Show that if $\alpha\in[0^\circ;45^\circ]:\sin(45^\circ+\alpha)=\cos(45^\circ-\alpha)$ and $\cos(45^\circ+\alpha)=\sin(45^\circ-\alpha).$
I tried to use the unit circle, but I am not sure how to draw the angles $\alpha, 45^\circ+\alpha$ and $45^\circ-\alpha.$ I have noticed $45^\circ+\alpha\le90^\circ$ and $45^\circ-\alpha\ge0^\circ.$
Can I use any other approaches? Why do we have restrictions on $\alpha$? Thank you in advance!
Best Answer
Hint: $$\sin \theta = \cos(90^\circ -\theta) \\ \cos\theta =\sin(90^\circ -\theta)$$ The range of $\alpha$ is irrelevant, as this identity holds for all $\theta\in\mathbb R$.