Math / Trigonometry

QuestionConsider the following triangles to determine the exact values of the trigonometric functions. (Hint: Leave as is, no need to simplify.) (a) $\cos (\alpha)=$ $\qquad$ (b) $\cos (\beta)=$ $\qquad$ (c) $\sin (\alpha)=$ $\qquad$ (d) $\sin (\beta)=$ $\qquad$ (d) $\sin (\alpha+\beta)=$ $\qquad$ (e) $\sin (2 \beta)=$ $\qquad$

Studdy Solution

STEP 1

What is this asking? We need to find the exact values of cosine and sine for angles $\alpha$ and $\beta$ , and then find $\sin(\alpha + \beta)$ and $\sin(2\beta)$ . Watch out! Remember SOH CAH TOA for right triangles!
Also, don't forget your trigonometric identities for $\sin(\alpha + \beta)$ and $\sin(2\beta)$ !

STEP 2

1. Find Cosine and Sine of Alpha
2. Find Cosine and Sine of Beta
3. Find Sine of Alpha Plus Beta
4. Find Sine of Two Beta

STEP 3

Let's start with $\cos(\alpha)$ .
Remember CAH: Cosine is Adjacent over Hypotenuse.
Looking at the triangle with angle $\alpha$ , the adjacent side is 8 and the hypotenuse is 10.

STEP 4

So, $\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{8}{10}$

STEP 5

Now for $\sin(\alpha)$ .
Remember SOH: Sine is Opposite over Hypotenuse.
The opposite side to $\alpha$ is 6 and the hypotenuse is still 10.

STEP 6

Therefore, $\sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{10}$

STEP 7

Time for $\cos(\beta)$ ! Adjacent over Hypotenuse.
For angle $\beta$ , the adjacent side is 3 and the hypotenuse is 5.

STEP 8

So, $\cos(\beta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{5}$

STEP 9

And for $\sin(\beta)$ , opposite over hypotenuse.
The opposite side is 4 and the hypotenuse is 5.

STEP 10

Thus, $\sin(\beta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5}$

STEP 11

Remember the sum formula: $\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$ .
We already found all these values, so let's plug them in!

STEP 12

We found $\sin(\alpha) = \frac{6}{10}$ , $\cos(\alpha) = \frac{8}{10}$ , $\sin(\beta) = \frac{4}{5}$ , and $\cos(\beta) = \frac{3}{5}$ .

STEP 13

Substituting these values, we get: $\sin(\alpha + \beta) = \left(\frac{6}{10}\right)\left(\frac{3}{5}\right) + \left(\frac{8}{10}\right)\left(\frac{4}{5}\right)$

STEP 14

$\sin(\alpha + \beta) = \frac{18}{50} + \frac{32}{50} = \frac{18 + 32}{50} = \frac{50}{50}$

STEP 15

For $\sin(2\beta)$ , we use the double angle formula: $\sin(2\beta) = 2\sin(\beta)\cos(\beta)$ .

STEP 16

We already know $\sin(\beta) = \frac{4}{5}$ and $\cos(\beta) = \frac{3}{5}$ , so let's substitute!

STEP 17

$\sin(2\beta) = 2 \cdot \left(\frac{4}{5}\right)\left(\frac{3}{5}\right)$

STEP 18

$\sin(2\beta) = \frac{2 \cdot 4 \cdot 3}{5 \cdot 5} = \frac{24}{25}$

STEP 19

(a) $\cos(\alpha) = \frac{8}{10}$ (b) $\cos(\beta) = \frac{3}{5}$ (c) $\sin(\alpha) = \frac{6}{10}$ (d) $\sin(\beta) = \frac{4}{5}$ (e) $\sin(\alpha + \beta) = \frac{50}{50}$ (f) $\sin(2\beta) = \frac{24}{25}$

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Studdy Solution

STEP 1

STEP 2

1. Find Cosine and Sine of Alpha2. Find Cosine and Sine of Beta3. Find Sine of Alpha Plus Beta4. Find Sine of Two Beta

STEP 3

Let's **start** with cos⁡(α)\cos(\alpha)cos(α).Remember **CAH**: Cosine is **Adjacent** over **Hypotenuse**.Looking at the triangle with angle α\alphaα, the **adjacent** side is **8** and the **hypotenuse** is **10**.

STEP 4

So, cos⁡(α)=AdjacentHypotenuse=810\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{8}{10}cos(α)=HypotenuseAdjacent​=108​

STEP 5

Now for sin⁡(α)\sin(\alpha)sin(α).Remember **SOH**: Sine is **Opposite** over **Hypotenuse**.The **opposite** side to α\alphaα is **6** and the **hypotenuse** is still **10**.

STEP 6

Therefore, sin⁡(α)=OppositeHypotenuse=610\sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{10}sin(α)=HypotenuseOpposite​=106​

STEP 7

Time for cos⁡(β)\cos(\beta)cos(β)! **Adjacent** over **Hypotenuse**.For angle β\betaβ, the **adjacent** side is **3** and the **hypotenuse** is **5**.

STEP 8

So, cos⁡(β)=AdjacentHypotenuse=35\cos(\beta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{5}cos(β)=HypotenuseAdjacent​=53​

STEP 9

And for sin⁡(β)\sin(\beta)sin(β), **opposite** over **hypotenuse**.The **opposite** side is **4** and the **hypotenuse** is **5**.

STEP 10

Thus, sin⁡(β)=OppositeHypotenuse=45\sin(\beta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5}sin(β)=HypotenuseOpposite​=54​

STEP 11

Remember the **sum formula**: sin⁡(α+β)=sin⁡(α)cos⁡(β)+cos⁡(α)sin⁡(β)\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)sin(α+β)=sin(α)cos(β)+cos(α)sin(β).We already found all these values, so let's **plug them in**!

STEP 12

We found sin⁡(α)=610\sin(\alpha) = \frac{6}{10}sin(α)=106​, cos⁡(α)=810\cos(\alpha) = \frac{8}{10}cos(α)=108​, sin⁡(β)=45\sin(\beta) = \frac{4}{5}sin(β)=54​, and cos⁡(β)=35\cos(\beta) = \frac{3}{5}cos(β)=53​.

STEP 13

Substituting these values, we get: sin⁡(α+β)=(610)(35)+(810)(45)\sin(\alpha + \beta) = \left(\frac{6}{10}\right)\left(\frac{3}{5}\right) + \left(\frac{8}{10}\right)\left(\frac{4}{5}\right)sin(α+β)=(106​)(53​)+(108​)(54​)

STEP 14

sin⁡(α+β)=1850+3250=18+3250=5050\sin(\alpha + \beta) = \frac{18}{50} + \frac{32}{50} = \frac{18 + 32}{50} = \frac{50}{50}sin(α+β)=5018​+5032​=5018+32​=5050​

STEP 15

For sin⁡(2β)\sin(2\beta)sin(2β), we use the **double angle formula**: sin⁡(2β)=2sin⁡(β)cos⁡(β)\sin(2\beta) = 2\sin(\beta)\cos(\beta)sin(2β)=2sin(β)cos(β).

STEP 16

We already know sin⁡(β)=45\sin(\beta) = \frac{4}{5}sin(β)=54​ and cos⁡(β)=35\cos(\beta) = \frac{3}{5}cos(β)=53​, so let's **substitute**!

STEP 17

sin⁡(2β)=2⋅(45)(35)\sin(2\beta) = 2 \cdot \left(\frac{4}{5}\right)\left(\frac{3}{5}\right)sin(2β)=2⋅(54​)(53​)

STEP 18

sin⁡(2β)=2⋅4⋅35⋅5=2425\sin(2\beta) = \frac{2 \cdot 4 \cdot 3}{5 \cdot 5} = \frac{24}{25}sin(2β)=5⋅52⋅4⋅3​=2524​

STEP 19

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1. Find Cosine and Sine of Alpha
2. Find Cosine and Sine of Beta
3. Find Sine of Alpha Plus Beta
4. Find Sine of Two Beta

Let's start with $\cos(\alpha)$ .
Remember CAH: Cosine is Adjacent over Hypotenuse.
Looking at the triangle with angle $\alpha$ , the adjacent side is 8 and the hypotenuse is 10.

So, $\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{8}{10}$

Now for $\sin(\alpha)$ .
Remember SOH: Sine is Opposite over Hypotenuse.
The opposite side to $\alpha$ is 6 and the hypotenuse is still 10.

Therefore, $\sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{10}$

Time for $\cos(\beta)$ ! Adjacent over Hypotenuse.
For angle $\beta$ , the adjacent side is 3 and the hypotenuse is 5.

So, $\cos(\beta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{5}$

And for $\sin(\beta)$ , opposite over hypotenuse.
The opposite side is 4 and the hypotenuse is 5.

Thus, $\sin(\beta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5}$

Remember the sum formula: $\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$ .
We already found all these values, so let's plug them in!

We found $\sin(\alpha) = \frac{6}{10}$ , $\cos(\alpha) = \frac{8}{10}$ , $\sin(\beta) = \frac{4}{5}$ , and $\cos(\beta) = \frac{3}{5}$ .

Substituting these values, we get: $\sin(\alpha + \beta) = \left(\frac{6}{10}\right)\left(\frac{3}{5}\right) + \left(\frac{8}{10}\right)\left(\frac{4}{5}\right)$

$\sin(\alpha + \beta) = \frac{18}{50} + \frac{32}{50} = \frac{18 + 32}{50} = \frac{50}{50}$

For $\sin(2\beta)$ , we use the double angle formula: $\sin(2\beta) = 2\sin(\beta)\cos(\beta)$ .

We already know $\sin(\beta) = \frac{4}{5}$ and $\cos(\beta) = \frac{3}{5}$ , so let's substitute!

$\sin(2\beta) = 2 \cdot \left(\frac{4}{5}\right)\left(\frac{3}{5}\right)$

$\sin(2\beta) = \frac{2 \cdot 4 \cdot 3}{5 \cdot 5} = \frac{24}{25}$