Suppose that
θ
is an angle in standard position whose terminal side intersects the unit circle at
, 35/37−12/37
.
sketch a right-angled triangle.
label the horizontal base as x
the vertical side as y
the hypotenuse as r, and the base angle as Ø
memorize these ratios:
sinØ = y/r
cosØ = x/r
tanØ = y/x
and now their reciprocals:
cscØ = 1/sinØ = r/y
secØ = 1/cosØ = r/x
cotØ = 1/tanØ = x/y
You will not be able to handle trig questions without knowing these.
ok.
It appears to be in QIV, and is a 12-35-37 right triangle.
If you draw it correctly, you should be able just to read off the values of the trig functions of that angle.
ok, I am supposing Ø to be as you stated, now what?
did you notice that 35^2 + (-12)^2 = 37^2 ?
Do you know the basic trig ratios in terms of the x, y, and r of the corresponding right-angled triangle?
no that why I ask
what will sin , cot, csc be for that >
To find the angle θ in standard position, we can use the coordinates of the point where the terminal side of the angle intersects the unit circle.
The coordinates of the given point are (35/37, -12/37). Let's call the x-coordinate of this point, 35/37, as x, and the y-coordinate, -12/37, as y.
We can use the trigonometric functions cosine (cos) and sine (sin) to find the angle θ.
First, calculate the value of cosine (cosθ) using the x-coordinate:
cosθ = x/1
cosθ = 35/37
Next, calculate the value of sine (sinθ) using the y-coordinate:
sinθ = y/1
sinθ = -12/37
Now, we have the values of cosθ and sinθ. We can use the inverse trigonometric functions to find θ.
Using the inverse cosine (arccos) function, we can find the value of θ:
θ = arccos(cosθ)
θ = arccos(35/37)
Similarly, we can use the inverse sine (arcsin) function to find the value of θ:
θ = arcsin(sinθ)
θ = arcsin(-12/37)
Depending on the context or the desired form of the answer, you may need to use a calculator or a table of inverse trigonometric functions to find the actual value of θ.