Does the trigonometric identity $\cos^2(\theta)+\sin^2(\theta)=1$ apply even when $\theta$ is not in radians or degrees but simply a fraction?

Question

I have been trying to solve this question but have so far been unable to do so as the question does not seem to be "cohesive throughout". Here is my reasoning:

The question is: given that $\cos A=−3/5$, $\sin B=−5/13$, and both $A$ and $B$ are in the 3rd quadrant, find $\cos^2(A)+\sin^2(A)$.

I know of the trigonometric identity $\cos^2(\theta)+\sin^2(\theta)=1$. In this identity however, $\theta$ is in place of $A$. cos $A$ is a fraction in the case of the question, however I often see $\theta$ as in the radian or degree form. Does this mean that the trigonometric identity does not apply to the question or is my assumption based on familiarity incorrect?

Also, if the identity were to apply to the question, does the question have an answer or not? When I attempted to solve this question, I did not get $1$ as the answer.

Answer 1

The fraction is never in Radians or degrees it is a dimensionless quantity. The input to the trigonometric function is in radians or degrees. So the "identity" $\sin^2 x+\cos^2x =1$ always holds which is why it is called an identity (something that always holds).

$$\begin{aligned}\sin\theta &=\dfrac{\overbrace{\text{side opposite}}^{\text{units}}}{\underbrace{\text{hypotenuse}}_{\text{units}}}\\\cos\theta&=\dfrac{\overbrace{\text{side adjacent}}^{\text{units}}}{\underbrace{\text{hypotenuse}}_{\text{units}}}\end{aligned}$$

Answer 2

The question tells you that $\cos A$ is a fraction, not that $A$ itself is a fraction. Like many of the cases I imagine you've encountered, $A$ is just some angle that can be expressed in degrees or radians.

While you could calculate the value of $\cos^2\hspace{-0.9mm}A+\sin^2\hspace{-0.9mm}A$ by hand, I can assure you that it will be $1$ in this case. In general, $\cos^2(\text{something})+\sin^2(\text{something})=1$ will be true unless the "something" happens to be an expression that involves division by zero, taking the square root of a negative number, or something else similarly problematic.

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Here's how you could calculate the value of $\cos^2\hspace{-0.9mm}A+\sin^2\hspace{-0.9mm}A$ without relying on that identity:

Since you know that $A$ is some angle in quadrant III, and you know that $\cos A=-3/5$, consider the triangle I've drawn in Desmos here, and let $A$ be the angle at the origin. By design, $$\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{-3}{5}.$$ From looking at the triangle, it's also evident that $$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{-4}{5}.$$ Both of those negative signs come from the fact that we're moving in the negative $x$ and $y$ directions, since of course it's impossible for a side of a triangle to have a negative length. At this point it's just calculation: $$\cos^2\hspace{-0.9mm}A+\sin^2\hspace{-0.9mm}A=\left(\frac{-3}{5}\right)^2+\left(\frac{-4}{5}\right)^2=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1$$