Hey everyone, in this problem we're asked to use our double angle formulas in order to verify the identity. The sine of 2θ over the tangent of θ is equal to the cosine of 2θ plus 1. Now you should go ahead and try this problem on your own before checking back in with me, but let's go ahead and get started. Now remember when verifying an identity, you're going to want to start with your more complicated side and it's not always going to be really obvious which side is more complicated. So remember that if you really don't know, starting with either side is really fine. Now here, I'm going to start with this left side because it has a fraction in it. So let's go ahead and get started here using our simplifying strategies. Now remember that we want to be constantly scanning for identities and the first thing I notice here is that I'm taking the sine of 2θ, which tells me that I can use my double angle formula for sine to rewrite this as 2 sine θ cosine θ. Now I'm going to keep my denominator the same for now the tangent of θ. Now I can't really do anything yet here but remember one of our strategies is to break down everything in terms of sine and cosine. So let's go ahead and break down this tangent in terms of sine and cosine because everything else already is. I'm gonna keep my numerator the same for now again, 2 sine θ cosine θ, and then I'm going to rewrite that tangent as the sine of θ over the cosine of θ. Now, where do we go from here? Well, remember when we're dividing by a fraction, we're really just multiplying by the reciprocal. So I can think of this as 2 sine θ cosine θ times the reciprocal of that fraction on the bottom, cosine θ over sine θ. Now from here those sine θs will cancel out and I'm left with 2 times the cosine squared of θ since those cosines are multiplying each other. Now where do I go from here? Because remember our goal is to make this side equal to the cosine of 2θ plus 1. Now remember we're scanning for identities and here I see that I have the 2 times the cosine squared of θ, which I recognize as a part of one of my cosine double angle formulas. So let's go ahead and use that here. Now the cosine of 2θ, if I go ahead and add 1 to both sides here, I'm left with exactly the expression that I have over here. So I can replace this with the cosine of 2θ plus 1. Now let's remember what that right side of our equation is. It is the cosine of 2θ plus 1. So from here I'm done and I have fully verified this identity. The cosine of 2θ plus 1 is equal to the cosine of 2θ plus 1. Now we're done here. Please let me know if you have any questions. Thanks for watching.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
6. Trigonometric Identities and More Equations
Double Angle Identities
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Double Angle Identities practice set
