Trig Simplification

**rtblue** · February 4th 2011, 05:57 PM

Hey guys, I have a pretty botchy understanding of trigonomety. My class has not started trigonometry yet, but I was interested so I tried to learn some. The question I have is this:

$Evaluate: \sin(\frac{\pi}{4}) - \cos(\frac{\pi}{3})+\tan(\frac{3\pi}{4})-\csc(\frac{5\pi}{4})+\sec(\frac{4\pi}{3})$

My Attempt:

1. I see that the first term(sine of 45) will be $\displaystyle \frac{\sqrt{2}}{2}$

2. I see that the second term(cosine of 60) will be $\displaysyle\frac{1}{2}$

3. Now, I get a bit confused. I realize that the third term is the tangent of 135. I kind of took a stab at this and said that it was -tan45 (this is purely guesswork). That gave me -1.

4. And the confusion grows. I evaluate the fourth term as 1/sin(225). I'm pretty sure sin225= -sin45, and so this gave me $\displaysyle -\sqrt{2}$ .

5. So i take the fifth term as 1/cos(240), which i think is 1/-cos60. I get this to equal -2.

Once I compile all of this, I get $\displaystyle \frac{3\sqrt{2}-7}{2}$

This is one of the answer choices, but I'm not sure if I messed up somewhere. Any help is appreciated.

And one quick side question:
Is tan(x)= -tan(180-x)?

Thank you for any help.

**Prove It** · February 4th 2011, 06:38 PM

First of all, you should know that sine is positive in the first and second quadrants, cosine is positive in the first and fourth quadrants, and tangent is positive in the first and third quadrants.

Hints:

$\displaystyle \tan{\frac{3\pi}{4}} = \tan{\left(\pi - \frac{\pi}{4}\right)} = -\tan{\frac{\pi}{4}}$ since it is in the second quadrant.

$\displaystyle \csc{\frac{5\pi}{4}} = \frac{1}{\sin{\frac{5\pi}{4}}} = \frac{1}{\sin{\left(\pi + \frac{\pi}{4}\right)}} = \frac{1}{-\sin{\frac{\pi}{4}}}$ since it is in the third quadrant.

See if you can do the others...

**rtblue** · February 4th 2011, 06:55 PM

Thanks Prove It, that first sentence of yours was very informative. What are the quadrants in which cotangent, cosecant, and secant are positive/negative? Would it just be the opposite of their respective inverses?

**Prove It** · February 4th 2011, 06:59 PM

No, it's the same as their reciprocals. Since sine is positive in the first and second quadrants, so is 1/sine, etc...

**Ithaka** · February 4th 2011, 07:01 PM

Originally Posted by rtblue

Hey guys, I have a pretty botchy understanding of trigonomety. My class has not started trigonometry yet, but I was interested so I tried to learn some. The question I have is this:

$Evaluate: \sin(\frac{\pi}{4}) - \cos(\frac{\pi}{3})+\tan(\frac{3\pi}{4})-\csc(\frac{5\pi}{4})+\sec(\frac{4\pi}{3})$

My Attempt:

1. I see that the first term(sine of 45) will be $\displaystyle \frac{\sqrt{2}}{2}$

2. I see that the second term(cosine of 60) will be $\displaysyle\frac{1}{2}$

3. Now, I get a bit confused. I realize that the third term is the tangent of 135. I kind of took a stab at this and said that it was -tan45 (this is purely guesswork). That gave me -1.

4. And the confusion grows. I evaluate the fourth term as 1/sin(225). I'm pretty sure sin225= -sin45, and so this gave me $\displaysyle -\sqrt{2}$ .

5. So i take the fifth term as 1/cos(240), which i think is 1/-cos60. I get this to equal -2.

Once I compile all of this, I get $\displaystyle \frac{3\sqrt{2}-7}{2}$

This is one of the answer choices, but I'm not sure if I messed up somewhere. Any help is appreciated.

And one quick side question:
Is tan(x)= -tan(180-x)?

Thank you for any help.

These 2 are correct:

$\sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$

$\cos(\frac{\pi}{3})=\frac{1}{2}$

Then:
$\tan(\frac{3\pi}{4})=\tan(\pi-\frac{\pi}{4})$

$\tan(\pi-\frac{\pi}{4})=-\tan(\frac{\pi}{4})=-1$

$\csc(\frac{5\pi}{4})=\frac{1}{\sin(\frac{5\pi}{4}) }$

$\sin(\frac{5\pi}{4})=\sin(\pi+\frac{\pi}{4})=-\frac{\sqrt2}{2}$

$\csc(\frac{5\pi}{4})=-\frac{2}{\sqrt2}$

$\sec(\frac{4\pi}{3})=\frac{1}{\cos(\frac{4\pi}{3}) }$

$\cos(\frac{4\pi}{3})=\cos(\pi+\frac{\pi}{3})=-\frac{1}{2}$

$\sec(\frac{4\pi}{3})=-\frac{2}{1}=-2$

Replacing this results, you get:

$\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{3})+\tan(\frac{3\pi}{4})-\csc(\frac{5\pi}{4})+\sec(\frac{4\pi}{3})$

$=\frac{\sqrt2}{2} -\frac{1}{2}+ (-1) -(-\frac{2}{\sqrt2}) +(-2)$

$=\frac{\sqrt2}{2} -\frac{7}{2}+\sqrt2$

$=\frac{\sqrt2-7+2\sqrt2}{2}$

$=\frac{3\sqrt2-7}{2}$

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