1. The expression sinπ is equal to 0, while the expression 1/cscπ is undefined. Why is sinø=1/cscø still an identity?
2. Prove cos(ø + π/2)= - sin ø
1. since cscØ = 1/sinØ
if sin Ø = 0 , the cscØ = 1/0
since we cannot divide by zero, and 1/0 is undefined
csc 0 is undefined
Here is a graph of y = csc x for about 4 periods
the vertical lines are asymptotes, and shows where the cscx is undefined. the sine curve is super-imposed. notice that cscx is undefined when sinx = 0
http://www.wolframalpha.com/input/?i=plot+y+%3D+csc%28x%29+%2C+y+%3D+sinx+%2C+-13%3Cx%3C13
how neat is that ?
2. cos(Ø + π/2)
= cosØcosπ/2 - sinØsinπ/2
= cosØ x (0) - sinØ x (1)
= -sinxØ
1. The expression sin π is equal to 0 because the sine function represents the ratio of the length of the side opposite to an angle in a right triangle to the length of the hypotenuse. In the case of π radians (180 degrees), the angle is at the x-axis, so the opposite side has a length of 0.
On the other hand, the expression 1/csc π is undefined because the cosecant function, csc(θ), represents the reciprocal of the sine function. Since sin π is equal to 0, the reciprocal of 0 is undefined.
However, even though sin ø=1/csc ø is not defined at π, it still holds true as an identity for other values of ø. This is because sin ø and csc ø are inverse functions of each other, except for when sin ø equals 0 at certain values like π. In general, for all other values of ø, the reciprocal of sin ø will give a valid result, and sin ø will be equal to 1 divided by csc ø.
2. To prove that cos(ø + π/2) = - sin ø, we can use the trigonometric identities and the properties of cosine and sine functions.
Starting with the left-hand side (LHS) cos(ø + π/2):
cos(ø + π/2) = cos ø * cos(π/2) - sin ø * sin(π/2)
Since cos(π/2) = 0 and sin(π/2) = 1:
cos(ø + π/2) = cos ø * 0 - sin ø * 1
cos(ø + π/2) = 0 - sin ø
cos(ø + π/2) = - sin ø
Therefore, we have proved that cos(ø + π/2) = - sin ø.
To understand why sin(ø)=1/csc(ø) is still considered an identity, let's break it down and analyze each part separately.
First, let's recap some trigonometric definitions:
- The sine function (sin) is defined as the ratio of the length of the side opposite the angle ø in a right triangle to the hypotenuse of the triangle.
- The cosecant function (csc) is the reciprocal of the sine function. Therefore, csc(ø) is equal to 1/sin(ø).
Now, looking at sin(ø)=1/csc(ø):
1. When ø = π, sin(π) is equal to 0. This means that the length of the side opposite the angle π in a right triangle is 0, so there is no elevation in that direction.
2. csc(π) is equal to 1/sin(π), which becomes 1/0. Division by zero is undefined in mathematics, so 1/csc(π) is also undefined.
However, we can approach this identity from a different angle. Let's consider the unit circle, which is a circle with a radius of 1 centered at the origin. Each point on the unit circle corresponds to an angle in radians.
1. At the angle ø = π, the unit circle reaches the point (-1, 0). Looking at the coordinates, we can see that this point has an x-coordinate of -1. The y-coordinate is 0, indicating no elevation in the y-direction.
2. Now, let's consider sin(ø) and csc(ø) in terms of the unit circle. sin(ø) is equal to the y-coordinate of the point on the unit circle corresponding to the angle ø, while csc(ø) is equal to 1 divided by the y-coordinate of that same point.
At the angle ø = π, sin(π) is 0, and therefore 1/csc(π) is undefined, as we discussed earlier. However, looking at this geometric interpretation, we can ignore the undefined value at this specific angle and consider the identity as a whole.
For all other angles, sin(ø) and csc(ø) are well-defined and have a reciprocal relationship. Therefore, sin(ø) is indeed equal to 1/csc(ø), and the identity holds true in general, except for the specific angle π.
Moving on to the second question, let's prove cos(ø + π/2) = -sin(ø):
We can use the trigonometric identities for the cosine and sine of the sum of two angles:
cos(ø + π/2) = cos(ø)cos(π/2) - sin(ø)sin(π/2)
In trigonometry, we know that cos(π/2) = 0 and sin(π/2) = 1. Plugging in these values:
cos(ø + π/2) = cos(ø) * 0 - sin(ø) * 1
cos(ø + π/2) = -sin(ø)
Therefore, cos(ø + π/2) = -sin(ø), and we have proven the identity.