Question

I need help with this practice Having trouble solving it The subject is trigonometry

Answer 1

To solve the problem, we will make use of the identity:

`\cos (\alpha-\beta)=\cos (\alpha)\cos (\beta)+\sin (\alpha)\sin (\beta)_{}`

ANGLE α

The angle lies in the second quadrant. The only positive ratio is the sine.

If we have that:

`\tan \alpha=-(12)/(5)`

Displaying this on a triangle for ease of working, we have:

Therefore, the length of the hypotenuse will be:

`\begin{gathered} x=\sqrt[]{12^2+5^2}=\sqrt[]{144+25}=\sqrt[]{169} \\ x=13 \end{gathered}`

Therefore, we have that:

`\begin{gathered} \sin \alpha=(12)/(13) \\ \cos \alpha=-(5)/(13) \end{gathered}`

ANGLE β

This angle lies in the fourth quadrant. Only the cosine ratio is positive in this quadrant.

We are given in the question:

`\cos \beta=(3)/(5)`

Displaying this on a triangle for ease of working, we have:

Therefore, using the Pythagorean Triplets, we have that:

`y=4`

Therefore, we have that:

`\sin \beta=-(4)/(5)`

SOLVING THE IDENTITY

Applying the identity quoted earlier, we have:

`\begin{gathered} \cos (\alpha-\beta)=\cos (\alpha)\cos (\beta)+\sin (\alpha)\sin (\beta)_{} \\ \cos (\alpha-\beta)=(-(5)/(13))((3)/(5))+((12)/(13))(-(4)/(5)) \\ \cos (\alpha-\beta)=-(63)/(65) \end{gathered}`