How do you evaluate the expression #cos(u+v)# given #cosu=4/7# with #0

Question

How do you evaluate the expression #cos(u+v)# given #cosu=4/7# with #0<u<pi/2# and #sinv=-9/10# with #pi<v<(3pi)/2#?

Elijah Alexander · Accepted Answer

#(9sqrt(33)-4sqrt(19))/70#

Since #cos(u+v)=cosucosv-sinusinv#

Let's compute (1) cos v and (2) sin u:

(1) #sinu=+-sqrt(1-cos^2u)#;

in the first quadrant #(0< u < pi/2) sinu>0#, then

#sinu=+sqrt(1-(4/7)^2)=sqrt(1-16/49)=sqrt((49-16)/49)=sqrt(33)/7#

(2) #cos v=+-sqrt(1-sin^2v)#

in the third quadrant #(pi < v < (3pi)/2) cosv<0#, then

#cosv=-sqrt(1-(-9/10)^2)=-sqrt(1-81/100)=-sqrt((100-81)/100)=-sqrt(19)/10#

Then

#cos(u+v)=cosucosv-sinusinv=# #=4/7(-sqrt(19)/10)-sqrt(33)/7(-9/10)=# #=-(4sqrt(19))/70+(9sqrt(33))/70# #=(9sqrt(33)-4sqrt(19))/70#

Dylan Arnold · Answer

undefined Given $ \cos(u) = \frac{4}{7} $ with $ 0 < u < \frac{\pi}{2} $ and $ \sin(v) = -\frac{9}{10} $ with $ \pi < v < \frac{3\pi}{2} $, we need to evaluate $ \cos(u + v) $.

We'll use the trigonometric identities:

1. $ \cos(u + v) = \cos(u)\cos(v) - \sin(u)\sin(v) $

First, we need to find $ \sin(u) $. Since $ \cos(u) = \frac{4}{7} $ and $ 0 < u < \frac{\pi}{2} $, we can use the Pythagorean identity:

$$ \sin^2(u) + \cos^2(u) = 1 $$
$$ \sin^2(u) + \left(\frac{4}{7}\right)^2 = 1 $$
$$ \sin^2(u) + \frac{16}{49} = 1 $$
$$ \sin^2(u) = 1 - \frac{16}{49} = \frac{33}{49} $$
$$ \sin(u) = \pm \sqrt{\frac{33}{49}} = \pm \frac{\sqrt{33}}{7} $$

Since $ 0 < u < \frac{\pi}{2} $, $ \sin(u) $ must be positive. So, $ \sin(u) = \frac{\sqrt{33}}{7} $.

Now, we have $ \cos(u) $, $ \sin(u) $, and $ \sin(v) $. We need to find $ \cos(v) $.

Since $ \sin(v) = -\frac{9}{10} $ and $ \pi < v < \frac{3\pi}{2} $, we can use the fact that $ \cos(v) $ is negative in the third quadrant:

$$ \cos^2(v) + \sin^2(v) = 1 $$
$$ \cos^2(v) + \left(-\frac{9}{10}\right)^2 = 1 $$
$$ \cos^2(v) + \frac{81}{100} = 1 $$
$$ \cos^2(v) = 1 - \frac{81}{100} = \frac{19}{100} $$
$$ \cos(v) = -\sqrt{\frac{19}{100}} = -\frac{\sqrt{19}}{10} $$

Now, substitute the values into the formula for $ \cos(u + v) $:

$$ \cos(u + v) = \frac{4}{7} \left( -\frac{\sqrt{19}}{10} \right) - \frac{\sqrt{33}}{7} \left( -\frac{9}{10} \right) $$

$$ \cos(u + v) = -\frac{4\sqrt{19}}{70} + \frac{9\sqrt{33}}{70} $$

$$ \cos(u + v) = \frac{-4\sqrt{19} + 9\sqrt{33}}{70} $$

So, $ \cos(u + v) = \frac{-4\sqrt{19} + 9\sqrt{33}}{70} $.

How do you evaluate the expression #cos(u+v)# given #cosu=4/7# with #0&lt;u&lt;pi/2# and #sinv=-9/10# with #pi&lt;v&lt;(3pi)/2#?

How do you evaluate the expression #cos(u+v)# given #cosu=4/7# with #0<u<pi/2# and #sinv=-9/10# with #pi<v<(3pi)/2#?