Question

Use the given information to find (a) sin (s +t), (b) tan (s+t), and (c) the quadrant of s+t cos s=3/5 and cos t= 15/17, s and t quadrant I

Answer 1

Step-by-step explanation:

Let find the other trig function first, just sin s and sin c.

Use pythagoren identity to find sin a.

$1 - \cos {}^{2} (s) = \sin {}^{2} (s)$

cos s=3/5 so cos^2(s)=9/25

$1 - \frac{9}{25} = \sin {}^{2} (s)$

$\frac{16}{25} = \sin {}^{2} (s)$

$\frac{4}{5} = \sin(s)$

Since this is in the first quadrant, it is positve.

Let do the the same thing for sin t.

$1 - \frac{169}{289} = \sin {}^{2} ( {}^{} t)$

$\frac{120}{289} = \sin {}^{2} (t)$

$\frac{2 \sqrt{30} }{17} = \sin(t)$

Now, we can compute

a.

$\sin(s + t) = \sin(s) \cos(t) + \cos(s) \sin(t)$

$\frac{4}{5} \times \frac{13}{17} + \frac{3}{5} \times ( \frac{2 \sqrt{30} }{17} )$

$\frac{52}{85} + \frac{6 \sqrt{30} }{85}$

$\sin(s + t) = \frac{52 + 6 \sqrt{30} }{85}$

b

$\tan(s + t) = \frac{ \tan(s) + \tan(t) }{1 - \tan(s) \tan(t) }$

Tan s is 4/3

Tan t is 2 times root of 30/ 13

So we have

$\frac{ \frac{4}{3} + \frac{2 \sqrt{30} }{13} }{1 - \frac{8 \sqrt{30} }{39} }$

$\tan(s + t) = \frac{52 + 6 \sqrt{30} }{39 - 8 \sqrt{30} }$

c. They are all in 1st quadrant