Chapter 7, Problem 4T

For the angles α and β in the figures, find cos(α + β).

To determine

To find: The exact value of expression $\cos \left(\alpha +\beta \right)$.

Answer to Problem 4T

The value of $\cos \left(\alpha +\beta \right)$ is $\cos \left(\alpha +\beta \right)=\frac{2\sqrt{5}-1}{3\sqrt{5}}$.

Explanation of Solution

Formula used:

The product sum formula $\cos \left(x+y\right)=\cos x\cos y-\sin x\sin y$.

Calculation:

From the given figures for the first triangle , let the hypotenuse be x.

Using Pythagoras theorem the value of x is as follows.

$\begin{array}{c}x=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}\\ =\sqrt{1+4}\\ =\sqrt{5}\end{array}$

Therefore, hypotenuse of the first triangle is $\sqrt{5}$.

Similarly for the second triangle the base be y.

Using Pythagoras theorem the base is calculated as follows.

$\begin{array}{c}y=\sqrt{{\left(3\right)}^2-{\left(2\right)}^2}\\ =\sqrt{9-4}\\ =\sqrt{5}\end{array}$

Therefore, for the second triangle the base is $\sqrt{5}$.

Using sum-product formula the value of $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$.

From the figure obtain the value of $\cos \beta$ as follows.

$\begin{array}{c}\cos \beta =\frac{\text{adjecent side}}{\text{hypotenuse}}\\ =\frac{\sqrt{5}}{3}\end{array}$

Similarly obtain the value of sin $\beta$ as follows.

$\begin{array}{c}\sin \beta =\frac{\text{opposite side}}{\text{hypotenuse}}\\ =\frac{2}{3}\end{array}$

Using the first triangle obtain the value of sin $\alpha$ as follows.

$\begin{array}{c}\sin \alpha =\frac{\text{opposite side}}{\text{hypotenuse}}\\ =\frac{1}{\sqrt{5}}\end{array}$

Similarly, obtain the value of cos $\alpha$ as follows.

$\begin{array}{c}\cos \alpha =\frac{\text{adjecent side}}{\text{hypotenuse}}\\ =\frac{2}{\sqrt{5}}\end{array}$

Use above values in sum-difference formula and proceed as follows.

$\begin{array}{c}\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ =\left(\frac{2}{\sqrt{5}}\right)\left(\frac{\sqrt{5}}{3}\right)-\left(\frac{1}{\sqrt{5}}\right)\left(\frac{2}{3}\right)\\ =\frac{2}{3}\left(1-\frac{1}{\sqrt{5}}\right)\\ =\frac{2\sqrt{5}-1}{3\sqrt{5}}\end{array}$

Therefore, $\cos \left(\alpha +\beta \right)=\frac{2\sqrt{5}-1}{3\sqrt{5}}$.