I'm having some trouble with this question. I don't really understand what it is asking.
$\theta$ and $\alpha$ are acute angles in standard position. $\sin\theta=\frac{3}{5}$ and $\cos\alpha=\frac{12}{13}$
What is the exact value of $\cos(\alpha+\theta)$?
I've tried to find a trig identity that will allow us to solve this and haven't had any success. All Help is appreciated! :)
I'm somewhat familiar with the cosine addition formula.
So far, I have
$\cos(\alpha+\theta) = \cos(\frac{12}{13})\cdot\cos(y)-\sin(\frac{3}{5})\cdot\sin(y)$
I now have
$\cos^2\theta+\sin^2\theta=1$
$\cos^2\theta=1-\frac{9}{25}$
$\sqrt{\cos^2\theta}=\sqrt{\frac{16}{25}}$
$\cos\theta=\frac{4}{5}$
Edit: Thanks to everyone who helped out! Here is the answer I came up with:
(cont. from above)
$\sin^2\theta=1-\cos^2\theta\\\sin^2\theta=1-\frac{144}{169}\\\sqrt{sin^2\theta}=\sqrt{\frac{25}{169}}\\\sin\theta=\frac{5}{13}\\\cos(\alpha+\theta)=(\frac{12}{13})(\frac{4}{5})-(\frac{5}{13})(\frac{3}{5})\\\therefore\cos(\alpha+\theta)=\frac{33}{65}$
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$\begingroup$Thanks everyone! Here is the answer I got:
$\cos^2\theta+\sin^2\theta=1\\\cos^2\theta=1-\sin^2\theta\\\cos^2\theta=1-(\frac{3}{5})^{2}\\\cos^2\theta=1-\frac{9}{25}\\\sqrt{\cos^{2}\theta}=\sqrt{\frac{16}{25}}\\\cos\theta=\frac{4}{5}$
$\sin^2\theta=1-\cos^2\theta\\\sin^2\theta=1-\frac{144}{169}\\\sqrt{sin^2\theta}=\sqrt{\frac{25}{169}}\\\sin\theta=\frac{5}{13}\\\cos(\alpha+\theta)=(\frac{12}{13})(\frac{4}{5})-(\frac{5}{13})(\frac{3}{5})\\\therefore\cos(\alpha+\theta)=\frac{33}{65}$
$\endgroup$ 1 $\begingroup$Hint: All you need to do to finish is to note that we always have $\sin^2x+\cos^2x=1$ (why?)
$\endgroup$ $\begingroup$Hint: using Pythagoras' identity, you can determine $|\sin\alpha|$ and $|\cos\theta|$. Can you determine their signs?
$\endgroup$ $\begingroup$Using the cosine addition formula, $\cos(\alpha+\theta)=\cos\alpha\cos\theta-\sin\alpha\sin\theta$.
Using the Pythagorean formula, $\cos\theta=\frac45$ and $\sin\alpha=\frac5{13}$.
Can you take it from here?
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