Need some help

[sweetthang]

I am just having such a hard time, can someone please help me with a few of these problems.

1. 3cos²(π/5) + 3sin²(π/5)... with this I dont even know where to start.

2. Sinplify the expression: (1-sin²x)(1+tan²x)... I can foil it to 1+tan²-sin²-tan²sin² then to 1+ sin/cos - sin² - tan²sin²
but then I get stuck and I dont know what to do after

Lastly, finding all solutions in the interval [0,2π) of the equation -2sin2a=1.

Im getting behind in my class and these are homework questions that I think if I know how to answer these then I should be ok with others. Thanks so much if anyone can help!

 

[galactus]

I am just having such a hard time, can someone please help me with a few of these problems.

1. 3cos²(π/5) + 3sin²(π/5)... with this I dont even know where to start.
Notice the important identity \(\displaystyle cos^{2}(x)+sin^{2}(x)=1\)

2. Sinplify the expression: (1-sin²x)(1+tan²x)... I can foil it to 1+tan²-sin²-tan²sin² then to 1+ sin/cos - sin² - tan²sin²
but then I get stuck and I dont know what to do after

Lastly, finding all solutions in the interval [0,2π) of the equation -2sin2a=1.

Im getting behind in my class and these are homework questions that I think if I know how to answer these then I should be ok with others. Thanks so much if anyone can help!

 

[Unco]

1. 3cos²(π/5) + 3sin²(π/5)... with this I dont even know where to start.
Hint: acos²x + asin²x = a(cos²x + sin²x) = a(1) = 1.

2. Sinplify the expression: (1-sin²x)(1+tan²x)... I can foil it to 1+tan²-sin²-tan²sin² then to 1+ sin/cos - sin² - tan²sin²
but then I get stuck and I dont know what to do after
As above, cos²(x) + sin²x =1 <-> 1 - sin²x = cos²x
1+tan²x = sec²x

Lastly, finding all solutions in the interval [0,2π) of the equation -2sin2a=1.


\(\displaystyle \sin{2a} = -\frac{1}{2}\)

Which method have you used in class? Formulae or c.a.s.t or ? (Just so we can be of better help.)

Edit: Sorry, Galactus!

 

[sweetthang]

Thank you so much I think Im starting to actually understand it. My problem is Im taking this class online, bad mistake. Im getting a B, and usually people are fine with that, but I really want to understand what Im doing.

Thanks again.

 

[sweetthang]

Ok so another question, it was shown up above the cos + sin = 1, so what happens with you subtract?

cos²(5a)-sin²(5a)

Or when you divide:

cos2a/2sin(a)cos(a)

Thanks for any help.

 

[stapel]

...it was shown up above the cos + sin = 1...

Sine of what? Cosine of what?

(Since sin(x) + cos(x) does not, in general, equal 1, you will need to specify what angles you are using. It might help if you spelled out where you saw this; I cannot find it "above".)

Thank you.

Eliz.

 

[sweetthang]

im sorry it was shown up above in the first reply.. i didnt right out the angle i apologize.

 

[stapel]

it was shown up above in the first reply.

In the first reply, I only see "cos²(x) + sin²(x) = 1", which is a very different thing, being the Pythagorean Identity (probably one of the first identities you learned in trigonometry). The subtraction identity related to "cos²(x) - sin²(x)" is, I believe, one of the double-angle identities. Check your trig book.

Eliz.