# Proving Trig identities Problems that need correction

• Mar 24th 2011, 01:52 PM
MajorJohnson
Proving Trig identities Problems that need correction
Here are some more problems that need checking from my other thread I made.

1) $\displaystyle sin x * tan x + cos x$

$\displaystyle sin x * (sin x ) / (cos x) + cos x$

$\displaystyle (sin x ^ 2) / (cos x) + cos x = Answer: sin x ^ 2$

2) $\displaystyle (1)/ (1 + cos x) + (1 ) / (1 - cos x) =$

$\displaystyle (1 - cos x) / (1 + cos x) + (1 + cos x) / (1 - cos x = Answer: (2)/(2) = 1$
)

For this problem, does it matter if the right side's answer is the same value on the left side but flipped?

3) $\displaystyle sin t * tan t = 1 - cos ^ 2 t / cos t$

Doing the right side:
$\displaystyle sin ^ 2 t/ cos t$

$\displaystyle (sin t) / (cos t) * sin t$
$\displaystyle Answer: tan t * sec t$
• Mar 24th 2011, 02:06 PM
dwsmith
Quote:

Originally Posted by MajorJohnson
Here are some more problems that need checking from my other thread I made.

1) $\displaystyle sin x * tan x + cos x$

$\displaystyle sin x * (sin x ) / (cos x) + cos x$

$\displaystyle (sin x ^ 2) / (cos x) + cos x = Answer: sin x ^ 2$

$\displaystyle \displaystyle \frac{\sin^2(x)}{\cos(x)}+\cos(x)\cdot\frac{cos(x) }{\cos(x)}=\frac{\sin^2(x)+\cos^2(x)}{\cos(x)}=\cd ots$

2 is wrong. Try obtaining the same common denominator.

3 is correct.
• Mar 24th 2011, 02:07 PM
pickslides
What are you trying to do with these?

for the first one, I get,

$\displaystyle \displaystyle \sin x \tan x +\cos x$

$\displaystyle \displaystyle \sin x \frac{\sin x }{\cos x} +\cos x$

$\displaystyle \displaystyle \frac{\sin^2 x }{\cos x} +\cos x$

$\displaystyle \displaystyle \frac{\sin^2 x }{\cos x} +\frac{\cos^2 x}{\cos x}$

$\displaystyle \displaystyle \frac{\sin^2 x+\cos^2 x}{\cos x}$

$\displaystyle \displaystyle \frac{1}{\cos x}$

$\displaystyle \displaystyle \sec x$
• Mar 26th 2011, 06:42 AM
MajorJohnson
My objective is to find the other side.

Quote:

Originally Posted by dwsmith
$\displaystyle \displaystyle \frac{\sin^2(x)}{\cos(x)}+\cos(x)\cdot\frac{cos(x) }{\cos(x)}=\frac{\sin^2(x)+\cos^2(x)}{\cos(x)}=\cd ots$

2 is wrong. Try obtaining the same common denominator.

3 is correct.

$\displaystyle 2 / (1- cos x) (1 + cos x)$

I also want to know if my method for this problem is legal:

$\displaystyle sec^2 x * csc^2 x = sec ^ 2 x + csc ^ 2 x$

$\displaystyle (1 + tan ^ 2 x ) (1 + cot ^ 2 x)$

$\displaystyle 1 + cot ^ 2 x + tan ^ 2 x + (tan ^ 2 x) (cos ^ 2 x)$

$\displaystyle (sin x / cos x) (sin x / cos x) * (cos x / sin x ) (cos x / sin x)$

Cross multipling ( sin x / cos x) * (cos x / sin x) would give me one.

$\displaystyle 1 + cot ^ 2 x + tan ^ 2 x + 1$

Here i canceled the ones out and got ' cot ^ 2 x + tan ^ 2 x '.

$\displaystyle csc ^ 2 x - 1 + sec ^ 2 - 1$

Canceling the ones again would give me csc^ 2 x + sec^2x.
• Mar 26th 2011, 08:22 AM
Quacky
Edit: I misunderstood your approach, I think it works, other than the fact that you cancel the ones at the end. When you reach the stage $\displaystyle 1+Cot^2(x)+Tan^2(x)+1$, you can just use identities to show that it is the same as $\displaystyle Sec^2(x)+Cosec^2(x)$. I'll leave my method here, but after rereading, I think your proof is otherwise fine.

$\displaystyle Sec^2(x)Cosec^2(x)=Sec^2(x)+Cosec^2(x)$

Generally, you'd start with the side which is easiest to simplify. Here, that's the right hand side. Start by putting everything over a common denominator:

$\displaystyle \dfrac{1}{Cos^2(x)}+\dfrac{1}{Sin^2(x)}$

$\displaystyle =\dfrac{Sin^2(x)}{Sin^2(x)Cos^2(x)}+\dfrac{Cos^2(x )}{Sin^2(x)Cos^2(x)}$

$\displaystyle =\dfrac{Sin^2(x)+Cos^2(x)}{Sin^2(x)Cos^2(x)}$

$\displaystyle =\dfrac{1}{Sin^2(x)Cos^2(x)}$

$\displaystyle =Sec^2(x)Cosec^2(x)$ which is the left hand side.

You also haven't finished 2) - you can simplify the fraction further using the identity $\displaystyle Sin^2(x)+Cos^2(x)=1$
• Mar 27th 2011, 09:04 AM
MajorJohnson
Thank you. In solving these problems in general, can there be more than one possible answer?
• Mar 27th 2011, 01:07 PM
Quacky
Quote:

Originally Posted by MajorJohnson
Thank you. In solving these problems in general, can there be more than one possible answer?

Depends on the problem. If you've been asked to simplify, usually not, but if it's a proof, there will often be several approaches, as we've shown here.
• Mar 27th 2011, 01:20 PM
pickslides
Quote:

Originally Posted by MajorJohnson
Thank you. In solving these problems in general, can there be more than one possible answer?

In a lot of trig identity problems you can complete the proof using a number of different methods as the common identitiies themselves are all linked together.
• Mar 27th 2011, 04:16 PM
MajorJohnson
Okay, but even if i have to find the other side of an identity, such as with some of the other ones i had posted in my other thread, there can only be one answer right?
• Mar 31st 2011, 02:59 PM
MajorJohnson
How would i go about solving this problem?

$\displaystyle cos (PI/2 - x) sec x$

So far I have:

cosPI/2 - cos x 1/ cos x
• Mar 31st 2011, 03:14 PM
topsquark
Quote:

Originally Posted by MajorJohnson
How would i go about solving this problem?

$\displaystyle cos (PI/2 - x) sec x$

So far I have:

cosPI/2 - cos x 1/ cos x

Hint: $\displaystyle cos(a - b) = cos(a) \cdot cos(b) + sin(a) \cdot sin(b)$

-Dan
• Apr 1st 2011, 03:52 AM
MajorJohnson
Okay that helped me a little bit, I also have another question unrelated to this topic but with dealing with Solving Right triangles. When going about solving the outer sides, how do I know whether or not I should take the given outside side and times it by whatever degree given inside the triangle, as opposed to dividing the outside side by the inside? (Example $\displaystyle 3 * sin (45.3), 3 \div sin(45.3)$
• Apr 1st 2011, 04:20 AM
masters
Quote:

Originally Posted by MajorJohnson
Okay that helped me a little bit, I also have another question unrelated to this topic but with dealing with Solving Right triangles. When going about solving the outer sides, how do I know whether or not I should take the given outside side and times it by whatever degree given inside the triangle, as opposed to dividing the outside side by the inside? (Example $\displaystyle 3 * sin (45.3), 3 \div sin(45.3)$

Hi MajorJohnson,

Your question is a little vague (at least to me, it is). Might I suggest you start a new thread with your new problem and be a little more specific.

This thread is getting a bit too long.