Step-1

Title: Trigonometry function( sine, cosine, tangent)

Grade: 1400-a Lesson: S3-L1

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step Type Explanation Answer

1

Problem

If sin(α) = \$ 3/5 \$ and cos(β) = \$ 4/5\$, where α and β are acute angles, find tan(α + β).

2

Step

The given values are

sin(α) = \$ 3/5 \$, cos(β) = \$ 4/5\$

3

Formula:

\$ tan(α+β) = (tan(α) + tan(β) )/( 1 - tan(α) + tan(β) ) \$

4

Hint

We can find cos(α) and sin(β) using the Pythagorean identity:

\$ cos(α) = \sqrt ( 1 - sin^2(α) ) \$

\$ sin(β) = \sqrt ( 1 - cos^2(β) ) \$

5

Step

Let’s calculate:

\$ cos(α) = \sqrt ( 1 - (3/5)^2 ) \$

\$ sin(β) = \sqrt ( 1 - (4/5)^2 ) \$

6

Step

After simplification

\$ cos(α) = \sqrt ( 1 - 9/25 ) = 4/5 \$

\$ sin(β) = \sqrt ( 1 - 16/25 ) = 3/5 \$

7

Step

Now, we can find tan(α) and tan(β):

\$ tan(α) = sin(α) / cos(α) = (3/5) / (4/5) = 3/4 \$

\$ tan(β) = sin(β) / cos(β) = (3/5) / (4/5) = 3/4 \$

8

Step

Now, substitute these values into the formula for tan(α+β):

\$ tan(α+β) = (3/4 + 3/4 )/( 1 - (3/4 * 3/4) ) \$

\$ tan(α+β) = ( 6/4 )/( 1 - 9/16 ) \$

9

Step

After simplification

\$ tan(α+β) = ( 3/2 )/( (16 - 9) /16 ) \$

\$ tan(α+β) = ( 3/2 )/( 7/16 ) \$

10

Step

To simplify, multiply the numerator and denominator by 16:

\$ tan(α+β) = 24/7 \$

11

Step

Therefore, \$ tan(α+β) = 24/7 \$

12

Sumup

Can you summarize what you’ve understood in the above steps?

13

Choice.A

This is not correct because it does not match the correct result for tan(α+β)

\$ 18/13 \$

14

Choice.B

This is not correct because it does not match the correct result for tan(α+β)

\$ 7/12 \$

15

Choice.C

This is not correct because it does not match the correct result for tan(α+β)

\$ 13/9 \$

16

Choice.D

This is correct because it does match the correct result for tan(α+β)

\$ 24/7 \$

17

Answer

Option

D

18

Sumup

Can you summarize what you’ve understood in the above steps?


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