Solving the Trigonometric Expression: cos 36° − cos 72°

This problem asks us to find the value of the trigonometric expression $\cos 36^\circ - \cos 72^\circ$. To solve this, we need to know the standard values of $\cos 36^\circ$ and $\cos 72^\circ$ or use trigonometric identities to simplify the expression.

The standard values for $\cos 36^\circ$ and $\cos 72^\circ$ are derived using geometric methods (like properties of a regular pentagon) or algebraic techniques (solving trigonometric equations). These values are:

• $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$
• $\cos 72^\circ = \frac{\sqrt{5}-1}{4}$

Note that $\cos 72^\circ = \cos (90^\circ - 18^\circ) = \sin 18^\circ$, so this is equivalent to stating that $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$.

Step-by-Step Calculation

Now, we can substitute these values into the given expression:

The expression is $\cos 36^\circ - \cos 72^\circ$.

Substitute the values:

$\cos 36^\circ - \cos 72^\circ = \frac{\sqrt{5}+1}{4} - \frac{\sqrt{5}-1}{4}$

Combine the fractions since they have a common denominator:

$\cos 36^\circ - \cos 72^\circ = \frac{(\sqrt{5}+1) - (\sqrt{5}-1)}{4}$

Carefully remove the parentheses in the numerator. Remember to distribute the minus sign to both terms inside the second parenthesis:

$\cos 36^\circ - \cos 72^\circ = \frac{\sqrt{5}+1 - \sqrt{5} + 1}{4}$

Combine like terms in the numerator. The $\sqrt{5}$ terms cancel out ($\sqrt{5} - \sqrt{5} = 0$), and the constant terms add up ($1 + 1 = 2$):

$\cos 36^\circ - \cos 72^\circ = \frac{2}{4}$

Simplify the fraction:

$\cos 36^\circ - \cos 72^\circ = \frac{1}{2}$

Thus, the value of $\cos 36^\circ - \cos 72^\circ$ is $\frac{1}{2}$.

Alternative Approach using Identities (Difference to Product)

We could also try using the difference-to-product trigonometric identity:

$\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

Let $A = 36^\circ$ and $B = 72^\circ$.

$\frac{A+B}{2} = \frac{36^\circ + 72^\circ}{2} = \frac{108^\circ}{2} = 54^\circ$

$\frac{A-B}{2} = \frac{36^\circ - 72^\circ}{2} = \frac{-36^\circ}{2} = -18^\circ$

Substituting these into the identity:

$\cos 36^\circ - \cos 72^\circ = -2 \sin(54^\circ) \sin(-18^\circ)$

Using the property $\sin(-\theta) = -\sin(\theta)$:

$\cos 36^\circ - \cos 72^\circ = -2 \sin(54^\circ) (-\sin(18^\circ))$

$\cos 36^\circ - \cos 72^\circ = 2 \sin(54^\circ) \sin(18^\circ)$

Using the complementary angle identity $\sin(90^\circ - \theta) = \cos \theta$, we have $\sin(54^\circ) = \sin(90^\circ - 36^\circ) = \cos 36^\circ$.

So, the expression becomes:

$\cos 36^\circ - \cos 72^\circ = 2 \cos 36^\circ \sin 18^\circ$

We know $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$ and $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$.

Substitute these values:

$\cos 36^\circ - \cos 72^\circ = 2 \left(\frac{\sqrt{5}+1}{4}\right) \left(\frac{\sqrt{5}-1}{4}\right)$

$\cos 36^\circ - \cos 72^\circ = 2 \left(\frac{(\sqrt{5}+1)(\sqrt{5}-1)}{16}\right)$

Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ for the numerator:

$(\sqrt{5}+1)(\sqrt{5}-1) = (\sqrt{5})^2 - (1)^2 = 5 - 1 = 4$

Substitute this back:

$\cos 36^\circ - \cos 72^\circ = 2 \left(\frac{4}{16}\right)$

$\cos 36^\circ - \cos 72^\circ = 2 \left(\frac{1}{4}\right)$

$\cos 36^\circ - \cos 72^\circ = \frac{2}{4} = \frac{1}{2}$

Both methods yield the same result, $\frac{1}{2}$.

Revision Table: Key Trigonometric Values

Additional Information: Deriving cos 36° and cos 72°

The values of $\cos 36^\circ$ and $\cos 72^\circ$ (or $\sin 18^\circ$ and $\sin 54^\circ$) are considered special angles in trigonometry. Their derivation often involves geometric properties of a regular pentagon or solving trigonometric equations.

• Geometric Method: A regular pentagon has interior angles of $108^\circ$. By drawing diagonals, we can form isosceles triangles with angles $36^\circ$, $72^\circ$, $72^\circ$ or $36^\circ$, $36^\circ$, $108^\circ$. Using the properties of similar triangles and the golden ratio ($\phi = \frac{1+\sqrt{5}}{2}$), these trigonometric values can be found. For instance, $\cos 36^\circ = \frac{\phi}{2}$ and $\cos 72^\circ = \frac{1}{2\phi}$.
• Algebraic Method: Let $\theta = 18^\circ$. Then $5\theta = 90^\circ$. We can write this as $2\theta = 90^\circ - 3\theta$. Taking the sine of both sides gives $\sin(2\theta) = \sin(90^\circ - 3\theta)$, which simplifies to $2\sin\theta\cos\theta = \cos(3\theta)$. Using the triple angle formula $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, we get $2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$. Since $\theta=18^\circ$, $\cos\theta \ne 0$, so we can divide by $\cos\theta$: $2\sin\theta = 4\cos^2\theta - 3$. Substitute $\cos^2\theta = 1-\sin^2\theta$: $2\sin\theta = 4(1-\sin^2\theta) - 3$. This leads to a quadratic equation in $\sin\theta$: $4\sin^2\theta + 2\sin\theta - 1 = 0$. Solving this quadratic equation for $\sin\theta$ (where $\theta=18^\circ$ is in the first quadrant, so $\sin\theta > 0$) gives $\sin 18^\circ = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$. Since $\sin 18^\circ > 0$, we take the positive root: $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$. From this, we can find $\cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \sqrt{1-\sin^2 36^\circ} = \sqrt{1-\cos^2 54^\circ}$. Also, $\cos 36^\circ = 1 - 2\sin^2 18^\circ = 1 - 2\left(\frac{\sqrt{5}-1}{4}\right)^2 = 1 - 2\left(\frac{5 - 2\sqrt{5} + 1}{16}\right) = 1 - 2\left(\frac{6 - 2\sqrt{5}}{16}\right) = 1 - \frac{6 - 2\sqrt{5}}{8} = \frac{8 - (6 - 2\sqrt{5})}{8} = \frac{8 - 6 + 2\sqrt{5}}{8} = \frac{2 + 2\sqrt{5}}{8} = \frac{1 + \sqrt{5}}{4}$.