Sum, Difference, and Product Identities

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Trigonometry › Sum, Difference, and Product Identities

Questions 1 - 10

Simplify by applying the compound angle formula:

Explanation

Using the compound angle formula, we can rewrite each half of the non-coefficient terms in the given expression. Given that $\cos(x - y) = sinxsiny + cosxcosy$ and $\cos(x + y) = sinxsiny - cosxcosy$ , substitution yields the following:

$(1/2)(\cos(x - y) + \cos(x + y))$

This is the formula for the product of two cosines, $cosxcosy = (1/2)(\cos(x - y) + \cos(x + y))$ .

Solve for the following using the correct identity:

$cos(45 + 30) = \frac{\sqrt{6} - 2\sqrt{2}}{4}$

$cos(45 + 30) = \sqrt{6} - 2\sqrt{2}$

$cos(45 + 30) = \frac{\sqrt{2} - 2\sqrt{6}}{4}$

$cos(45 + 30) = - 2\sqrt{2}$

Explanation

The correct identity to use for this kind of problem is

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$ . We will let $\alpha = 45 = \frac{\pi}{4}$ and $\beta = 30 = \frac{\pi}{6}$ .

$cos(45 + 30) = cos(\frac{\pi}{4})cos(\frac{\pi}{6}) - sin(\frac{\pi}{4})sin(\frac{\pi}{6})$

$cos(45 + 30) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$cos(45 + 30) = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}$

$cos(45 + 30) = \frac{\sqrt{6} - 2\sqrt{2}}{4}$

Solve for the following using the correct identity:

$cos(45 + 30) = \frac{\sqrt{6} - 2\sqrt{2}}{4}$

$cos(45 + 30) = \sqrt{6} - 2\sqrt{2}$

$cos(45 + 30) = \frac{\sqrt{2} - 2\sqrt{6}}{4}$

$cos(45 + 30) = - 2\sqrt{2}$

Explanation

The correct identity to use for this kind of problem is

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$ . We will let $\alpha = 45 = \frac{\pi}{4}$ and $\beta = 30 = \frac{\pi}{6}$ .

$cos(45 + 30) = cos(\frac{\pi}{4})cos(\frac{\pi}{6}) - sin(\frac{\pi}{4})sin(\frac{\pi}{6})$

$cos(45 + 30) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$cos(45 + 30) = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}$

$cos(45 + 30) = \frac{\sqrt{6} - 2\sqrt{2}}{4}$

Simplify by applying the compound angle formula:

Explanation

$(1/2)(\cos(x - y) + \cos(x + y))$

This is the formula for the product of two cosines, $cosxcosy = (1/2)(\cos(x - y) + \cos(x + y))$ .

Which of the following correctly demonstrates the compound angle formula?

$\cos(x + y) = sinxcosy - cosxsiny$

$\cos(x + y) = sinxsiny - cosxcosy$

$\cos(x - y) = sinxsiny - cosxcosy$

$\cos(x - y) = sinxcosy - cosxsiny$

Explanation

The compound angle formula for cosines states that $\cos(x +- y) = sinxsiny -+ cosxcosy$ .

Which of the following correctly demonstrates the compound angle formula?

$\cos(x + y) = sinxcosy - cosxsiny$

$\cos(x + y) = sinxsiny - cosxcosy$

$\cos(x - y) = sinxsiny - cosxcosy$

$\cos(x - y) = sinxcosy - cosxsiny$

Explanation

The compound angle formula for cosines states that $\cos(x +- y) = sinxsiny -+ cosxcosy$ .

Which of the following completes the identity $cos(\alpha + \beta) =$

$sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)$

$sin(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

$cos(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

$cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

Explanation

This is a known trigonometry identity and has been proven to be true. It is often helpful to solve for the quantity within a cosine function when there are unknowns or if the quantity needs to be simplified

Solve for the following given that $x = 2\pi$ . Use the formula for the sum of two sines.

$sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x)$

$\sqrt{3}$

$\pi$

$\frac{1}{2}$

Explanation

We begin by considering our formula for the sum of two sines

$sin(x) + sin(y) = 2sin\frac{x + y}{2}cos\frac{x-y}{2}$

We will let $x = \frac{\pi}{3} + x$ and $y = \frac{\pi}{3} - x$ and plug these values into our formula.

$sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x) = 2sin\frac{(\frac{\pi}{3} + x) + (\frac{\pi}{3} - x)}{2}cos\frac{(\frac{\pi}{3} + x) - (\frac{\pi}{3} - x)}{2}$