Trigonometry - When do I know that SINE of an Angle became Cosine

Question

I am doing a trigonometry exercise and I am struggling to understand one thing, the exercise is pretty elementar, but I am still in the basics of it... The exercise is as follows: Let 90 < A < 180 and tgA = -1/2

What is: sin(270 - A) cos(270 - A)

The picture is attached. My doubt is: I know that sin(270 - A) is an angle and that this angle is on the second quadrant because [270 - (180 - A)] is equal to 90 + A Then I drew up a right triangle and since I know that tgA = -1/2, I drew adjacent = 2 , opposite = 1 and hypotenuse = square root of 5 So far so good, but then the exercise asked sin(270-A), then I simply evaluated the triangle I drew by getting the opposite value over hypotenuse, which in this case is (square roof of 5 / 5) I have access to the answer sheet to see if I am understanding it right, it turns out that -2/5 / 5 It seems as though the adjacent turned into opposite and vice versa Why do they change places? What is the logic behind it?

I am sorry if my explanation was poor, I can give more examples and further details if you guys want so

Thank you a lot in advance

Answer 1

Comments have alluded to the identities$$cos(A\pm B)=\cos A\cos B\mp\sin A\sin B,\,\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,$$which you can prove e.g. by seeing where $\binom{\cos A}{\sin A}$ goes when rotated anticlockwise by $B$. So $\sin 2A=2\sin A\cos A$ and $$\sin(270^\circ-A)\cos(270^\circ-A)=\frac12\sin(540^\circ-2A)=\frac12\sin(180^\circ-2A)=\frac12\sin 2A.$$Or perhaps you wanted the factors individually? $$\sin(270^\circ-A)=\sin270^\circ\cos A-\cos270^\circ\sin A=-\cos A;$$similarly, $\cos(270^\circ-A)=-\sin A.$

Answer 2

First, whenever a question involves angles outside the range $0<\theta<90^\circ$, I would avoid completely any use of the words “adjacent” and “opposite.” The angle $\theta$ is measured counterclockwise from the positive $x$ axis, we find the point $(x,y)$ where the ray at that angle intersects the unit circle, and then $\sin(\theta)=y$ and $\cos(\theta)=x.$

I agree that if $\theta$ is in the second quadrant then $270^\circ-\theta$ is also in the second quadrant. But your particular angle $\theta$ produced a point farther from the $y$ axis than from the $x$ axis, and $270^\circ-\theta$ should give a point farther from the $x$ axis than the $y$ axis, hence the $y$ coordinate has the larger magnitude, hence the sine has greater magnitude than the cosine.

But the sine should be positive in the second quadrant. So I get $2\sqrt5/5.$

I would rather use trig identities rather than all these different angles and quadrants. The relevant identity in this case is $\sin(270^\circ-\theta)=-\cos(\theta)$.