Evaluate cos of 260 times cos of 70 minus sin of 260 times sin of 70.
First, let’s recall the cos compound angle formula for cos of 𝐴 plus 𝐵. We have that cos of 𝐴 plus 𝐵 is equal to cos of 𝐴 times cos of 𝐵 minus sin of 𝐴 times sin of 𝐵. Here, we notice that the right-hand side of this equation is very similar to the expression given in the question. And in order to make the right-hand side of our new equation the same as the expression in the question, we can simply set 𝐴 equal to 260 and 𝐵 equal to 70.
Now let’s substitute these values into the equation. This gives us that cos of 260 plus 70 is equal to cos of 260 times cos of 70 minus sin of 260 times sin of 70. And we can rewrite the left-hand side of this equation as cos of 330. Now 330 is not a nice angle to work with. So let’s find another angle, 𝜃, such that cos of 𝜃 is equal to cos of 330. In order to do this, let’s draw a cos graph.
Here we have a graph of 𝑦 equals cos 𝜃. Now let’s mark on cos of 330. Here, the horizontal dashed line represents 𝑦 is equal to cos of 330. Now any other point where this line crosses the 𝑦 equals cos 𝜃 line will give us a value of 𝜃 such that cos of 𝜃 is equal to cos of 330. We can see that the two lines cross at some point between zero and 90. Now in order to find the value of this point, we can use the fact that the graph is symmetric about 180 degrees.
And what this means is that the distance from 360 to 330 will be the same as the distance from zero to the point we’re trying to find. Now the distance from 360 to 330 is 30. So therefore, the distance from zero to the point we’re trying to find will also be 30. That means that this point is 30 degrees. And what this tells us is that cos of 330 is equal to cos of 30. And now since cos of 330 is equal to the expression we’re trying to evaluate, we can write that cos of 30 is equal to cos of 260 times cos of 70 minus sin of 260 times sin of 70.
And so all that remains is to find the value of cos of 30. In order to find this value, let’s draw an equilateral triangle with sides of length two. Now let’s cut the triangle in half vertically. So now, we have a right triangle with an angle of 30 degrees. Let’s draw this triangle out separately. Now since we know the length of the hypotenuse and one of the missing sides, we can find the missing length using the Pythagorean theorem. This gives us a length of root three.
In order to find the value of cos of 30, we can use SOHCAHTOA. And since we’re trying to find cos of the angle, we’ll use CAH. This tells us that cos of 𝜃 is equal to the adjacent over the hypotenuse. Now on our triangle, the angle we’re interested in is 30 degrees. So let’s call that angle 𝜃. And now we can label the opposite, adjacent, and hypotenuse of the triangle with respect to this angle 𝜃.
So the side of length root three is the adjacent. The side of length one is the opposite. And the side of length two is the hypotenuse. Now we have that cos of 30 is equal to the adjacent over the hypotenuse. So that’s root three over two. This gives our solution of cos of 260 times cos of 70 minus sin of 260 times sin of 70 is equal to root three over two.