Skip to 0 minutes and 12 secondsHello. It's your turn on trigonometry by the unit circle. OK. Now let us recall some laws that will permit us to reduce our computations to angles between 0 and pi over 2. OK. The laws that I wish to remind you are the antipodal law, which says that the sine of pi plus x is equal to minus the sine of x, and the cosine of pi plus x is equal to minus the cosine of x. Then we have another important fact, that the sine is an odd function. That is, the sine of minus x is equal to minus the sine of x. And the cosine is an even function.

Skip to 1 minute and 14 secondsThat is, the cosine of minus x is equal to the cosine of x. And finally, we remember the periodicity of these trigonometric functions. That is, the sine of x plus any integer multiple of 2 times pi is equal to sine of x.

Skip to 1 minute and 44 secondsAnd the cosine of x plus any integer multiple of 2 pi is equal the cosine of x.

Skip to 1 minute and 58 secondsOK. Now with these laws, let us see that it's very easy to compute these values. Good. We will compute these values applying these laws. You will see that when you have more practice, then you can maybe follow other quicker procedures. OK. But let us apply these laws. First of all, the cosine of 2 pi over 3, I want to reduce this computation to an angle between 0 and pi over 2. Good. I observe that this is equal to the cosine of pi plus minus pi over 3, because this is pi minus pi over 3. And then applying the antipodal law, I get that this is equal to minus the cosine of minus pi over 3. OK.

Skip to 3 minutes and 18 secondsBut the cosine is an even function. Therefore, this is equal to minus the cosine of pi over 3. And now the cosine of pi over 3 is equal to 1 over 2, therefore, minus the cosine is minus 1 over 2.

Skip to 3 minutes and 43 secondsgood. And now, let us consider the tangent of minus pi over 3. OK. By definition, this is equal to the sine of minus pi over 3 over the cosine of minus pi over 3. OK. Because the sine is an odd function, this is equal to minus the sine of pi over 3 over-- now, the cosine is an even function. And therefore, this is equal to the cosine of pi over 3. Good. We have reduced ourself to angles between 0 and pi over 2. And now, the sine of pi over 3 is equal to the square root of 3 over 2. Therefore, at the numerator we get minus the square root of 3 over 2.

Skip to 4 minutes and 48 secondsAnd at the denominator, the cosine of pi over 3 is 1 over 2. And then we get minus the square root of 3. Good. And finally now, we have to compute the sine of 19 pi over 4. OK. Clearly, this angle is greater than 2 times pi. Therefore, we can apply the periodicity to reduce this angle. What do we get? Observe that 19 pi over 4 is equal to 16 pi plus 3 pi over 4, which is the sine of 4 times pi plus 3 pi over 4. Good. Now, this is a multiple of 2 pi. Therefore, this is equal to the sine of 3 pi over 4. OK. And now this is the sine of what? OK.

Skip to 6 minutes and 18 secondsThis is exactly like pi minus pi over 4. That is, pi plus minus pi over 4. And by the antipodal laws, this is equal to what? To the minus sine of minus pi over 4. OK. But the sine is an odd function. Therefore, the sine of minus pi over 4 is minus the sine of pi over 4. We have another minus here. And therefore, we get the sine of pi over 4. OK. We have reduced ourselves to an angle between 0 and pi over 2. And the sine of pi over 4 is equal to the square root of 2 over 2. OK. Thank you very much for your attention.

# It's your turn on trigonometry by the unit circle

Do your best in trying to **solve the following problems**. In any case some of them are solved in the video and all of them are solved in the pdf file below.

### Exercise 1.

Compute \(\cos \dfrac {2\pi}3\), \(\tan\left(-\dfrac \pi 3\right)\) and \(\sin\dfrac{19\pi}4\).

### Exercise 2.

Compute the following values: \[1)\ \ \sin \dfrac 76\pi,\ \cos \dfrac {17}6\pi,\ \tan \dfrac 76\pi;\] \[2)\ \ \sin \dfrac 53\pi,\ \cos \dfrac {11}3\pi,\ \tan \dfrac 53\pi;\] \[3)\ \ \cos\left(- \dfrac 76\pi\right),\ \cos\left(- \dfrac \pi6\right),\ \sin\left(- \dfrac {2\pi}3\right). \]

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