Help doing math problem! if cosθ = 3/5 and sin θ<0, find: a. sinθ b.secθ c.cscθ d.tan θ e.cotθ?

if cosθ = 3/5 and sin θ%26lt;0, find: a. sinθ b.secθ c.cscθ d.tan θ e.cotθ


How do I do it??

Help doing math problem! if cosθ = 3/5 and sin θ%26lt;0, find: a. sinθ b.secθ c.cscθ d.tan θ e.cotθ?
cosθ = 3/5 --%26gt; secθ = 5/3 [secx=1/cosx]


sinθ = -4/5 --%26gt; csc θ= -5/4 [sin²x= 1- cos²x]


tanθ= -4/3 --%26gt; cot θ= -3/4 [tanx=sinx/cosx]





Saludos.
Reply:Use the Pythagorean trigonometric identity: sin^2 (theta) and cos^2 (theta) = 1.





(3/5)^2 + sin^2 (theta) = 1 = 9/25 + sin^2 (theta) = 1. sin^2 (theta) = 16/25. sin (theta) = 4/5.





sec (theta) = 5/3. csc (theta) = 5/4. tan (theta) = 4/3. cot (theta) = 3/4.
Reply:cos(x) = 3/5





squaring





cos^2(x) = 9/25





1 - sin^2(x) = 9/25 (since cos^2(x) = 1 - sin^2(x))





sin^2(x) = 1 - (9/25) = (25-9)/25





sin^2(x) = 16/25





a) sin(x) = -4/5 (since sinx %26lt;0)





b) secx = 1/cosx = 5/3





c) csc(x) = 1/sinx = -5/4





d) tan(x) = sinx/cosx = -(4/5)/(3/5) = -4/3





e) cotx = 1/tanx = -3/4
Reply:cos(x)=3/5 ----%26gt; sec(x)= 5/3





sin(x) = -4/5 ----%26gt; csc(x)= -5/4





tan(x) = -4/3 -----%26gt; cot(x) =-3/4
Reply:cos = adjacent side/ hypotenuse, so the hypotenuse is 5.


sin is opposite side / hpotenuse


By Pythogoren, theorem= 5^2=3^2+4^2


so sin =4/5, but it is negative. so sin(theta)=-4/5


b)sec(theta)=1/cos(theta)=5/3


c)csc(theta)=1/sin(theta)=-5/4


d)tan(theta)=sin/cos= (-4/5)(3/5)=-4/3


e)cot=1/tan=-3/4
Reply:The answers by lou h is right. This is how to get it


sin= y/r cos= x/r tan= y/x


csc= reciprocal of sin sec= reciprocal of cos cot= reciprocal of tan





since you have cos then you know that x=3 and r=5





just use pythagorean theorem to find y.


x squared plus y squared = r squared


when you do this you find that y=4





since cos%26amp;sec are positive, the 4 other functions all have to be the same. Since sin θ%26lt;0 that means it is neg, which means that all of the others are negative as well.