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# The Half-Angle Identities - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let's do another problem that involves a half-angle formula. Evaluate the sine of 105 degrees using a half-angle formula.

Now remember the half-angle formula for cosine is cosine of theta over 2 equals plus or minus the square root of 1 plus cosine theta over 2. Now here if we're evaluating cosine of 105, that's going to be theta over 2, what's theta? Well it has to be 210 degrees. So we have 1 plus the cosine of 210 degrees all over 2. Now what's the cosine of 210? Well let's draw the unit circle, 210 degrees is 30 degrees past 180, so about here.

Now there are a couple of ways we can go about finding the cosine of 210, but one way to do it is to recognize that the reference angle here is 30 degrees, so we compare to the cosine of 30 degrees. Now recall how this works, you have to make an adjustment for cosine of 210 degrees depending on which quadrant you're in, cosine of 210 degrees should be negative and since cosine of 30 is positive, we need to put a minus sign in front, so this is going to equal minus cosine 30. Now cosine 30 is root 3 over 2 rather and so that's our answer for cosine 210 and we plug in here. So plus or minus 1 minus root 3 over 2 all over 2.

Now we need to simplify this and the best way to do that is to multiply this by a fancy form of 1, 2 over 2 that gets rid of this little denominator here and so let's see what happens. Plus or minus now the 2 distributes over these two terms, 2 times 1 is 2, 2 times root 3 over 2 is root 3 and 2 times 2 in the bottom is 4. I don't need quite that much radical and then we've got plus or minus root 2 minus root 3 all over 2 and now we are with the plus or minus because our answer could be positive or negative.

We're dealing with 105 degrees, just really quickly 105 degrees is 15 degrees past 90 right about here. Now we're in the second quadrant cosine is negative here, so I need minus root 2 minus root 3 all over 2.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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