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

User Ivan Dokov
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1 Answer

6 votes

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{.}

User Csantanapr
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