The question says "Given that cos(theta)=(-7/25) and sin(theta)>0, calculate the exact values of tan(theta) and csc(theta).
We start out by drawing a triangle.
25 is on the side of the hypotenuse and -7 is on the adjacent side, because cos is adjacent over hypotenuse.
Somehow he gets to this halfway through: 225-49=176.
I can understand where he got the 49.
Probably from (-7)^2
I thought you had to use the equation
a^2+b^2=c^2 or
y^2+x^2=r^2
So it would be y^2 +(-7)^2=25^2
(-7)^2 is 49. Subtract this from 25^2, so we can get y^2 by itself.
Would that be going the right way?
I'm mainly just confused about how he got the 225?
He ends up with the answers:
tan(theta)= (-square root of 176)/7 and
csc(theta)= (25 x square root of 176)/176
Looks like either the hypotenuse is 15, so 15^2 = 225
or, he made a mistake and it should have been 625, rather than 225. I think this is the case, because then you have a 7-24-25 triangle.
I think its the second one, but how did you get 625?
He started out saying that cos = 7/25. That means that if the other leg i y, then
7^2 + y^2 = 25^2 = 625
y = 24
To calculate the exact values of tan(theta) and csc(theta), we need to find the missing side lengths of the triangle based on the given information.
Starting with the equation you mentioned, a^2 + b^2 = c^2, or y^2 + (-7)^2 = 25^2, is correct.
Now, let's solve for y^2:
(-7)^2 simplifies to 49, and 25^2 equals 625.
Subtracting 49 from 625 gives us 576.
So, y^2 = 576.
To find y, which represents the opposite side in the triangle, we take the square root of 576.
The square root of 576 is 24. Therefore, y = 24.
Now that we have the values for the adjacent side (x = -7), the opposite side (y = 24), and the hypotenuse (r = 25), we can calculate tan(theta) and csc(theta).
Recall that tan(theta) = opposite/adjacent and csc(theta) = hypotenuse/opposite.
So, tan(theta) = y/x = 24/(-7).
Simplifying, we get tan(theta) = -24/7.
And csc(theta) = r/y = 25/24.
In conclusion, the exact values of tan(theta) and csc(theta) are -24/7 and 25/24, respectively.