# tan=sec? test questions

1. If $\theta$ is in quadrant 1 and $\tan(\theta) = .6$ then $\sec(\theta) =$?
This seems pretty easy to me:
$\tan^2(\theta)-\sec^2(\theta)=1$
$-\sec^2(\theta)=.64$
$\sec(\theta)=8$

2. Another one, $\cos(\theta)=\sin(2\theta)$
Should that be $\cos^2(\theta)+\sin^2(\theta)=1/2$? Which would be .5

3. For all angles $\theta$, $\cos(-\theta)$ = $\cos(\theta)$
I said false how can I negative angle be equal to a positive one?

4. If $\sin^2(\theta)= .5$ then $(\sin(\theta))^2 = (\cos(\theta))^2$
I said false because $\sin$ and $\cos$ can never equal .5 together but it was wrong.
This answer is wrong but I don't understand why.

I am seriously considering changing majors, I know most people here will think I am an idiot, lazy or whatever else for taking such simple high school level math courses in college but I really am having trouble with it. I keep making mistakes no matter what I do I can never get better than a D or C on a test. So I am trying to evaluate the mistakes I made on the test so I don't make them again, but inevitably I will.

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$\cos(-45^\circ) = 1/\sqrt{2}$ and $\cos(45^\circ) = 1/\sqrt{2}$ . It doesn't mean $45 = -45$. –  GEdgar Jun 10 '11 at 21:42
Is that because the reference angle will be positive? –  Adam Jun 10 '11 at 21:46
The more effort you put into posting your question and making it easy to read, the more effort the reader is willing to put into answering your questions or giving advice. –  JavaMan Jun 10 '11 at 21:48
I've converted your equations to LaTeX. Apologies if I changed your intended meaning, feel free to change it if I did. –  Zev Chonoles Jun 10 '11 at 22:30
Thank you, that stuff is very hard to figure out. –  Adam Jun 10 '11 at 22:34

You need to read the questions more carefully. Checking the typing would make them easier to read, as there are many typos. Setting them in $\LaTeX$ (see the FAQ) would help a lot, too.

For the first, the identity is $1+tan^2\theta=sec^2\theta$. Your first equality is incorrect as $\tan^2\theta \neq 1$. The dash after $1$ looks like a minus sign, but it seems to be a separator. How about line feeds? Then you lost a decimal point at the end.

For the second, what is sin2theta? $\sin^2 \theta$ or $\sin 2\theta$?

Please review the question and make it legible.

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How do I format all that stuff? –  Adam Jun 10 '11 at 22:15
You enclose $\LaTeX$ in dollar signs. There are tutorials on the web for how to write $\LaTeX$. For anything on this site, you can right-click and choose Show Source to see how it was done. –  Ross Millikan Jun 10 '11 at 22:21
@Adam: math.harvard.edu/texman –  John Jun 10 '11 at 22:21

(2) If the problem asked you for the angle $\theta$ at which $\cos \theta = \sin(2\theta)$,

then think about what you know about a 30, 60, 90 right triangle: $$\cos(30^\circ) = \sin(2\cdot 30^\circ) = \sin(60^\circ) = \frac {\sqrt{3}}{2}$$

So theta would be $30^\circ$.

(4) If $\sin^2 \theta = .5$ then $\sin^2 \theta = \cos^2 \theta$.

This is true, since if $\sin^2 \theta = .5 = \frac 12$, then we know that $$\sin \theta = +- \sqrt {\frac 12} = +- \frac {\sqrt 2}{2} = +- \frac {1}{\sqrt 2} = +- \cos \theta$$ (That means the reference angle is $45^\circ$.)

Hence $\cos^2 \theta = \sin^2 \theta = 0.5$.

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Draw a triangle in the first quadrant with $\tan(\theta) = 3/5$. It will have height 3, base 5 and hypoteneuse $\sqrt{34}$. When I compute $\sec(\theta)$ I get $\sqrt{34}/5$. Drawing the appropriate picture makes this worlds simpler.

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How would I know to use 3/5? –  Adam Jun 10 '11 at 21:50
$.6=\frac{3}{5}$ –  Joe Johnson 126 Jun 10 '11 at 21:53
How do I calculate that? –  Adam Jun 10 '11 at 21:56
Adam, 3/5 = 0.60. Check it by dividing 3 by 5. Or, you can think of 0.6 as 6/10, which reduces to 3/5. –  amWhy Jun 10 '11 at 21:59
In exams, it will usually be some simple fraction. Otherwise, you can use the Pythagorean theorem or the $\arctan$ button on your calculator. –  Ross Millikan Jun 10 '11 at 22:23

# part 1:

Identity is $$\sec^2\theta -\tan^2\theta =1$$ $$\sec^2\theta -0.36 =1$$ $$\sec\theta =\sqrt{1.36}\implies\sec\theta=1.1662$$

# part 2:

$$\cos \theta=\sin2\theta\implies\cos\theta=\cos\left(\dfrac\pi2-2\theta\right)\implies\theta=\dfrac\pi2-2\theta\implies\theta=\dfrac\pi6$$ $$\cos^2(\dfrac \pi6)+\sin^2(\dfrac \pi6)=1$$

# part3:

$$\cos(-\theta)\implies\cos(0^\circ-\theta)\implies\cos 0^\circ\cdot\cos\theta+\sin 0^\circ\cdot\sin\theta\implies\cos\theta$$

so: $$\cos(-\theta)=\cos\theta$$

# part4:

if $\sin^2\theta=\dfrac12\implies\sin^2\theta=\left(\dfrac{1}{\sqrt{2}}\right)^2\implies\sin^2\theta=\sin^2\left(\dfrac{\pi}{4}\right)$

if

$\sin^2\theta=\sin^2\alpha$ then general solution of $\theta=n\pi\pm\alpha\;\;,n \in Z$.

so:

$$\theta=n\pi\pm\dfrac{\pi}{4}\;\;,n \in Z$$

so there are not $\sin\theta=\cos\theta=0.5\;\;while\;\;\sin^2\theta=\cos^2\theta=0.5$

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