Given tanØ =4/3 and sin(-4/5) find cosØ A. 1/3 B. - 1/3 C. 3/5 D. - 3/5 E. -√3/5
if cosØ = -3/5 and sinØ < 0 find tanØ (nearest fourth decimal digit) ------- If cos and sin are both negative theta is in QIII. ---- Since cos = x/r = -3/5, x = -3 and r = 5 ---- Solve for "y": y = sqrt[r^2 - x^2] = sqrt[25-9] = sqrt(16) = 4 --- But y is negative in QIII, so y = -4 ----- Tan is positive in QIII: tan(theta) = y/x = -4/-3 = 4/3
thats a dif question but im showing u how to work a problem like that
could you give an example where the value of sin isn't 0 so i can compare them easier
i follow you up until "solve for y"
question seems a little off
Did you mean: "Given tan(Ø)=4/3 and sin(Ø)=-4/5 find cos(Ø)"
here u go freckles
wow they made a huge type-o then
ive found so many errors in the basic math on plato it worrys me when i get to the harder stuff like this, especially when others see the mistakes too :(
I bet you they meant to write "Given tan(Ø)=4/3 and sin(Ø)=-4/5 find cos(Ø)"
it shouldn't say sin(-4/5) this tells us nothing
well in that case can you help me solve that problem and ill put the answer to that in?
\[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\] use this
replace tan(theta) with 4/3 and replace sin(theta) with -4/5 and solve for cos(theta)
if you want you can replace cos(theta) with u if that makes things easier to see and helps you solve
soooooo -3/5?
\[\frac{4}{3}=\frac{\frac{-4}{5}}{u} \\ \frac{4}{3}=\frac{-4}{5 u} \\ \\ \text{ cross multiply } \\ 20u=-12 \\ u=\frac{-12}{20}=\frac{-3}{5}\] yep
but seriously I hope you see why the sin(-4/5) is a type-o
thank you so much, and i do thats why the first answer seemed confusing. its sad that plato isnt formatted correctly though :/
very sad
you would think tons of people would have looked it over
and corrected the software
especially when we are expected to only use the knowledge given to us by plato, asking for help like i did is considered "cheating" lol
Join our real-time social learning platform and learn together with your friends!