Need some help with advanced Trigonometry Problems

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ursoweird123

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Jul 8, 2010
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I have a few questions that I need some help with. If you know how to do some, that's great; if you know how to do them all, that's even better. Show work please.

1) Find all [theta] in the interval [0,2[pi]) that satisfy the equation. sin2[theta] = 0.
I've got 2sin[theta]cos[theta] = 0, but I don't know what to do next.

2) Write the expression sqrt(x^2 + 4) in terms of [theta] when x = 2tan[theta].
I have no idea where to begin.

3) Find sin2A if sinA = 1/4 and 0<=A<=[pi]/2.
Once again, I don't know where to begin.

4) Find all [theta] in the interval [0,2[pi]) that satisfy the equation.
2cos[theta]tan[theta] + tan[theta] = 0.
I assume you subtract tan[theta] from both sides to make it 2cos[theta]tan[theta] = -tan[theta], but I don't know what to do next.

5) If cos2[theta] = 1/3 and 0<=2[theta]<=[pi], find cos[theta].
I got cos[theta] = sqrt(6) / 3, but I don't think that is right.

6) Rewrite the given equation using the substitutes x = rcos[theta] and y = rsin[theta]. Simplify your answer. x^2 + y^2 + 3x = 0.
I got (rcos[theta])^2 + (rsin[theta])^2 + 3(rcos[theta]) = 0, but I don't know what to do next.

7) Write the given expression in algebraic form. tan(arccos(x/3)). arccos = the inverse of cosine (cos^-1).

8) Compute arcsin(-1/2). arcsin = the inverse of sine (sin^-1).

Thank you in advance for your help. Hopefully you can help me with a few of these. We cannot use calculators.
 
You've posted an inordinate number of problems, but I can lend a hand on 2 of them.


2) Write the expression sqrt(x^2 + 4) in terms of [theta] when x = 2tan[theta]
Click to expand...
.

Just sub in what they give you, \(\displaystyle x=2tan{\theta}\)

\(\displaystyle \sqrt{(2tan{\theta})^{2}+4}=\sqrt{4tan^{2}{\theta}+4}=\sqrt{4(tan^{2}{\theta}+1)}\)

Now, finish?. Remember the identity \(\displaystyle tan^{2}{\theta}+1=sec^{2}{\theta}\)


6) Rewrite the given equation using the substitutes x = rcos[theta] and y = rsin[theta]. Simplify your answer. x^2 + y^2 + 3x = 0.
I got (rcos[theta])^2 + (rsin[theta])^2 + 3(rcos[theta]) = 0, but I don't know what to do next.
Click to expand...

Begin by factoring out the r^2. Then, you have \(\displaystyle r^{2}(cos^{2}{\theta}+sin^{2}{\theta})\). Note the identity in the parentheses. Look familiar?.
 
ursoweird123 said:
I have a few questions that I need some help with. If you know how to do some, that's great; if you know how to do them all, that's even better. Show work please.

1) Find all [theta] in the interval [0,2[pi]) that satisfy the equation. sin2[theta] = 0.

At what value/s of 'x' ? sin(x) = 0

I've got 2sin[theta]cos[theta] = 0, but I don't know what to do next.

2) Write the expression sqrt(x^2 + 4) in terms of [theta] when x = 2tan[theta].
start with
x[sup:303p1vi0]2[/sup:303p1vi0] = [ 2tan(?) ][sup:303p1vi0]2[/sup:303p1vi0] = 4tan[sup:303p1vi0]2[/sup:303p1vi0](?)
I have no idea where to begin.

3) Find sin2A if sinA = 1/4 and 0<=A<=[pi]/2.

sin(2A) = 2 sin(A)*cos(A)

cos(A) = ?[1 - sin[sup:303p1vi0]2[/sup:303p1vi0](A)]

Once again, I don't know where to begin.

4) Find all [theta] in the interval [0,2[pi]) that satisfy the equation.
2cos[theta]tan[theta] + tan[theta] = 0.
I assume you subtract tan[theta] from both sides to make it 2cos[theta]tan[theta] = -tan[theta], but I don't know what to do next.

5) If cos2[theta] = 1/3 and 0<=2[theta]<=[pi], find cos[theta].
I got cos[theta] = sqrt(6) / 3, but I don't think that is right.

6) Rewrite the given equation using the substitutes x = rcos[theta] and y = rsin[theta]. Simplify your answer. x^2 + y^2 + 3x = 0.
I got (rcos[theta])^2 + (rsin[theta])^2 + 3(rcos[theta]) = 0, but I don't know what to do next.

7) Write the given expression in algebraic form. tan(arccos(x/3)). arccos = the inverse of cosine (cos^-1).

8) Compute arcsin(-1/2). arcsin = the inverse of sine (sin^-1).

Thank you in advance for your help. Hopefully you can help me with a few of these. We cannot use calculators.
Click to expand...

Looks like you have problem with algebra - you need to review that.
 
Thanks for your help :D! It's summer work for my Calculus class and I took Pre-Calculus 1st Semester last year so I have kind of forgotten how to do a lot of this stuff. The only ones I have left are 4 and 7. I think I have 7...is it sqrt(9 - x^2)?
 
ursoweird123 said:
Now all I need to know is 2cos[theta]tan[theta] = -tan[theta].
Click to expand...
HINT: divide both sides by tan[theta] : 2cos[theta] = -1
 
\(\displaystyle 7) \ tan[arccos(x/3)] \ = \ ?\)

\(\displaystyle Let \ arccos(x/3) \ = \ \theta, \ then \ cos(\theta) \ = \ x/3\)

\(\displaystyle This \ implies \ that \ tan(\theta) \ = \ \frac{\sqrt{9-x^2}}{x} \ = \ tan[arccos(x/3)]\)
 
\(\displaystyle Basically \ a \ good \ knowledge \ of \ trig. \ identities \ and \ a \ little \ algebra \ will \ do \ the \ trick.\)

\(\displaystyle 3) \ sin(2A) \ = \ sin(A+A) \ = \ sin(A)cos(A)+cos(A)sin(A) \ = \ 2sin(A)cos(A)\)

\(\displaystyle Now, \ we \ are \ given \ that \ sin(A) \ = \ 1/4, \ hence \ by \ Pythagoras, \ cos(A) \ = \ \sqrt{15}/4\)

\(\displaystyle Ergo, \ sin(2A) \ = \ (2)(1/4)(\sqrt{15}/4) \ = \ \sqrt{15}/8\)

\(\displaystyle What \ is \ sin(3A)?\)
 
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