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 4

   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 5

   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]