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Rasek [7]
3 years ago
11

Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s an

d t in quadrant ll.
Mathematics
1 answer:
Anton [14]3 years ago
4 0

Answer:

Part a) sin(s + t) =-\frac{63}{65}    

Part b) tan(s + t) = -\frac{63}{16}

Part c) (s+t) lie on Quadrant IV

Step-by-step explanation:

[Part a) Find sin(s+t)

we know that

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

step 1

Find sin(s)

sin^{2}(s)+cos^{2}(s)=1

we have

cos(s)=-\frac{12}{13}

substitute

sin^{2}(s)+(-\frac{12}{13})^{2}=1

sin^{2}(s)+(\frac{144}{169})=1

sin^{2}(s)=1-(\frac{144}{169})

sin^{2}(s)=(\frac{25}{169})

sin(s)=\frac{5}{13} ---> is positive because s lie on II Quadrant

step 2

Find cos(t)

sin^{2}(t)+cos^{2}(t)=1

we have

sin(t)=\frac{4}{5}

substitute

(\frac{4}{5})^{2}+cos^{2}(t)=1

(\frac{16}{25})+cos^{2}(t)=1

cos^{2}(t)=1-(\frac{16}{25})

cos^{2}(t)=\frac{9}{25}

cos(t)=-\frac{3}{5} is negative because t lie on II Quadrant

step 3

Find sin(s+t)

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

we have

sin(s)=\frac{5}{13}

cos(t)=-\frac{3}{5}

sin(t)=\frac{4}{5}

cos(s)=-\frac{12}{13}

substitute the values

sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})

sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})

sin(s + t) =-\frac{63}{65}

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

sin(s)=\frac{5}{13}

cos(t)=-\frac{3}{5}

sin(t)=\frac{4}{5}

cos(s)=-\frac{12}{13}

step 1

Find tan(s)

tan(s)=sin(s)/cos(s)

substitute

tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}

step 2

Find tan(t)

tan(t)=sin(t)/cos(t)

substitute

tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}

step 3

Find tan(s+t)

tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))

substitute the values

tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))

tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})

tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})

tan(s + t) = -\frac{63}{16}

Part c) Quadrant of s+t

we know that

sin(s + t) =negative  ----> (s+t) could be in III or IV quadrant

tan(s + t) =negative ----> (s+t) could be in III or IV quadrant

Find the value of cos(s+t)

cos(s+t) = cos(s) cos(t) -sin (s) sin(t)

we have

sin(s)=\frac{5}{13}

cos(t)=-\frac{3}{5}

sin(t)=\frac{4}{5}

cos(s)=-\frac{12}{13}

substitute

cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})

cos(s+t) = (\frac{36}{65})-(\frac{20}{65})

cos(s+t) =\frac{16}{65}

we have that

cos(s+t)=positive -----> (s+t) could be in I or IV quadrant

sin(s + t) =negative  ----> (s+t) could be in III or IV quadrant

tan(s + t) =negative ----> (s+t) could be in III or IV quadrant

therefore

(s+t) lie on Quadrant IV

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