Changing Trigonometry Cofunction Identities Over the Equals Sign?

[jddoxtator]

Because you are a hard working student, I will give you this as a bonus. You can memorize them for now, but later when you get better in trigonometry, you will be able to derive them quickly from the unit circle.

$\displaystyle \sin(\theta) = \cos\left(\theta - \frac{\pi}{2}\right) = -\sin(\theta - \pi) = -\cos\left(\theta - \frac{3\pi}{2}\right) = \sin\left(\theta - 2\pi\right) = -\cos\left(\theta + \frac{\pi}{2}\right) = -\sin\left(\theta + \pi\right) = \cos\left(\theta + \frac{3\pi}{2}\right) = \sin\left(\theta + 2\pi\right)$

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$\displaystyle \cos(\theta) = -\sin\left(\theta - \frac{\pi}{2}\right) = -\cos\left(\theta - \pi\right) = \sin\left(\theta - \frac{3\pi}{2}\right) = \cos\left(\theta - 2\pi\right) = \sin\left(\theta + \frac{\pi}{2}\right) = -\cos\left(\theta + \pi\right) = -\sin\left(\theta + \frac{3\pi}{2}\right) = \cos\left(\theta + 2\pi\right)$

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Because the cosine function is even, we can do this: $\displaystyle \cos(-\theta) = \cos(\theta)$. This means $\displaystyle \cos\left(\theta - \frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2} - \theta\right)$.


Because the sine function is odd, we can do this: $\displaystyle \sin(-\theta) = -\sin(\theta)$. This means $\displaystyle \sin\left(\theta - \frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2} - \theta\right)$.

Some of this is covered in the textbook, but this expands it a bit.
From this, I am gleaming that the trigonometric function is the identity of the same degree in a different quadrant under the what ever sign applies to that quadrant for the degree, as the radians are clearly defining quadrants and the polarities of that function in the respective quadrant..

 

[mario99]

Some of this is covered in the textbook, but this expands it a bit.
From this, I am gleaming that the trigonometric function is the identity of the same degree in a different quadrant under the what ever sign applies to that quadrant for the degree, as the radians are clearly defining quadrants and the polarities of that function in the respective quadrant..

If you look at the graphs of the cosine and sine functions, you will notice that they are periodic and they are almost identical. It is only that one of them is shifted by $90$ degrees than the other. This is why, for example, when you shift the sine function by $90$ degrees to the left or you shift it to the right by $90$ degrees and flip it upside down, you get the cosine function. This is one way to get the identities. Another way is to draw a unit circle in the cartesian coordinate that will have four quadrants. You will consider the x-axis as the cosine function and the y-axis as the sine function. The angle is positive when it moves counterclockwise and it starts at the positive x-axis. Since we are in a unit circle and we consider the x-axis as a cosine, this gives $\cos( \ 0 \ \text{degrees} \ ) = 1$. The same concept applies to the sine function, since we consider the sine as a y-axis, it is clear $\sin( \ 0 \ \text{degrees} \ ) = 0$. This means if we know the angle, $\displaystyle \theta$, we can get any point on the unit circle from $(\cos\theta,\sin \theta)$. When the angle $\theta = 0$, we get the point $(\cos 0, \sin 0) = (1, 0)$ on the unit circle.

Now if we increase the angle from $\theta = 0$ to $\theta = 90$ degrees, we get the point $(\ \cos( \ 90 \ \text{degrees} \ ), \sin( \ 90 \ \text{degrees} \ ) \ ) = (0,1)$ on the unit circle. And it is better to work in radians, so $\displaystyle 90 \ \text{degrees} = \frac{\pi}{2}$. This means $\displaystyle \cos 0 = \sin \frac{\pi}{2}$ or $\displaystyle \cos \theta = \sin\left(\theta + \frac{\pi}{2}\right)$. You repeat the same process to the other two points $(-1,0)$ and $(0,-1)$ and by including a negative angle (clockwise rotation), you can derive all the identities I have given. It takes time to understand this method, so don't get frustrated if you get lost somewhere now. You will be able to do it one day.

 

[jddoxtator]

That actually helps a lot.
This explanation verified for me that the hypotenuse always goes through the origin which, when related to (x,y) as cosine and sine, make 100% more sense as they directly follow the 1 unit ratio's of the related function. When adding the polarity of (x,y) and the direction of the angle opening affecting function polarity, this all makes perfect sense. That also tells me where to place the angle when given the quadrant and the polarity of the degree.

 

[Steven G]

Consider sqrt(9) = sqrt(25-16). Since 25-16 =9 both sides are equal.
We know sqrt(9) = 3. Now sqrt(25)-sqrt(16) =5-4=1. Since 3 and 1 are not equal something went wrong, name sqrt(25-16) is not equal to sqrt(25)-sqrt(16)!!
If you look at Pythagoras' theorem which states that in any right triangle, a^2 + b^2 =c^2. Using your method, we'd get that a+b=c. Think about this: If in any right triangle we have a+b= c why would the statement of the theorem be a^2 + b^2 =c^2??
That alone, for me, tells me that sqrt(a^2 + b^2) is NOT a + b

 

[jddoxtator]

Consider sqrt(9) = sqrt(25-16). Since 25-16 =9 both sides are equal.
We know sqrt(9) = 3. Now sqrt(25)-sqrt(16) =5-4=1. Since 3 and 1 are not equal something went wrong, name sqrt(25-16) is not equal to sqrt(25)-sqrt(16)!!
If you look at Pythagoras' theorem which states that in any right triangle, a^2 + b^2 =c^2. Using your method, we'd get that a+b=c. Think about this: If in any right triangle we have a+b= c why would the statement of the theorem be a^2 + b^2 =c^2??
That alone, for me, tells me that sqrt(a^2 + b^2) is NOT a + b

Thank you for that in depth explanation of the rules of equality when square rooting equations.
My error was that I treated the binomial like separate terms when a binomial is in fact a single term.
Lapse of judgment that tend to happen when cramming in new information.