{\circ }\right)=\frac{\sqrt{3}}{2}[/latex]

   Now we can substitute these values into the equation and simplify.

   [latex]\begin{align}\sin \left({135}^{\circ }-{120}^{\circ }\right)&=\frac{\sqrt{2}}{2}\left(-\frac{1}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \\ &=\frac{-\sqrt{2}+\sqrt{6}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4} \end{align}[/latex]

The pattern displayed in this problem is [latex]\sin \left(\alpha +\beta \right)[/latex]. Let [latex]\alpha ={\cos }^{-1}\frac{1}{2}[/latex] and [latex]\beta ={\sin }^{-1}\frac{3}{5}[/latex]. Then we can write

[latex]\begin{align} \cos \alpha &=\frac{1}{2},0\le \alpha \le \pi\\ \sin \beta &=\frac{3}{5},-\frac{\pi }{2}\le \beta \le \frac{\pi }{2} \end{align}[/latex]

We will use the Pythagorean identities to find [latex]\sin \alpha[/latex] and [latex]\cos \beta[/latex].

[latex]\begin{align}\sin \alpha &=\sqrt{1-{\cos }^{2}\alpha } \\ &=\sqrt{1-\frac{1}{4}} \\ &=\sqrt{\frac{3}{4}} \\ &=\frac{\sqrt{3}}{2} \\ \cos \beta &=\sqrt{1-{\sin }^{2}\beta } \\ &=\sqrt{1-\frac{9}{25}} \\ &=\sqrt{\frac{16}{25}} \\ &=\frac{4}{5}\end{align}[/latex]

Using the sum formula for sine,

[latex]\begin{align}\sin \left({\cos }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{3}{5}\right)&=\sin \left(\alpha +\beta \right) \\ &=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ &=\frac{\sqrt{3}}{2}\cdot \frac{4}{5}+\frac{1}{2}\cdot \frac{3}{5} \\ &=\frac{4\sqrt{3}+3}{10}\end{align}[/latex]

Let’s first write the sum formula for tangent and substitute the given angles into the formula.

[latex]\begin{align}\tan \left(\alpha +\beta \right)&=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \tan \left(\frac{\pi }{6}+\frac{\pi }{4}\right)&=\frac{\tan \left(\frac{\pi }{6}\right)+\tan \left(\frac{\pi }{4}\right)}{1-\left(\tan \left(\frac{\pi }{6}\right)\right)\left(\tan \left(\frac{\pi }{4}\right)\right)} \end{align}[/latex]

Next, we determine the individual tangents within the formula:

[latex]\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}},\tan \left(\frac{\pi }{4}\right)=1[/latex]

So we have

[latex]\begin{align}\tan \left(\frac{\pi }{6}+\frac{\pi }{4}\right)&=\frac{\frac{1}{\sqrt{3}}+1}{1-\left(\frac{1}{\sqrt{3}}\right)\left(1\right)} \\ &=\frac{\frac{1+\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} \\ &=\frac{1+\sqrt{3}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}-1}\right) \\ &=\frac{\sqrt{3}+1}{\sqrt{3}-1} \end{align}[/latex]

We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

1. To find [latex]\sin \left(\alpha +\beta \right)[/latex], we begin with [latex]\sin \alpha =\frac{3}{5}[/latex] and [latex]0<\alpha <\frac{\pi }{2}[/latex]. The side opposite [latex]\alpha[/latex] has length 3, the hypotenuse has length 5, and [latex]\alpha[/latex] is in the first quadrant. Using the Pythagorean Theorem, we can find the length of side [latex]a:[/latex]

   [latex]\begin{gathered}{a}^{2}+{3}^{2}={5}^{2} \\ {a}^{2}=16 \\ a=4 \end{gathered}[/latex]

   

   Figure 6

   Since [latex]\cos \beta =-\frac{5}{13}[/latex] and [latex]\pi <\beta <\frac{3\pi }{2}[/latex], the side adjacent to [latex]\beta[/latex] is [latex]-5[/latex], the hypotenuse is 13, and [latex]\beta[/latex] is in the third quadrant. Again, using the Pythagorean Theorem, we have

2. [latex]\begin{align}{\left(-5\right)}^{2}+{a}^{2}&={13}^{2} \\ 25+{a}^{2}&=169 \\ {a}^{2}&=144\\ a&=\pm 12 \end{align}[/latex]

   Since [latex]\beta[/latex] is in the third quadrant, [latex]a=-12[/latex].

   

   Figure 7

   The next step is finding the cosine of [latex]\alpha[/latex] and the sine of [latex]\beta[/latex]. The cosine of [latex]\alpha[/latex] is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5: [latex]\cos \alpha =\frac{4}{5}[/latex]. We can also find the sine of [latex]\beta[/latex] from the triangle in Figure 5, as opposite side over the hypotenuse: [latex]\sin \beta =-\frac{12}{13}[/latex]. Now we are ready to evaluate [latex]\sin \left(\alpha +\beta \right)[/latex].

   [latex]\begin{align}\sin \left(\alpha +\beta \right)&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ &=\left(\frac{3}{5}\right)\left(-\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(-\frac{12}{13}\right) \\ &=-\frac{15}{65}-\frac{48}{65} \\ &=-\frac{63}{65} \end{align}[/latex]
3. We can find [latex]\cos \left(\alpha +\beta \right)[/latex] in a similar manner. We substitute the values according to the formula.
   [latex]\begin{align}\cos \left(\alpha +\beta \right)&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &=\left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right) \\ &=-\frac{20}{65}+\frac{36}{65} \\ &=\frac{16}{65} \end{align}[/latex]
4. For [latex]\tan \left(\alpha +\beta \right)[/latex], if [latex]\sin \alpha =\frac{3}{5}[/latex] and [latex]\cos \alpha =\frac{4}{5}[/latex], then
   [latex]\tan \alpha =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}[/latex]

   If [latex]\sin \beta =-\frac{12}{13}[/latex] and [latex]\cos \beta =-\frac{5}{13}[/latex],
   then

   [latex]\tan \beta =\frac{\frac{-12}{13}}{\frac{-5}{13}}=\frac{12}{5}[/latex]

   Then,

   [latex]\begin{align}\tan \left(\alpha +\beta \right)&=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ &=\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4}\left(\frac{12}{5}\right)} \\ &=\frac{\text{ }\frac{63}{20}}{-\frac{16}{20}} \\ &=-\frac{63}{16} \end{align}[/latex]
5. To find [latex]\tan \left(\alpha -\beta \right)[/latex], we have the values we need. We can substitute them in and evaluate.
   [latex]\begin{align}\tan \left(\alpha -\beta \right)&=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } \\ &=\frac{\frac{3}{4}-\frac{12}{5}}{1+\frac{3}{4}\left(\frac{12}{5}\right)} \\ &=\frac{-\frac{33}{20}}{\frac{56}{20}} \\ &=-\frac{33}{56}\end{align}[/latex]

Analysis of the Solution

A common mistake when addressing problems such as this one is that we may be tempted to think that [latex]\alpha[/latex] and [latex]\beta[/latex] are angles in the same triangle, which of course, they are not. Also note that

The cofunction of [latex]\tan \theta =\cot \left(\frac{\pi }{2}-\theta \right)[/latex]. Thus,

[latex]\begin{align}\tan \left(\frac{\pi }{9}\right)&=\cot \left(\frac{\pi }{2}-\frac{\pi }{9}\right) \\& =\cot \left(\frac{9\pi }{18}-\frac{2\pi }{18}\right) \\& =\cot \left(\frac{7\pi }{18}\right)\end{align}[/latex]