Find the value of two times cos of 45 degrees times sin of 30 degrees.
In this question, we need to find the trigonometric values of two special angles; they’re cos of 45 degrees and sin of 30 degrees. And luckily, we have a nice way to help us recall these. We draw a table, and on the left-hand side, we write sin 𝜃, cos 𝜃, and tan 𝜃. And on the top, we write 30 degrees, 45, and 60. The first two rows are really straightforward. We write one, two, three, then three, two, one. We then turn these all into a fraction with a denominator of two. And then the final step is to take the square root of all of our numerators.
Now, of course, the square root of one is simply one. So, we really don’t need the square root here. And we find that sin of 30 degrees and cos of 60 degrees are one-half. Tan of 𝜃 is not so straightforward. To find the values for tan of 𝜃, we divide the values for sin of 𝜃 by the values for cos of 𝜃. So, tan of 30 would be one-half divided by root three over two. But since the denominators are the same, we can simply divide the numerators. And we find that tan of 30 is one over root three. Tan of 45 is root two over root two, which is equal to one. And then, tan of 60 is root three over one. And of course we really don’t need that denominator.
So, we now know the values for sin, cos, and tan of 30 degrees, 45, and 60. We’ll begin then by finding cos of 45 degrees in the table. We can see quite clearly is the square root of two over two. And we can also find sin of 30. It’s one-half. This means that two cos of 45 times sin of 30 is two times the square root of two over two times one-half. And then since multiplication is communicative and we know that two times one-half is simply one, we find that two cos of 45 times sin of 30 is root two over two.