Use sum and difference formulas for sine
The sum and difference formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas.
sum and difference formula for sine
[latex]\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta[/latex]
[latex]\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta[/latex]
How To: Given two angles, find the sine of the difference between the angles.
Write the difference formula for sine.
Substitute the given angles into the formula.
Simplify.
Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b.
[latex]\sin \left({45}^{\circ }-{30}^{\circ }\right)[/latex]
[latex]\sin \left({135}^{\circ }-{120}^{\circ }\right)[/latex]
Show Solution
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]
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]
Find the exact value of [latex]\sin \left({\cos }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{3}{5}\right)[/latex].
Show Solution
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]