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Cos (a+b) sin a=-3/5 sin b=-5/13 a & b in Q3

Guest Jul 22, 2014

Best Answer 

 #1
avatar+91432 
+8

Sin(a) and sin(b) are both negative which means that a and b have to be in the 3rd or 4th quadrant.

 

 $$cos(a+b)\\\\
=cos(a)cos(b)-sin(a)sin(b)\\\\
=\pm\frac{4}{5}\times \pm \frac{12}{13}- \;\;\frac{-3}{5}\times \frac{-5}{13}\\\\
=\pm\frac{48}{65}- \;\;\frac{15}{65}\\\\
=\frac{-15-48}{65}\;\;or\;\;\frac{-15+48}{65}\\\\
=\frac{-63}{65}\;\;or\;\;\frac{33}{65}\\\\$$

Melody  Jul 23, 2014
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2+0 Answers

 #1
avatar+91432 
+8
Best Answer

Sin(a) and sin(b) are both negative which means that a and b have to be in the 3rd or 4th quadrant.

 

 $$cos(a+b)\\\\
=cos(a)cos(b)-sin(a)sin(b)\\\\
=\pm\frac{4}{5}\times \pm \frac{12}{13}- \;\;\frac{-3}{5}\times \frac{-5}{13}\\\\
=\pm\frac{48}{65}- \;\;\frac{15}{65}\\\\
=\frac{-15-48}{65}\;\;or\;\;\frac{-15+48}{65}\\\\
=\frac{-63}{65}\;\;or\;\;\frac{33}{65}\\\\$$

Melody  Jul 23, 2014
 #2
avatar+18827 
+5

Cos (a+b) sin a=-3/5 sin b=-5/13 a & b in Q3  ?

$$\boxed{\cos(a+b)=\;?} \quad \sin{(a)} =-{3\over5} \qquad \sin {(b)}=-{5\over13}$$

$$\cos{(a+b)}=\cos{(a)}*\cos(b)-sin{(a)}*\sin{(b)}$$

cos(a)=?   and   cos(b)=?

$$\textstyle{
\cos{(a)}=\sqrt{1-\sin^2{(a)}}=
\sqrt{1-({3\over5})^2}= {\sqrt{5^2-3^2}\over5}={\sqrt{16}\over5}={\pm4\over5}=\pm{4\over5}
}$$

$$\textstyle{
\cos{(b)}=\sqrt{1-\sin^2{(b)}}=
\sqrt{1-({5\over13})^2}= {\sqrt{13^2-5^2}\over13}={\sqrt{144}\over13}={\pm12\over13}=\pm{12\over13}
}$$

$$\cos{(a+b)}=\pm
({4\over5})
\times
({12\over13})
-
(-{3\over5})
\times
(-{5\over13})$$

$$\cos{(a+b)}=\pm
({4\over5})
\times
({12\over13})
-
({3\over5})
\times
({5\over13})$$

$$\cos{(a+b)}=({\pm(4*12)\over5*13})
-
({3*5\over5*13})$$

$$\cos{(a+b)}={\pm(4*12)-3*5\over5*13}$$

$$\cos{(a+b)}={\pm48-15\over65}$$

$$\text{1.) }\cos{(a+b)}={48-15\over65}={33\over65}=0.50769230769$$

$$\text{2.) }\cos{(a+b)}={-48-15\over65}=-{63\over65}=-0.96923076923$$

heureka  Jul 23, 2014

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