Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values.

Keep in mind that we can always check the answer using a graphing calculator in radian mode.

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

Keep in mind that we can always check the answer using a graphing calculator in degree mode.

1. Let’s begin by writing the formula and substitute the given angles.
   [latex]\begin{array}{ccc}\hfill \mathrm{sin}\left(\alpha -\beta \right)& =& \mathrm{sin}\,\alpha \,\mathrm{cos}\,\beta -\mathrm{cos}\,\alpha \,\mathrm{sin}\,\beta \hfill \\ \hfill \mathrm{sin}\left(45°-30°\right)& =& \mathrm{sin}\left(45°\right)\mathrm{cos}\left(30°\right)-\mathrm{cos}\left(45°\right)\mathrm{sin}\left(30°\right)\hfill \end{array}[/latex]

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

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

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

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

   Next, we find the values of the trigonometric expressions.

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

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

   [latex]\begin{array}{ccc}\hfill \mathrm{sin}\left(135°-120°\right)& =& \frac{\sqrt{2}}{2}\left(-\frac{1}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\hfill \\ & =& \frac{-\sqrt{2}+\sqrt{6}}{4}\hfill \\ & =& \frac{\sqrt{6}-\sqrt{2}}{4}\hfill \\ \phantom{\rule{4em}{0ex}}\mathrm{sin}\left(135°-120°\right)\hfill & =& \frac{\sqrt{2}}{2}\left(-\frac{1}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\hfill \\ & =& \frac{-\sqrt{2}+\sqrt{6}}{4}\hfill \\ & =& \frac{\sqrt{6}-\sqrt{2}}{4}\hfill \end{array}[/latex]

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

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

Using the sum formula for sine,

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

Next, we determine the individual function values within the formula:

So we have

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]\,\mathrm{sin}\left(\alpha +\beta \right),[/latex]we begin with[latex]\,\mathrm{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. See (Figure). Using the Pythagorean Theorem, we can find the length of side[latex]\,a\text{:}[/latex]
   [latex]\begin{array}{ccc}\hfill {a}^{2}+{3}^{2}& =& {5}^{2}\hfill \\ \hfill {a}^{2}& =& 16\hfill \\ \hfill a& =& 4\hfill \end{array}[/latex]
   

   Figure 4.

   Since[latex]\,\mathrm{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. See (Figure). Again, using the Pythagorean Theorem, we have

   [latex]\begin{array}{ccc}\hfill {\left(-5\right)}^{2}+{a}^{2}& =& {13}^{2}\hfill \\ \hfill 25+{a}^{2}& =& 169\hfill \\ \hfill {a}^{2}& =& 144\hfill \\ \hfill a& =& ±12\hfill \end{array}[/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):[latex]\,\mathrm{cos}\,\alpha =\frac{4}{5}.\,[/latex]We can also find the sine of[latex]\,\beta \,[/latex]from the triangle in (Figure), as opposite side over the hypotenuse:[latex]\,\mathrm{sin}\,\beta =-\frac{12}{13}.\,[/latex]Now we are ready to evaluate[latex]\,\mathrm{sin}\left(\alpha +\beta \right).[/latex]

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

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

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

   Then,

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

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

We see that the left side of the equation includes the sines of the sum and the difference of angles.

We can rewrite each using the sum and difference formulas.

We see that the identity is verified.

We can begin by rewriting the numerator on the left side of the equation.

We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.

Using the difference formula for tangent, this problem does not seem as daunting as it might.

Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that[latex]\,\mathrm{tan}\,\beta =\frac{47}{50},[/latex]and[latex]\,\mathrm{tan}\left(\beta -\alpha \right)=\frac{40}{50}=\frac{4}{5}.\,[/latex]We can then use difference formula for tangent.

Now, substituting the values we know into the formula, we have

Use the distributive property, and then simplify the functions.

Now we can calculate the angle in degrees.