# Wrong answers on test, need help

• July 19th 2009, 10:33 PM
pinnacle2009
Wrong answers on test, need help
Here are a few problems I missed on the test. The professor doesn't show where you missed the problem or anything. Would like to know how to work these:

1) Solve for x: $tan(pi(x)/2)=0$

2) Solve for x: $1-2cos(x)=0$

3) Calculate (give exact answers. i.e. no decimals)
a: $sin(105)$
b: $cos(7pi/12)$

4)Solve for x: $2sin^2x=1+cos^2x$

Thanks for any help!
• July 19th 2009, 11:08 PM
DeMath
Quote:

Originally Posted by pinnacle2009
Here are a few problems I missed on the test. The professor doesn't show where you missed the problem or anything. Would like to know how to work these:

1) Solve for x: $tan(pi(x)/2)=0$

2) Solve for x: $1-2cos(x)=0$

3) Calculate (give exact answers. i.e. no decimals)
a: $sin(105)$
b: $cos(7pi/12)$

4)Solve for x: $2sin^2x=1+cos^2x$

Thanks for any help!

For 1)

$\tan \left( {\frac{\pi }{2}x} \right) = 0 \Leftrightarrow \frac{\pi }{2}x = \underbrace {\arctan 0}_0 + \pi k \Leftrightarrow x = 2k{\text{ }}\left( {k \in \mathbb{Z}} \right).$

For 2)

$1 - 2\cos x = 0 \Leftrightarrow \cos x = \frac{1}{2} \Leftrightarrow x = \pm \arccos \frac{1}{2} + 2\pi k \Leftrightarrow x = \pm \frac{\pi }{3} + 2\pi k{\text{ }}\left( {k \in \mathbb{Z}} \right).$
• July 19th 2009, 11:15 PM
red_dog
3)

$\sin 105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin 60^{\circ}\cos 45^{\circ}+\sin 45^{\circ}\cos 60^{\circ}=$

$=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{\ sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2 }}{4}$
• July 19th 2009, 11:18 PM
red_dog
3) b)

$\cos\frac{7\pi}{12}=\cos\left(\frac{\pi}{3}+\frac{ \pi}{4}\right)=$

$=\cos\frac{\pi}{3}\cos\frac{\pi}{4}-\sin\frac{\pi}{3}\sin\frac{\pi}{4}=$

$=\frac{1}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\s qrt{2}-\sqrt{6}}{4}$
• July 19th 2009, 11:22 PM
red_dog
4)

$2\sin^2x=1+\cos^2x\Rightarrow 2\sin^2x=2-\sin^2x\Rightarrow3\sin^2x=2\Rightarrow\sin^2x=\fr ac{2}{3}\Rightarrow$

$\Rightarrow\sin x=\pm\sqrt{\frac{2}{3}}\Rightarrow x=(-1)^k\arcsin\left(\pm\sqrt{\frac{2}{3}}\right)+k\pi , \ k\in\mathbb{Z}$
• July 19th 2009, 11:28 PM
pinnacle2009
Thanks very much!
• July 20th 2009, 02:28 AM
mr fantastic
Quote:

Originally Posted by pinnacle2009
Here are a few problems I missed on the test. The professor doesn't show where you missed the problem or anything.
[snip]

Then you should arrange an appointment with your professor and ask him/her.
• July 22nd 2009, 08:00 AM
pinnacle2009
Somehow 2 and 4 he still says are wrong. I think on 4 he wanted us to use the power reducing formulas and on 2, I have no idea, unless he wanted us to use another formula.

I tried working 4 using the power reducing formula and am still having trouble.

Could someone look at these problems again please? (Speechless)

I may be taking pre-calculus again lol (Thinking)
• July 22nd 2009, 03:23 PM
yeongil
Quote:

Originally Posted by pinnacle2009
Somehow 2 and 4 he still says are wrong. I think on 4 he wanted us to use the power reducing formulas and on 2, I have no idea, unless he wanted us to use another formula.

I tried working 4 using the power reducing formula and am still having trouble.

Could someone look at these problems again please? (Speechless)

I may be taking pre-calculus again lol (Thinking)

There's no formula to use in #2, per se. You have to recognize the trig functions of special angles. The answer provided by DeMath is correct, but you could also state your answers as
$x = \frac{\pi}{3} + 2\pi k$ and
$x = \frac{5\pi}{3} + 2\pi k$, where k is an integer.
(Remember that $\frac{5\pi}{3}$ is coterminal to $-\frac{\pi}{3}$).

If the answers are supposed to be in $0 \le x < 2\pi$, then shame on you for not stating that in the first place. The answers would then be $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

01
• July 22nd 2009, 03:39 PM
yeongil
Quote:

Originally Posted by pinnacle2009
4)Solve for x: $2sin^2x=1+cos^2x$

Use the power-reducing identities? Ok.

\begin{aligned}
2\sin^2 x &= 1 + \cos^2 x \\
2\left(\frac{1 - \cos 2x}{2}\right) &= 1 + \frac{1 + \cos 2x}{2} \\
1 - \cos 2x &= 1 + \frac{1 + \cos 2x}{2} \\
2 - 2\cos 2x &= 2 + 1 + \cos 2x \\
3\cos 2x &= -1 \\
\cos 2x &= -\frac{1}{3}
\end{aligned}

Does the professor want answers approximated in decimal? Then,
$2x \approx 1.911 + 2\pi k\;\Rightarrow\;x \approx 0.955 + k\pi$
$2x \approx 4.372 + 2\pi k\;\Rightarrow\;x \approx 2.186 + k\pi$
$2x \approx 8.194 + 2\pi k\;\Rightarrow\;x \approx 4.097 + k\pi$
$2x \approx 10.656 + 2\pi k\;\Rightarrow\;x \approx 5.328 + k\pi$

01
• July 22nd 2009, 10:14 PM
pinnacle2009
Quote:

Originally Posted by yeongil
There's no formula to use in #2, per se. You have to recognize the trig functions of special angles. The answer provided by DeMath is correct, but you could also state your answers as
$x = \frac{\pi}{3} + 2\pi k$ and
$x = \frac{5\pi}{3} + 2\pi k$, where k is an integer.
(Remember that $\frac{5\pi}{3}$ is coterminal to $-\frac{\pi}{3}$).

If the answers are supposed to be in $0 \le x < 2\pi$, then shame on you for not stating that in the first place. The answers would then be $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

01

No range was stated on the test.

Thanks Yeongil. I had a friend work this one for me (she's a math major with the same professor). She got that same answer.