Trig. (need help with 5 problems)

[Guest]

1.) Verify the identity tan(x) + cot(y) / tan(x) * cot(y) = tan(y) + cot(x)

2.) Rewrite the expression sin^2(x) * cos^2(x) in terms of the first power of cosine.
This is what I did when I attempted it:
(1 - cos(2x))/2 * (1+cos(2x))/2 =
1 - cos^2(2x)/4 =
1/4 (1 - cos^2(2x)) =
1/4(1 - 1 + cos(4x)/2) =
1/4 - 1/8 + 1/8cos(4x) =
Then following the example in the book, I guess it ends at:
1/8(3 + cos(4x)
If this is correct, why 3? (1/8)*3 doesn't equal 1/4...

3.) Find all solutions to the equation 2sin^2(x) = 2 + cos(x) in the interval [0,2pie)
I attempted it and did this:
2sin^2(x) - 2 = cos(x)
-2(1 - sin^2(x)) = cos(x)
-2(cos^2(x)) = cos(x)
I'm stuck at this point.

4.) Use a difference identity to find the exact value of tan(15)
I did this:
tan(u-v) = tan(u) - tan(v) / 1+tan(u)*tan(v)
tan(60-45) = tan(60) - tan(45) / 1+tan(60)*tan(45)
= sqrt(3)-1 / 1+sqrt(3)
Then I did this:
(sqrt(3)-1)/1 * 1/1+sqrt(3) =
sqrt(3) + 3 - 1 - sqrt(3) = 2
tan(15)=2? My calculator disagrees.

5.) Use a half angle identity to find the exact value of sin(u/2) given that sec(u) = -5/3 and pie/2 < u < pie

:?

 

[galactus]

1.) Verify the identity tan(x) + cot(y) / tan(x) * cot(y) = tan(y) + cot(x)
Use \(\displaystyle \frac{sin(x)}{cos(x)}=tan(x)\) and \(\displaystyle \frac{cos(x)}{sin(x)}=cot(x)\)

2.) Rewrite the expression sin^2(x) * cos^2(x) in terms of the first power of cosine.
This is what I did when I attempted it:
(1 - cos(2x))/2 * (1+cos(2x))/2 =
1 - cos^2(2x)/4=
1/4 (1 - cos^2(2x)) =
1/4(1 - 1 + cos(4x)/2) =\(\displaystyle \frac{(1-cos(4x))}{8}\)
1/4 - 1/8 + 1/8cos(4x) =
Then following the example in the book, I guess it ends at:
1/8(3 + cos(4x)
If this is correct, why 3? (1/8)*3 doesn't equal 1/4...

3.) Find all solutions to the equation 2sin^2(x) = 2 + cos(x) in the interval [0,2pi)
I attempted it and did this:
2sin^2(x) - 2 = cos(x)
-2(1 - sin^2(x)) = cos(x)
-2(cos^2(x)) = cos(x)
I'm stuck at this point.

\(\displaystyle 2sin^{2}x-cosx-2=0\)

\(\displaystyle 2(1-cos^{2}x)-cosx-2=0\)

\(\displaystyle 2-2cos^{2}x-cosx-2=0\)


Multiply by -1:
\(\displaystyle 2cos^{2}x+cosx=0\)

Factor:

\(\displaystyle cosx(2cosx+1)=0\);

\(\displaystyle cosx=0\) and \(\displaystyle cosx=\frac{-1}{2}\)

Since cosine has period \(\displaystyle 2{\pi}\), we can find all solutions by adding

multiples of\(\displaystyle 2{\pi}\) that are in the interval \(\displaystyle [0,2{\pi}]\).

if \(\displaystyle cosx=\frac{-1}{2}\), then x=\(\displaystyle \frac{2{\pi}}{3}\) or

\(\displaystyle x=2{\pi}-\frac{2{\pi}}{3}=\frac{4{\pi}}{3}\)

Also, \(\displaystyle cosx=0\), then \(\displaystyle x=\frac{\pi}{2}\) or \(\displaystyle \frac{3{\pi}}{2}\)


4.) Use a difference identity to find the exact value of tan(15)
I did this:
tan(u-v) = tan(u) - tan(v) / 1+tan(u)*tan(v)
tan(60-45) = (tan(60) - tan(45)) / (1+tan(60)*tan(45))This is correct
= sqrt(3)-1 / 1+sqrt(3) Something ain't right.
Then I did this:
(sqrt(3)-1)/1 * 1/1+sqrt(3) .
sqrt(3) + 3 - 1 - sqrt(3) = 2
tan(15)=? My calculator disagrees.

5.) Use a half angle identity to find the exact value of sin(u/2) given that sec(u) = -5/3 and pie/2 < u < pie What kind of pie?. Apple?. Cherry?.

:?

 

[soroban]

Hello, bloodelf_ella!

1.) Verify the identity \(\displaystyle \frac{\tan(x)\,+\,\cot(y)}{\tan(x)\cdot\cot(y)}\;=\;\tan(y)\,+\,\cot(x)\)

We have: .\(\displaystyle \frac{\tan(x)}{\tan(x)\cot(y)}\,+\,\frac{\cot(y)}{\tan(x)\cot(y)} \;=\;\frac{1}{\cot(y)}\,+\,\frac{1}{\tan(x)}\;=\;\tan(y)\,+\,\cot(x)\)