Using the Double Angle Formula to Solve for Trigonometric Functions

[teme92]

Homework Statement

Given that cos(\pi/6) =\sqrt{}3/2, use the double angle formula for the cosine function to find cos(\pi/12) and sin(\pi/12) explicitly.

Homework Equations

cos(2x)=cos²x - sin²x
cos²x + sin²x = 1

The Attempt at a Solution

So it wants me to find cos(\pi/12) which is half the angle of cos(\pi/6). So I called these cosx and cos 2x.

I then said \sqrt{}3/2 = cos²x - sin²x

I used cos²x + sin²x = 1 and got sin²x on its own and subbed into the first formula and then got cosx on its own.

For sin(\pi/12) I subbed in sin²x = 1- cos²x and got sinx on its own.

Is this the correct method for finding the answers?

The inverse of cosx and sinx were \pi/12 so I assume I am but not sure. I'd be thankful to anyone who could clear this up.

 

[HallsofIvy]

Yes, cos(2x)= cos^2(x)- sin^2(x) so that cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1 and cos(2x)= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x). Set cos(2x) equal to \sqrt{3}/2 and solve for sin(x) and cos(x). Of course, you take the positive root.

I am not clear why you would question your reasoning.

 

[LCKurtz]

Homework Statement

Given that cos(\pi/6) =\sqrt{}3/2, use the double angle formula for the cosine function to find cos(\pi/12) and sin(\pi/12) explicitly.

Homework Equations

cos(2x)=cos²x - sin²x
cos²x + sin²x = 1

The Attempt at a Solution

So it wants me to find cos(\pi/12) which is half the angle of cos(\pi/6). So I called these cosx and cos 2x.

I then said \sqrt{3}/2 = cos²x - sin²x

I used cos²x + sin²x = 1 and got sin²x on its own and subbed into the first formula and then got cosx on its own.

For sin(\pi/12) I subbed in sin²x = 1- cos²x and got sinx on its own.

Is this the correct method for finding the answers?

The inverse of cosx and sinx were \pi/12 so I assume I am but not sure. I'd be thankful to anyone who could clear this up.

That looks correct. Did you get exact radical values? You have$$
\cos(2x) = \cos^2x -\sin^2 x = 2\cos^2x - 1 = 1-2\sin^2 x$$You are just using the last two equations to solve for $\cos x$ and $\sin x$ in terms of $\cos(2x)$.

 

[teme92]

Thanks guys! HallsofIvy its just sometimes when I do these questions I think I'm right and then I only get it partly correct or not correct at all. I was just wanting to be sure as these type of questions may come up in my finals.

 

[xiavatar]

@teme92. You have to get into the habit of not doubting your reasoning. Yes, self-criticism will help you improve your abilities at problem solving, but when it gets in the way of you being confident in your answers it can be a problem.

 

[teme92]

Hey xiavatar, when it comes down to exams I will go with my instincts unquestionably but I just want to be safe in the run up to them. As you said, self-criticism has improved my understanding of a lot of topics in mathematics so I'd prefer to be safe than sorry in this instant :)