Expand sine of 𝜃 plus 𝜋 by six using the sum and product formulae.
Let’s recall the compound angle formula for sin of 𝐴 plus 𝐵. We have that sin of 𝐴 plus 𝐵 is equal to sin of 𝐴 cos of 𝐵 plus cos of 𝐴 times sin of 𝐵. Now we are trying to find sin of 𝜃 plus 𝜋 by six. So let’s let 𝐴 equal 𝜃 and 𝐵 equal 𝜋 by six. Now we can substitute these into the formula to give us, this gives us sin of 𝜃 plus 𝜋 by six is equal to sin of 𝜃 cos of 𝜋 by six plus cos of 𝜃 times sin of 𝜋 by six.
Now we can find the values of cos of 𝜋 by six and sin of 𝜋 by six. In order to do this, let’s start by drawing an equilateral triangle with sides of length two. Next, let’s cut this triangle in half vertically, creating two right triangles both containing an angle of 𝜋 by six. Now let’s draw the right triangle on the left here separately. Here we have a right triangle with a hypotenuse of length two and another side of length one. In order to find the missing length, we can use the Pythagorean theorem.
So the missing length is the square root of two squared minus one squared. And this is equivalent to the square root of four minus one. So the length of the missing side is just root three. Next, we can use SOHCAHTOA in order to find the values of cos of 𝜋 by six and sin of 𝜋 by six. Since we’re trying to find sin of an angle and cos of an angle, the two we’re interested in is SOH and CAH. These tell us that sin of an angle 𝜃 is equal to the opposite over the hypotenuse and cos of an angle 𝜃 is equal to the adjacent over the hypotenuse.
So now if we let 𝜋 by six equal 𝜃, we can label the opposite, adjacent, and hypotenuse on our triangle accordingly. So the adjacent is the side of length root three. The opposite is the side of length one. And the hypotenuse is the side length two. Now we can calculate the value of sin of 𝜋 by six. And this is equal to the opposite over the hypotenuse. So that’s one over two or one-half. And then cos of 𝜋 by six is equal to the adjacent over the hypotenuse. So that’s equal to root three over two.
Now we have the values sin of 𝜋 by six and cos of 𝜋 by six. And we can substitute them back into our original equation. And we get that sin of 𝜃 plus 𝜋 by six is equal to root three over two sin of 𝜃 plus a half cos of 𝜃. Next, we can factor out a half to get a solution of sin of 𝜃 plus 𝜋 by six is equal to a half times root three sin 𝜃 plus cos 𝜃.