Precalculus Examples

$u = \cos (\frac{π}{4}) i + \sin (\frac{π}{4}) j$ , $v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

Step 1

Simplify $\cos (\frac{π}{4}) i + \sin (\frac{π}{4}) j$ .

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Simplify each term.

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The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$u = \frac{\sqrt{2}}{2} \cdot i + \sin (\frac{π}{4}) j$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

Combine $\frac{\sqrt{2}}{2}$ and $i$ .

$u = \frac{\sqrt{2} i}{2} + \sin (\frac{π}{4}) j$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$u = \frac{\sqrt{2} i}{2} + \frac{\sqrt{2}}{2} j$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

Combine $\frac{\sqrt{2}}{2}$ and $j$ .

$u = \frac{\sqrt{2} i}{2} + \frac{\sqrt{2} j}{2}$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

$u = \frac{\sqrt{2} i}{2} + \frac{\sqrt{2} j}{2}$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

Reorder $\frac{\sqrt{2} i}{2}$ and $\frac{\sqrt{2} j}{2}$ .

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$

Step 2

Simplify $\cos (\frac{2 π}{3}) i + \sin (\frac{2 π}{3}) j$ .

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Simplify each term.

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Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$v = - \cos (\frac{π}{3}) i + \sin (\frac{2 π}{3}) j$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$v = - \frac{1}{2} \cdot i + \sin (\frac{2 π}{3}) j$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

Combine $i$ and $\frac{1}{2}$ .

$v = - \frac{i}{2} + \sin (\frac{2 π}{3}) j$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$v = - \frac{i}{2} + \sin (\frac{π}{3}) j$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$v = - \frac{i}{2} + \frac{\sqrt{3}}{2} j$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

Combine $\frac{\sqrt{3}}{2}$ and $j$ .

$v = - \frac{i}{2} + \frac{\sqrt{3} j}{2}$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = - \frac{i}{2} + \frac{\sqrt{3} j}{2}$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

Reorder $- \frac{i}{2}$ and $\frac{\sqrt{3} j}{2}$ .

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

$u = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

Step 3

Solve the equation for $π$ .

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Rewrite the equation as $\cos (\frac{π}{4}) i + \sin (\frac{π}{4}) j = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$ .

$\cos (\frac{π}{4}) i + \sin (\frac{π}{4}) j = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Simplify the left side.

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Simplify each term.

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The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$\frac{\sqrt{2}}{2} \cdot i + \sin (\frac{π}{4}) j = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Combine $\frac{\sqrt{2}}{2}$ and $i$ .

$\frac{\sqrt{2} i}{2} + \sin (\frac{π}{4}) j = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$\frac{\sqrt{2} i}{2} + \frac{\sqrt{2}}{2} j = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Combine $\frac{\sqrt{2}}{2}$ and $j$ .

$\frac{\sqrt{2} i}{2} + \frac{\sqrt{2} j}{2} = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

$\frac{\sqrt{2} i}{2} + \frac{\sqrt{2} j}{2} = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

$\frac{\sqrt{2} i}{2} + \frac{\sqrt{2} j}{2} = \frac{\sqrt{2} j}{2} + \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Move all terms containing $j$ to the left side of the equation.

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Subtract $\frac{\sqrt{2} j}{2}$ from both sides of the equation.

$\frac{\sqrt{2} i}{2} + \frac{\sqrt{2} j}{2} - \frac{\sqrt{2} j}{2} = \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Combine the numerators over the common denominator.

$\frac{\sqrt{2} i + \sqrt{2} j - \sqrt{2} j}{2} = \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Combine the opposite terms in $\sqrt{2} i + \sqrt{2} j - \sqrt{2} j$ .

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Subtract $\sqrt{2} j$ from $\sqrt{2} j$ .

$\frac{\sqrt{2} i + 0}{2} = \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Add $\sqrt{2} i$ and $0$ .

$\frac{\sqrt{2} i}{2} = \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

$\frac{\sqrt{2} i}{2} = \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

$\frac{\sqrt{2} i}{2} = \frac{\sqrt{2} i}{2}$

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Since $\frac{\sqrt{2} i}{2} = \frac{\sqrt{2} i}{2}$ , the equation will always be true.

Always true

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Always true

$v = \frac{\sqrt{3} j}{2} - \frac{i}{2}$

Step 4

Solve the equation for $i$ .

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Simplify each term.

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Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \cos (\frac{π}{3}) i + \sin (\frac{2 π}{3}) j$

Always true

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \frac{1}{2} \cdot i + \sin (\frac{2 π}{3}) j$

Always true

Combine $i$ and $\frac{1}{2}$ .

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \frac{i}{2} + \sin (\frac{2 π}{3}) j$

Always true

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \frac{i}{2} + \sin (\frac{π}{3}) j$

Always true

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \frac{i}{2} + \frac{\sqrt{3}}{2} j$

Always true

Combine $\frac{\sqrt{3}}{2}$ and $j$ .

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \frac{i}{2} + \frac{\sqrt{3} j}{2}$

Always true

$\frac{\sqrt{3} j}{2} - \frac{i}{2} = - \frac{i}{2} + \frac{\sqrt{3} j}{2}$

Always true

Move all terms containing $i$ to the left side of the equation.

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Add $\frac{i}{2}$ to both sides of the equation.

$\frac{\sqrt{3} j}{2} - \frac{i}{2} + \frac{i}{2} = \frac{\sqrt{3} j}{2}$

Always true

Combine the opposite terms in $\frac{\sqrt{3} j}{2} - \frac{i}{2} + \frac{i}{2}$ .

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Add $- \frac{i}{2}$ and $\frac{i}{2}$ .

$\frac{\sqrt{3} j}{2} + 0 = \frac{\sqrt{3} j}{2}$

Always true

Add $\frac{\sqrt{3} j}{2}$ and $0$ .

$\frac{\sqrt{3} j}{2} = \frac{\sqrt{3} j}{2}$

Always true

$\frac{\sqrt{3} j}{2} = \frac{\sqrt{3} j}{2}$

Always true

$\frac{\sqrt{3} j}{2} = \frac{\sqrt{3} j}{2}$

Always true

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$\sqrt{3} j = \sqrt{3} j$

Always true

Move all terms containing $j$ to the left side of the equation.

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Subtract $\sqrt{3} j$ from both sides of the equation.

$\sqrt{3} j - \sqrt{3} j = 0$

Always true

Subtract $\sqrt{3} j$ from $\sqrt{3} j$ .

$0 = 0$

Always true

$0 = 0$

Always true

Since $0 = 0$ , the equation will always be true.

Always true