I want to solve the following task:
Which angle between 0° and 360° has a cosine (or sine, or tangens) of 0.5?
Same task, but for an angle between 540° and 720°?
I want to solve it without calculator of course, and so long I was wondering if a solution is only possible by drawing the (unit) circle?
Thanks
Best Answer
I think that it's well know property of a right triangle that if one angle is $30^{\circ}$, then the opposite side of that angle is half the length of the hypotenuse.
From formula for sine we have:
$$\sin \alpha = \frac{\text{opposite side}}{\text{hypotenuse}}$$
Using the property we mentioned earlier wh obtain:
$$\sin 30^{\circ} = \frac 12 = \sin 150^{\circ}$$
These are the only angles from $0^{\circ}$ to $360^{\circ}$ which sine has value of $\frac 12$
We also know that the following rule holds in the first quadrant.
$$\sin \alpha = \cos (90^{\circ} - \alpha)$$ $$\sin 30^{\circ} = \cos 60^{\circ} = \cos 300^{\circ} = \frac 12$$
These are the only angles from $0^{\circ}$ to $360^{\circ}$ which cosine has value of $\frac 12$
And for angles of $540^{\circ}$ to $720^{\circ}$ no angle has sine value of $\frac 12$, beacuse in this region sine function has negative values. While the only angle which has cosine value of $\frac 12$ is $\alpha = 660^{\circ}$