Find the Exact Value cos(21)

Problem

$\cos {21}^{\circ }$

Solution

1. Identify the task as finding the exact trigonometric value of $\cos {21}^{\circ }$

2. Recognize that ${21}^{\circ }$ is not a standard angle (like ${30}^{\circ }$ ${45}^{\circ }$ or ${60}^{\circ }$ and cannot be easily expressed as a simple sum or difference of standard angles using integers.

3. Express the angle using the sum of ${45}^{\circ }$ and ${72}^{\circ }$ or other combinations involving $3^{\circ }$ increments, as $3^{\circ }={18}^{\circ }-{15}^{\circ }$ Note that ${21}^{\circ }={36}^{\circ }-{15}^{\circ }$

4. Apply the cosine difference identity $\cos (A-B)=\cos \left(A\right)\cos B+\sin \left(A\right)\sin B$ where $A={36}^{\circ }$ and $B={15}^{\circ }$

5. Substitute the known exact values:

$\cos {36}^{\circ }=\frac{1+\sqrt{5}}{4}$

$\sin {36}^{\circ }=\frac{\sqrt{10-2\sqrt{5}}}{4}$

$\cos {15}^{\circ }=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin {15}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{4}$

6. Calculate the product:

$\cos {21}^{\circ }=\left(\frac{1+\sqrt{5}}{4}\right)\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)+\left(\frac{\sqrt{10-2\sqrt{5}}}{4}\right)\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)$

7. Simplify the expression into a single fraction.

Final Answer

$\cos {21}^{\circ }=\frac{(\sqrt{6}+\sqrt{2})(1+\sqrt{5})+(\sqrt{6}-\sqrt{2})\sqrt{10-2\sqrt{5}}}{16}$

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