In the following problem, the expression is the right side of the formula for cos (alpha - beta) with particular values for alpha and beta. cos (79 degree) cos (19 degree) + sin (79 degree) sin (19 degree)
Identify alpha and beta in each expression.
The value for alpha: degree
The value for beta: degree
Write the expression as the cosine of an angle. cos degree
Find the exact value of the expression. (Type an exact answer, using fraction, radicals and a rationalized denominator.)

Respuesta :

Answer:

1.    [tex]\alpha = 79[/tex] and [tex]\beta = 19[/tex]

2.   [tex]cos(60)[/tex]

3.   [tex]cos(60) = \frac{1}{2}[/tex]

Step-by-step explanation:

Given

[tex]cos(\alpha - \beta )[/tex]

[tex]cos(79)cos(19) + sin(79)sin(19)[/tex]

Solving for [tex]\alpha[/tex] and [tex]\beta[/tex]

In trigonometry;

[tex]cos(\alpha - \beta ) = cos\alpha\ cos\beta + sin\alpha\ sin\beta[/tex]

Equate the above expression to [tex]cos(79)cos(19) + sin(79)sin(19)[/tex]

[tex]cos(\alpha - \beta ) = cos\alpha\ cos\beta + sin\alpha\ sin\beta[/tex] and [tex]cos(\alpha - \beta ) = cos(79)cos(19) + sin(79)sin(19)[/tex]

By comparison

[tex]cos\alpha\ cos\beta + sin\alpha\ sin\beta = cos(79)cos(19) + sin(79)sin(19)[/tex]

Compare expression on the right hand side to the left hand side

[tex]cos\alpha\ cos\beta = cos(79)cos(19) \\\\ sin\alpha\ sin\beta = sin(79)sin(19)[/tex]

This implies that

[tex]cos\alpha\ = cos(79)\\cos\beta = cos(19) \\\\ and\\\\sin\alpha\ = sin(79)\\sin\beta = sin(19)[/tex]

By further comparison

[tex]\alpha = 79[/tex] and [tex]\beta = 19[/tex]

Substitute [tex]\alpha = 79[/tex] and [tex]\beta = 19[/tex] in [tex]cos(\alpha - \beta )[/tex]

[tex]cos(\alpha - \beta ) = cos(79 - 19)[/tex]

[tex]cos(\alpha - \beta ) = cos(60)[/tex]

Hence, the expression is [tex]cos(60)[/tex]

Solving for the exact values;

Express [tex]cos(60)[/tex] as a difference of angles

[tex]cos(60) = cos(90 - 30)[/tex]

Recall that [tex]cos(\alpha - \beta ) = cos\alpha\ cos\beta + sin\alpha\ sin\beta[/tex]

So;

[tex]cos(90- 30 ) = cos(90) cos(30) + sin(90) sin(30)[/tex]

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In trigonometry;

[tex]cos(90) = 0[/tex]; [tex]cos(30) = \frac{\sqrt{3}}{{2}}[/tex]; [tex]sin(90) = 1[/tex]; [tex]sin(30) = \frac{1}{2}[/tex];

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[tex]cos(90- 30 ) = cos(90) cos(30) + sin(90) sin(30)[/tex] becomes

[tex]cos(90- 30 ) = 0 * \frac{\sqrt{3}}{2} + 1 * \frac{1}{2}[/tex]

[tex]cos(90- 30 ) = 0 + \frac{1}{2}[/tex]

[tex]cos(90- 30 ) = \frac{1}{2}[/tex]

Hence;

[tex]cos(60) = \frac{1}{2}[/tex]

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