Question

The value of cos(alpha+beta) given sin(alpha)=40/41 and sin(beta)=15/17

Answer 1

Since

`\begin{gathered} \sin (\alpha)=(40)/(41), \\ \sin (\beta)=(15)/(17), \end{gathered}`

then

`\begin{gathered} \cos (\alpha)=\frac{\sqrt[]{41^2-40^2}}{41}=(9)/(41), \\ \cos (\beta)=\frac{\sqrt[]{17^2-15^2}}{17}=(8)/(17)\text{.} \end{gathered}`

Now, recall that:

`\begin{gathered} \cos (\alpha+\beta)=\cos (\alpha)\cos (\beta)-\sin (\alpha)\sin (\beta), \\ \sin (\alpha+\beta)=\sin (\alpha)\cos (\beta)+\cos (\alpha)\sin (\beta), \\ \tan (\alpha+\beta)=\frac{\tan (\alpha)+\tan (\beta)}{1-\text{tan(}\alpha)\tan (\beta)}\text{.} \end{gathered}`

Therefore:

`\cos (\alpha+\beta)=(9)/(41)*(8)/(17)-(40)/(41)*(15)/(17)=(72)/(697)-(600)/(697)=-(528)/(697)\text{.}`

Answer:

`\cos (\alpha+\beta)=-(528)/(697)\text{.}`