Question

I need help with this practice from my trigonometry prep bookI have attempted this problem previously but I am not too sure if my answer is correct so I’d like more clarification on this

Answer 1

The function is given as,

`\tan \alpha\text{ = }(-12)/(5)\text{ }(\text{ }\alpha\text{ lies in the second quadrant )}`

Calculating the value of h ,

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

Calculating the values of the remaining functions :

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

Now,

calculating the value of x ,

`\begin{gathered} x\text{ = }\sqrt[]{5^2-3^2} \\ x\text{ = }\sqrt[]{25\text{ -9 }} \\ x\text{ = }\sqrt[]{16} \\ x\text{ = 4} \end{gathered}`

`\sin \text{ }\beta=\text{ }(-4)/(5)`

Calculating the required functions ,

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

Therefore,

`\begin{gathered} \cos (\text{ }\alpha-\beta)\text{ =( }(-5)/(13))((3)/(5))\text{ + (}(12)/(13))((-4)/(5)) \\ \cos (\text{ }\alpha-\beta)\text{ = }(-15)/(65)\text{ }\frac{-\text{ 48}}{65} \\ \cos (\text{ }\alpha-\beta)\text{ = }\frac{-15\text{ - 48 }}{65} \\ \cos (\text{ }\alpha-\beta)\text{ = }(-63)/(65) \end{gathered}`

Thus the required answer is ,

`\cos (\text{ }\alpha-\beta)\text{ = }(-63)/(65)`