Question

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

Answer 1

Answer:

$cos(\alpha + \beta)=cos(68.4\°) \approx 0.37$

Step-by-step explanation:

Little triangle.

We know both legs, we can use the tangent trigonometric reason to find the angle.

$tan\alpha =\frac{2}{4}\\ tan \alpha=\frac{1}{2}\\ \alpha=tan^{-1}(\frac{1}{2} )\\ \alpha \approx 26.6\°$

Larger triangle.

We know the hypothenuse and the opposite leg. We can use the sin trigonometric reason to find the angle

$sin\beta =\frac{4}{6}\\ sin\beta=\frac{2}{3}\\ \beta=sin^{-1} (\frac{2}{3} )\\\beta= 41.8\°$

So, the sum of them is

$\alpha + \beta = 26.6+41.8= 68.4\°$

Then,

$cos(\alpha + \beta)=cos(68.4\°) \approx 0.37$

Therefore,

$cos(\alpha + \beta)=cos(68.4\°) \approx 0.37$

Answer 2

Answer:

$\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})$

Step-by-step explanation:

Let the hypotenuse of the smaller triangle be h units.

Then; from the Pythagoras Theorem.

$h^2=4^2+2^2$

$h^2=16+4$

$h^2=20$

$h=\sqrt{20}$

$h=2\sqrt{5}$

From the smaller triangle;

$\cos (\alpha)=\frac{4}{2\sqrt{5} }=\frac{2}{\sqrt{5} }$ and $\sin(\alpha)=\frac{2}{2\sqrt{5} }=\frac{1}{\sqrt{5} }$

From the second triangle, let the other other shorter leg of the second triangle be s units.

Then;

$s^2+4^2=6^2$

$s^2+16=36$

$s^2=36-16$

$s^2=20$

$s=\sqrt{20}$

$s=2\sqrt{5}$

$\cos(\beta)=\frac{2\sqrt{5} }{6}=\frac{\sqrt{5} }{3}$

and

$\sin(\beta)=\frac{4}{6}=\frac{2}{3}$

We now use the double angle property;

$\cos(\alpha +\beta)=\cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta)$

we plug in the values to obtain;

$\cos(\alpha +\beta)=\frac{2}{\sqrt{5} }\times \frac{\sqrt{5} }{3}-\frac{1}{\sqrt{5} }\times \frac{2}{3}$

$\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})$