For positive acute angles AA and B,B, it is known that \tan A=\frac{28}{45}tanA= 45 28 ​ and \cos B=\frac{3}{5}.CosB= 5 3 ​ . Find the value of \cos(A+B)cos(A+B) in simplest form.

Answer:

[tex]\cos(A + B) = \frac{23}{265}[/tex]

Step-by-step explanation:

Given

[tex]\tan A=\frac{28}{45}[/tex]

[tex]\cos B=\frac{3}{5}[/tex]

Required

[tex]\cos(A+B)[/tex]

In trigonometry:

[tex]\cos(A+B) = \cos\ A\ cosB - sinA\ sinB[/tex]

We need to solve for cosA, sinA and sinB

Given that:

[tex]\tan A=\frac{28}{45}[/tex] and the tangent of an angle is a ratio of the opposite side (x) to the adjacent side (y),

So, we have:

[tex]x =28[/tex] and [tex]y = 45[/tex]

For this angle A, we need t calculate its hypotenuse (z).

Using Pythagoras,

[tex]z^2 = x^2 + y^2[/tex]

[tex]z = \sqrt{x^2 + y^2[/tex]

[tex]z = \sqrt{28^2 + 45^2[/tex]

[tex]z = \sqrt{2809[/tex]

[tex]z = 53[/tex]

From here, we can calculate sin and cos A

[tex]\sin A= \frac{opposite}{hypotenuse}[/tex] and [tex]\cos A= \frac{adjacent}{hypotenuse}[/tex]

[tex]\sin A= \frac{x}{z}[/tex]

[tex]\sin A= \frac{28}{53}[/tex]

[tex]\cos A= \frac{y}{z}[/tex]

[tex]\cos A= \frac{45}{53}[/tex]

To angle B.

Give that

[tex]\cos B=\frac{3}{5}[/tex] and the cosine of an angle is a ratio of the adjacent side (a) to the hypotenuse side (c),

So, we have:

[tex]a = 3[/tex] and [tex]b = 5[/tex]

For this angle B, we need to calculate its opposite (b).

Using Pythagoras,

[tex]c^2 = a^2 + b^2[/tex]

[tex]5^2 = 3^2 + b^2[/tex]

[tex]25 = 9 + b^2[/tex]

Collect like terms

[tex]b^2 = 25 - 9[/tex]

[tex]b^2 = 16[/tex]

[tex]b = \sqrt{16[/tex]

[tex]b = 4[/tex]

From here, we can calculate sin B

[tex]\sin B= \frac{opposite}{hypotenuse}[/tex]

[tex]\sin B = \frac{b}{c}[/tex]

[tex]\sin B = \frac{4}{5}[/tex]

Recall that:

[tex]\cos(A+B) = \cos\ A\ cosB - sinA\ sinB[/tex]

and

[tex]\cos B=\frac{3}{5}[/tex]      [tex]\sin B = \frac{4}{5}[/tex]       [tex]\sin A= \frac{28}{53}[/tex]      [tex]\cos A= \frac{45}{53}[/tex]

[tex]\cos(A + B) = \frac{45}{53} * \frac{3}{5} - \frac{28}{53} * \frac{4}{5}[/tex]

[tex]\cos(A + B) = \frac{135}{265} - \frac{112}{265}[/tex]

[tex]\cos(A + B) = \frac{135-112}{265}[/tex]

[tex]\cos(A + B) = \frac{23}{265}[/tex]

The value of [tex]\cos (A+B)[/tex] is [tex]-\frac{96}{265}[/tex].

Trigonometrically speaking, the sine, cosine and tangent of an angle are described by the following formulas:

[tex]\sin \theta = \frac{y}{\sqrt{x^{2}+y^{2}}}[/tex] (1)

[tex]\cos \theta = \frac{x}{\sqrt{x^{2}+y^{2}}}[/tex] (2)

[tex]\tan \theta = \frac{y}{x}[/tex] (3)

Where:

And we have the following trigonometric identity for the cosine of the sum of angles:

[tex]\cos (A+B) = \cos A\cdot \cos B - \sin A \cdot \sin B[/tex] (4)

By (1) and (2), we rewrite (4):

[tex]\cos (A+B) = \frac{x_{A}\cdot x_{B}}{\sqrt{x_{A}^{2}+y_{A}^{2}}\cdot \sqrt{x_{B}^{2}+y_{B}^{2}}} - \frac{y_{A}\cdot y_{B}}{\sqrt{x_{A}^{2}+y_{A}^{2}}\cdot \sqrt{x_{B}^{2}+y_{B}^{2}}}[/tex]

If we know that [tex]x_{A} = 28[/tex], [tex]y_{A} = 45[/tex], [tex]x_{B} = 3[/tex] and [tex]y_{B} = 4[/tex], then the cosine for the sum of angles is:

[tex]\cos (A+B) = \frac{28\cdot 3}{\sqrt{28^{2}+45^{2}}\cdot \sqrt{3^{2}+4^{2}}}-\frac{45\cdot 4}{\sqrt{28^{2}+45^{2}}\cdot \sqrt{3^{2}+4^{2}}}[/tex]

[tex]\cos (A+B) = -\frac{96}{265}[/tex]

The value of [tex]\cos (A+B)[/tex] is [tex]-\frac{96}{265}[/tex].

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