Question

Given sine of alpha equals 2/3 and cosine of alpha is less than zero, find the exact value of the other five trigonometric functions.

Answer 1

there is a picture of \(\alpha\) an angle with \(\sin(\alpha)=\frac{2}{3}\)
what you need is the third side, which you find via pythagoras
\[\sqrt{3^2-2^2}=\sqrt5\]

Answer 2

then you can find all the trig ratios you want, just make sure to keep in mind that \(\cos(\alpha)\) is negative, so for example
\[\cos(\alpha)=-\frac{\sqrt5}{3}\]

Answer 3

uhm, thank you alot. But I litteraly dont know anything about this stuff, so i have no idea what you mean or how to do it:o

Answer 4

i guess that is why you have to do this exercise, to learn or memorize what the various trig ratios are
for example, the cosine of an angle is the "adjacent over the hypotenuse" which is why in this case it is \(\cos(\alpha)=-\frac{\sqrt5}{3}\)

Answer 5

i see.......

Answer 6

Sine Alpha = Opposite side over hypotenuse.
Cosine Alpha = Adjacent side over hyp.
Tangent = Sine/Cosine =Opposite/Adjacent.

Co secant = 1/ sin
Sec= 1/cos
cot= 1/tan

Also, Sin^2(X) + cos ^(X) = 1
(2/3)^ + cos^2(X) = 1
4/9 + cos^2(x) = 1
cos^2(x) = 1- 4/9 =5/9 (<0)

You know two sides of the triangle, you can find the third using Pythagorean theorem, and the rest using the formulae mentioned above.