Sum and Difference Identities - Pre-Calculus

Card 0 of 124

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha + \beta)$ .

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Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt ${r^2$ - $y^2$} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2$\sqrt{6}$}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt ${r^2$ - $x^2$} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {$\sqrt{15}$}{4}$ .

Using the sine sum formula, we see:

$\ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \ \* = \frac {1}{5} \cdot \frac {1}{4} + \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 + 6 \sqrt {10}}{20}$

Find $\sin\bigg($\frac{\pi }{3}$ + $\frac{\pi }{6}$\bigg)$ using the sum identity.

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Using the sum formula for sine,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where,

$a=\frac{\pi}{3}$$ , $b=\frac{\pi}{6}$$

yeilds:

$\sin \bigg($\frac{\pi}{3}$+\sin$\frac{\pi}{6}$\bigg)=\sin\bigg($\frac{\pi }{3}$\bigg)\cdot\cos\bigg($\frac{\pi }{6}$\bigg) +\cos \bigg($\frac{\pi}{3}$\bigg)\cdot\sin\bigg($\frac{\pi }{6}$\bigg)$

$=\frac{\sqrt{3}$}2{}\cdot$\frac{\sqrt{3}$}2{}+\frac{1}{2}$\cdot$\frac{1}{2}$$

$=\frac{3}{4}$+\frac{1}{4}$=1$ .

Calculate $\sin(75)$ .

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Notice that $\sin(75)$ is equivalent to $\sin(30+45)$ . With this conversion, the sum formula can be applied using,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where

, .

Therefore the result is as follows:

$\sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)$

$=\frac{1}{2}$\cdot $\frac{\sqrt{2}$}{2}+\frac{\sqrt{3}$}{2}\cdot$\frac{\sqrt{2}$}2{}$

$=\frac{\sqrt2}{4}$+\frac{\sqrt6}{4}$=\frac{\sqrt2 +\sqrt6}{4}$$ .

Evaluate the exact value of:

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In order to solve , two special angles will need to be used to solve for the exact values.

The angles chosen are and degrees, since:

Write the formula for the cosine additive identity.

Substitute the known variables.

$cos(30)cos(45)-sin(30)sin(45)=\left($\frac{\sqrt3}{2}\right)\left($\frac{\sqrt2}{2}\right)-\left($\frac{1}{2}\right)\left($\frac{\sqrt2}{2}\right)$

$=\frac{\sqrt6}{4}$-$\frac{\sqrt2}{4}$=\frac{\sqrt6-\sqrt2}{4}$$

$2cos(75)=2\left($\frac{\sqrt6-\sqrt2}{4}\right)=\frac{\sqrt6-\sqrt2}{2}$$

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\cos (\alpha + \beta)$ .

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Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt ${r^2$ - $y^2$} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2$\sqrt{6}$}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt ${r^2$ - $x^2$} = \sqrt {16-1} = \sqrt {15}$ . So $\sin \beta = \frac {$\sqrt{15}$}{4}$ .

Using the cosine sum formula, we then see:

$\ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\ \* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} - \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 $\sqrt{6}$ - \sqrt {15}}{20}$ .

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

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

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Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt ${r^2$ - $y^2$} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2$\sqrt{6}$}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt ${r^2$ - $x^2$} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {$\sqrt{15}$}{4}$ .

Using the cosine difference formula, we see:
$\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\ \* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 $\sqrt{6}$ + \sqrt {15}}{20}$

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

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

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Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt ${r^2$ - $y^2$} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2$\sqrt{6}$}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt ${r^2$ - $x^2$} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {$\sqrt{15}$}{4}$ .

Using the sine difference formula, we see:

$\ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \ \* = \frac {1}{5} \cdot \frac {1}{4} - \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 - 6 \sqrt {10}}{20}$

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\tan (\alpha + \beta)$ .

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Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt ${r^2$ - $y^2$} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So
$\tan \alpha = \frac {1}{2$\sqrt{6}$} = \frac {$\sqrt{6}$}{12}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt ${r^2$ - $x^2$} = \sqrt {16-1} = \sqrt {15}$ .

So $\tan \beta = $\sqrt{15}$$ .

Using the tangent sum formula, we see:

$\ \tan(\alpha+\beta) = $\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ \ \ \* = $\frac{\frac{\sqrt{6}$}{12}+$\sqrt{15}$}{1-$\frac{\sqrt{6}$}{12} \cdot $\frac{\sqrt{15}$}{1}}}} = $\frac{\frac{\sqrt{6}$+12$\sqrt{15}$}{12}}{$\frac{12-3\sqrt{10}$}{12}} \ \ \* = $\frac{\sqrt{6}$+12$\sqrt{15}$}{12-3$\sqrt{10}$} \cdot $\frac{12+3\sqrt{10}$}{12+3$\sqrt{10}$} \ \ \* = $\frac{12\sqrt{6}$+6$\sqrt{15}$+144$\sqrt{15}$+180$\sqrt{6}$}{144-90} \ \* = $\frac{192\sqrt{6}$+150$\sqrt{15}$}{54} = $\frac{32\sqrt{6}$+25$\sqrt{15}$}{9}$

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\tan(\alpha - \beta)$ .

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Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt ${r^2$ - $y^2$} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\tan \alpha = \frac {1}{2$\sqrt{6}$} = \frac {$\sqrt{6}$}{12}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt ${r^2$ - $x^2$} = \sqrt {16-1} = \sqrt {15}$ .

So $\tan \beta = $\sqrt{15}$$ .

Using the tangent sum formula, we see:

$\ \tan(\alpha-\beta) = $\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$ \ \ \* = $\frac{\frac{\sqrt{6}$}{12}-$\sqrt{15}$}{1+\frac{\sqrt{6}$}{12} \cdot $\frac{\sqrt{15}$}{1}}}} = $\frac{\frac{\sqrt{6}$-12$\sqrt{15}$}{12}}{$\frac{12+3\sqrt{10}$}{12}} \ \ \* = $\frac{\sqrt{6}$-12$\sqrt{15}$}{12+3$\sqrt{10}$} \cdot $\frac{12-3\sqrt{10}$}{12-3$\sqrt{10}$} \ \ \* = $\frac{12\sqrt{6}$-6$\sqrt{15}$-144$\sqrt{15}$+180$\sqrt{6}$}{144-90} \ \ \* = $\frac{192\sqrt{6}$-150$\sqrt{15}$}{54} = $\frac{32\sqrt{6}$-25$\sqrt{15}$}{9}$

Find the value of .

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To solve , we will need to use both the sum and difference identities for cosine.

Write the formula for these identities.

$cos(A\pm B)= cos(A)cos(B)\pm sin(A)sin(B)$

To solve for and , find two special angles whose difference and sum equals to the angle 15 and 75, respectively. The two special angles are 45 and 30.

Substitute the special angles in the formula.

$cos(45\pm 30)= cos(45)cos(30)\pm sin(45)sin(30)$

Evaluate both conditions.

$=\frac{\sqrt2}{2}$\left($\frac{\sqrt3}{2}\right)+\frac{\sqrt2}{2}$\left($\frac{1}{2}\right)=\frac{\sqrt6}{4}$+\frac{\sqrt2}{4}$$

$=\frac{\sqrt2}{2}$\left($\frac{\sqrt3}{2}\right)-$\frac{\sqrt2}{2}$\left($\frac{1}{2}\right)=\frac{\sqrt6}{4}$-$\frac{\sqrt2}{4}$$

Solve for .

$=\frac{\sqrt6}{4}$+\frac{\sqrt2}{4}$-\left[$\frac{\sqrt6}{4}$-$\frac{\sqrt2}{4}\right]$

$=\frac{\sqrt6}{4}$+\frac{\sqrt2}{4}$-$\frac{\sqrt6}{4}$+\frac{\sqrt2}{4}$$

$= $\frac{2\sqrt2}{4}$$

$= $\frac{\sqrt2}{2}$$

Given that $\tan x = 2$ and $\tan y = 3$ , find $\tan (x+y)$ .

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Jump straight to the tangent sum formula:

$\ \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}$

From here plug in the given values and simplify.

$\ \ = \frac {2+3}{1-2\cdot3} \ \= \frac {5}{-5} \ \= -1$

Find the exact value for: $4\sin $(15^{\circ}$)$

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In order to solve this question, it is necessary to know the sine difference identity.

The values of and must be a special angle, and their difference must be 15 degrees.

A possibility of their values that match the criteria are:

Substitute the values into the formula and solve.

$sin(15)=($\frac{\sqrt2}{2}$)($\frac{\sqrt3}{2}$)-($\frac{\sqrt2}{2}$)($\frac{1}{2}$)= $\frac{\sqrt6}{4}$-$\frac{\sqrt2}{4}$$

Evaluate .

$=4($\frac{\sqrt6}{4}$-$\frac{\sqrt2}{4}$)= \sqrt6-\sqrt2$

Find the exact value of:

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In order to find the exact value of , the sum identity of sine must be used. Write the formula.

The only possibilites of and are 45 and 60 degrees interchangably. Substitute these values into the equation and evaluate.

$sin(105)=($\frac{\sqrt2}{2}$)($\frac{1}{2}$)+($\frac{\sqrt2}{2}$)($\frac{\sqrt3}{2}$)=\frac{\sqrt2+\sqrt6}{4}$$

Which of the following expressions best represents $\tan\left ( $60^{\circ}$\right )$ ?

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Write the identity for $tan(2\theta)$ .

Set the value of the angle equal to .

Substitute the value of into the identity.

Evaluate

$\sin \left($\frac{5 \pi}{12}\right)$ .

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$\frac{5 \pi }{12}$ is equivalent to $\frac{2 \pi }{12 }$ + $\frac{3 \pi }{12 }$ or more simplified $\frac{\pi}{6 }$ + $\frac{\pi }{4}$ .

We can use the sum identity to evaluate this sine:

$\sin \left($\frac{\pi }{6}$ + $\frac{\pi }{4 }$ \right) = \sin\left ($\frac{\pi }{6}$ \right) \cos \left(\frac { \pi }{4} \right) + \cos \left($\frac{\pi }{6}\right) \sin\left( $\frac{ \pi }{4}$ \right)$

From the unit circle, we can determine these measures:

$\frac{1}{2}$ \cdot $\frac{1}{\sqrt 2 }$ + $\frac{ \sqrt 3 }{2}$ \cdot $\frac{1}{ \sqrt 2 }$ = $\frac{1}{2 \sqrt 2 }$+ $\frac{\sqrt 3 }{2 \sqrt 2 }$ = $\frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }$

Is the following equation an identity?

$\cos \left ( x-\pi \right )=\cos \left ( x \right )$

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$\cos \left ( x-\pi \right )=\cos \left ( x \right )\rightarrow \cos x\cos \pi +\sin x\sin \pi =\cos x\rightarrow \cos x(-1)+\sin x(0)=\cos x\rightarrow -\cos x\neq \cos x$

Due to this inequality, this is not an identity.

Evaluate

$\sin(285 ^ o)$ .

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The angle or .

Using the first one:

We can find these values in the unit circle:

$\frac{\sqrt 3 }{2 }$ \cdot $\frac{-1}{\sqrt 2 }$ + $\frac{1}{2}$ \cdot $\frac{-1}{\sqrt 2 }$ = $\frac{-\sqrt3 }{2 \sqrt 2 }$ + $\frac{-1}{2\sqrt2}$ = $\frac{-\sqrt 3 - 1 }{2 \sqrt 2 }$

Use the sum or difference identity to find the exact value: $cos255^\circ$

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Using the identity, we can break up the 255 into and then solve: $cos300\cdot cos45+sin300\cdot sin45\rightarrow $\frac{1}{2}$\cdot $\frac{\sqrt{2}$}{2}+\frac{-\sqrt{3}$}{2}\cdot $\frac{\sqrt{2}$}{2}\rightarrow $\frac{\sqrt{2}$-$\sqrt{6}$}{4}$

Is the following equation an identity? $cos(x-\pi)=cos(x)$

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$cos(x-\pi)=cos(x)\rightarrow cosxcos\pi+sinxsin\pi=cosx\rightarrow cosx(-1)+sinx(0)=cosx\rightarrow -cosx\neq cosx$

and due to this inequality, this is not an identity

Use the sum or difference identity to find the exact value: $\cos $255^{\circ}$$

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Using the identity, we can break up the into and then solve: $\cos 300\cdot \cos 45+\sin 300\cdot \sin 45\rightarrow $\frac{1}{2}$\cdot $\frac{\sqrt{2}$}{2}+\frac{-\sqrt{3}$}{2}\cdot $\frac{\sqrt{2}$}{2}\rightarrow $\frac{\sqrt{2}$-$\sqrt{6}$}{4}$ and so the correct answer is $$\frac{\sqrt{2}$-$\sqrt{6}$}{4}$ .