Respuesta :
Answer:
Part a) [tex]sin(s + t) =-\frac{63}{65}[/tex]
Part b) [tex]tan(s + t) = -\frac{63}{16}[/tex]
Part c) (s+t) lie on Quadrant IV
Step-by-step explanation:
[Part a) Find sin(s+t)
we know that
[tex]sin(s + t) = sin(s) cos(t) + sin(t)cos(s)[/tex]
step 1
Find sin(s)
[tex]sin^{2}(s)+cos^{2}(s)=1[/tex]
we have
[tex]cos(s)=-\frac{12}{13}[/tex]
substitute
[tex]sin^{2}(s)+(-\frac{12}{13})^{2}=1[/tex]
[tex]sin^{2}(s)+(\frac{144}{169})=1[/tex]
[tex]sin^{2}(s)=1-(\frac{144}{169})[/tex]
[tex]sin^{2}(s)=(\frac{25}{169})[/tex]
[tex]sin(s)=\frac{5}{13}[/tex] ---> is positive because s lie on II Quadrant
step 2
Find cos(t)
[tex]sin^{2}(t)+cos^{2}(t)=1[/tex]
we have
[tex]sin(t)=\frac{4}{5}[/tex]
substitute
[tex](\frac{4}{5})^{2}+cos^{2}(t)=1[/tex]
[tex](\frac{16}{25})+cos^{2}(t)=1[/tex]
[tex]cos^{2}(t)=1-(\frac{16}{25})[/tex]
[tex]cos^{2}(t)=\frac{9}{25}[/tex]
[tex]cos(t)=-\frac{3}{5}[/tex] is negative because t lie on II Quadrant
step 3
Find sin(s+t)
[tex]sin(s + t) = sin(s) cos(t) + sin(t)cos(s)[/tex]
we have
[tex]sin(s)=\frac{5}{13}[/tex]
[tex]cos(t)=-\frac{3}{5}[/tex]
[tex]sin(t)=\frac{4}{5}[/tex]
[tex]cos(s)=-\frac{12}{13}[/tex]
substitute the values
[tex]sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})[/tex]
[tex]sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})[/tex]
[tex]sin(s + t) =-\frac{63}{65}[/tex]
Part b) Find tan(s+t)
we know that
tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]
we have
[tex]sin(s)=\frac{5}{13}[/tex]
[tex]cos(t)=-\frac{3}{5}[/tex]
[tex]sin(t)=\frac{4}{5}[/tex]
[tex]cos(s)=-\frac{12}{13}[/tex]
step 1
Find tan(s)
[tex]tan(s)=sin(s)/cos(s)[/tex]
substitute
[tex]tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}[/tex]
step 2
Find tan(t)
[tex]tan(t)=sin(t)/cos(t)[/tex]
substitute
[tex]tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}[/tex]
step 3
Find tan(s+t)
[tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]
substitute the values
[tex]tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))[/tex]
[tex]tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})[/tex]
[tex]tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})[/tex]
[tex]tan(s + t) = -\frac{63}{16}[/tex]
Part c) Quadrant of s+t
we know that
[tex]sin(s + t) =negative[/tex] ----> (s+t) could be in III or IV quadrant
[tex]tan(s + t) =negative[/tex] ----> (s+t) could be in III or IV quadrant
Find the value of cos(s+t)
[tex]cos(s+t) = cos(s) cos(t) -sin (s) sin(t)[/tex]
we have
[tex]sin(s)=\frac{5}{13}[/tex]
[tex]cos(t)=-\frac{3}{5}[/tex]
[tex]sin(t)=\frac{4}{5}[/tex]
[tex]cos(s)=-\frac{12}{13}[/tex]
substitute
[tex]cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})[/tex]
[tex]cos(s+t) = (\frac{36}{65})-(\frac{20}{65})[/tex]
[tex]cos(s+t) =\frac{16}{65}[/tex]
we have that
[tex]cos(s+t)=positive[/tex] -----> (s+t) could be in I or IV quadrant
[tex]sin(s + t) =negative[/tex] ----> (s+t) could be in III or IV quadrant
[tex]tan(s + t) =negative[/tex] ----> (s+t) could be in III or IV quadrant
therefore
(s+t) lie on Quadrant IV
a) The function [tex]\sin (s+t)[/tex] is [tex]-\frac{63}{65}[/tex].
b) The function [tex]\tan (s+t)[/tex] is [tex]-\frac{63}{928}\sqrt{313} - \frac{567}{928}[/tex].
c) [tex]s+t[/tex] is located in quadrant IV.
How to determine angle values by trigonometric expressions
In this question we must find the values of the sine and tangent functions of the sum of two angles, as well as the quadrant to which it belongs.
a) In this case we need to apply trigonometric expressions to determine the value of the sine function:
[tex]\sin (s + t) = \sin s\cdot \cos t + \cos s\cdot \sin t[/tex]
[tex]\sin (s+t) = -\sqrt{1-\sin^{2}t}\cdot \sqrt{1-\cos^{2}s}+\cos s \cdot \sin t[/tex] (1)
If we know that [tex]\cos s = -\frac{12}{13}[/tex] and [tex]\sin t = \frac{4}{5}[/tex], then the sine of the sum of the two angles is:
[tex]\sin (s+t) = -\sqrt{1-\left(\frac{4}{5} \right)^{2}}\cdot \sqrt{1-\left(-\frac{12}{13} \right)^{2}}+\left(-\frac{12}{13} \right)\cdot \left(\frac{4}{5} \right)[/tex]
[tex]\sin (s+t) = -\frac{63}{65}[/tex]
The value of the function [tex]\sin (s+t)[/tex] is [tex]-\frac{63}{65}[/tex]. [tex]\blacksquare[/tex]
b) The tangent function of the sum of the two angles is found by means of the following trigonometric expressions:
[tex]\cos (s+t) = \cos s \cdot \cos t +\sin s \cdot \sin t[/tex]
[tex]\cos (s+t) = -\cos s \cdot \sqrt{1-\sin^{2}t}+\sin t \cdot \sqrt{1+\cos^{2}s}[/tex]
If we know that [tex]\cos s = -\frac{12}{13}[/tex] and [tex]\sin t = \frac{4}{5}[/tex], then the tangent of the sum of the two angles is:
[tex]\cos (s+t) = -\left(-\frac{12}{13} \right)\cdot \sqrt{1-\left(\frac{4}{5} \right)^{2}}+\left(\frac{4}{5} \right)\cdot \sqrt{1+\left(-\frac{12}{13} \right)^{2}}[/tex]
[tex]\cos (s+t) = \frac{4\cdot \sqrt{313}-36}{65}[/tex]
The tangent function is:
[tex]\tan (s+t) = \frac{\sin (s+t)}{\cos (s+t)}[/tex]
[tex]\tan (s+t) = -\frac{63}{4\sqrt{313}-36}[/tex]
[tex]\tan (s+t) = -\frac{63\cdot (4\sqrt{313}+36)}{(4\sqrt{313}-36)\cdot (4\sqrt{313}+36)}[/tex]
[tex]\tan (s+t) = -\frac{252\sqrt{313}+2268}{3712}[/tex]
[tex]\tan (s+t) = -\frac{63}{928}\sqrt{313} - \frac{567}{928}[/tex]
The value of the function [tex]\tan (s+t)[/tex] is [tex]-\frac{63}{928}\sqrt{313} - \frac{567}{928}[/tex]. [tex]\blacksquare[/tex]
c) Since [tex]\cos (s+t) > 0[/tex] and [tex]\sin (s+t) < 0[/tex], then [tex]s+t[/tex] is located in quadrant IV. [tex]\blacksquare[/tex]
To learn more on trigonometric expressions, we kindly invite to check this verified question: https://brainly.com/question/6904750
