# Finding the angles for a certain value of sine, cosine or tangens

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

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$\cos ^2+\sin ^2=1$ and complex numbers are your friends. – walcher Sep 9 '13 at 15:20
in how far?..... – TestGuest Sep 9 '13 at 15:21
If $\cos=\frac 12$ what is $\sin$? What is $\cos\theta+i\sin\theta$? – walcher Sep 9 '13 at 15:22
we don't use complex numbers yet – TestGuest Sep 9 '13 at 15:24
These are well known angles. Try drawing right angle triangles with the sin and tan values you want. – Warren Hill Sep 9 '13 at 15:40

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 $\alpha = 660^{\circ}$

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