Leaving Cert Mathematics Trigonometry: a) Show that cos 2θ = 1

a) Show that cos 2θ = 1 - 2 sin² θ - Leaving Cert Mathematics - Question 4 - 2019

Question 4

a) Show that cos 2θ = 1 - 2 sin² θ. b) Find the cosine of the acute angle between two diagonals of a cube.

Worked Solution & Example Answer:a) Show that cos 2θ = 1 - 2 sin² θ - Leaving Cert Mathematics - Question 4 - 2019

Step 1

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Answer

To show that $ext{cos}\,2 heta = 1 - 2\text{sin}^2\theta$ , we start with the known identity:

$\text{cos}\,2\theta = \text{cos}^2\theta - \text{sin}^2\theta$

Using the Pythagorean identity, we know:

$\text{sin}^2\theta + \text{cos}^2\theta = 1 \Rightarrow \text{sin}^2\theta = 1 - \text{cos}^2\theta$

Substituting this into our identity gives:

$\text{cos}\,2\theta = \text{cos}^2\theta - (1 - \text{cos}^2\theta)$

This simplifies to:

$\text{cos}\,2\theta = 2\text{cos}^2\theta - 1$

Next, rearranging terms allows us to express $ext{cos}\,2\theta$ in terms of $ext{sin}^2\theta$ :

$\text{cos}\,2\theta = 1 - 2\text{sin}^2\theta$

Thus, we have shown the required relationship.

Step 2

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Answer

Let the length of a side of the cube be $x$ .

An internal diagonal of the cube can be expressed as:

$d = \sqrt{x^2 + x^2 + x^2} = x\sqrt{3}$

The diagonals intersect at the center of the cube, and using the cosine rule:

$\cos A = \frac{x^2 + x^2 - \left(\sqrt{3}x\right)^2}{2\cdot x\cdot x}$

Substituting gives:

$\cos A = \frac{2x^2 - 3x^2}{2x^2} = \frac{-x^2}{2x^2} = -\frac{1}{2}$

Hence, the absolute value of the cosine angle we are after (since it’s acute) is:

$\cos A = -\frac{1}{2} \Rightarrow \text{which means } A = 120^{\circ}.$ So, to find cosine, we consider only the sign, resulting in:

$\cos A = \frac{1}{2}$

The angle between the diagonals is therefore determined:

$\cos A = \frac{1}{3}$