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November 1st, 2017, 06:25 PM   #1
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Help Solving/Verifying Trig Identities

Hi! In need of some help proving several identities. I'm so confused.

1) Simplify

cos theta / sin theta ^ 3 + sin theta * cos ^ 2 theta

2) Verify:
a) 1 + tan ^ 2 / csc theta = sec theta * tan theta <=== With this problem I'm able to get either secant or tangent, but not both.

b) (tan theta + cot theta) ^ 2 = sec ^ 2 theta + csc ^ 2 theta

Thank you for any help!!!
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November 1st, 2017, 07:19 PM   #2
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Quote:
Originally Posted by Bluebird8 View Post
Hi! In need of some help proving several identities. I'm so confused.

1) Simplify

cos theta / sin theta ^ 3 + sin theta * cos ^ 2 theta
I assume you mean ...

$\dfrac{\cos{\theta}}{\sin^3{\theta} + \sin{\theta}\cos^2{\theta}}$

if that is correct, enclose the terms in the denominator with parentheses, like so ...

cos theta / (sin theta ^ 3 + sin theta * cos ^ 2 theta)

back to the expression ...

$\dfrac{\cos{\theta}}{\sin{\theta}(\sin^2{\theta}+ \cos^2{\theta})}$

what now?


Quote:
2) Verify:
a) 1 + tan ^ 2 / csc theta = sec theta * tan theta <=== With this problem I'm able to get either secant or tangent, but not both.
note $1+\tan^2{\theta} = \sec^2{\theta}$ so the left side becomes

$\dfrac{\sec^2{\theta}}{\csc{\theta}} = \dfrac{\sin{\theta}}{\cos^2{\theta}} = \dfrac{1}{\cos{\theta}} \cdot \dfrac{\sin{\theta}}{\cos{\theta}}$

... finish it

Quote:
b) (tan theta + cot theta) ^ 2 = sec ^ 2 theta + csc ^ 2 theta
expand (square) the left side ...

$\tan^2{\theta} + 2\tan{\theta}\cot{\theta} + \cot^2{\theta}$

$\tan^2{\theta} + 2 + \cot^2{\theta}$

$(\tan^2{\theta} + 1) + (1 + \cot^2{\theta}) = ?$
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November 2nd, 2017, 09:22 AM   #3
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Oh my gosh, THANK YOU SO MUCH!!

I can't believe how easy those were, and how MUCH I complicated things!

Sorry for not writing the first problem in correct format. Once again, though, THANKS!
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November 14th, 2017, 12:20 PM   #4
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Back with more verification problems (because I am clearly no good at verifying trig identities):

1) sin beta * cos beta / (1 - 2cos^2 beta) = - 1/2 tan2 beta

2) cos^4 x - sin^4 x = cos2x

3) [sin 2A / (sin^2 A - 1)] + 3 tan A = tan A

I'd appreciate any help. I really don't know why these are so difficult for me.
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November 14th, 2017, 02:34 PM   #5
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(1) sin B cos B / (1 - 2cos^2 beta) = -2sin B cos B / ((1/2)(2cos^B - 1)) = - sin 2B/(2cos 2B)
= -1/2 tan2 beta

(2) cos^4 y - sin^4 y = (cos^2 y + sin^2 y)(cos^2 y - sin^2 y) = 1(cos 2y)

(3) This one is just as easy, so try to do it again, please.
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November 14th, 2017, 04:41 PM   #6
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Ok, I'm embarrassed that I didn't get the second one. However, the first one - where did the 1/2 come from and why is the denominator now 2cos^2 beta - 1? One of the things that troubles about verifying trig identities is that I want to "switch" things around so I can have an identity to work with; but I'm not sure I can. You just did in this problem, and I don't understand how?

For the third one, all I come up with is sin 2A / - cos2A. That would lead to - tan 2A + 3tanA.
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November 14th, 2017, 05:06 PM   #7
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1. $2\cdot\frac12\sin(A)\cos(A)=\sin(2A)$

2. $\frac{\sin(2x)}{\sin^2(x)-1}+3\tan(x)=-\frac{2\sin(x)\cos(x)}{\cos^2(x)}=-2\tan(x)+3\tan(x)=\tan(x)$
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November 16th, 2017, 06:07 AM   #8
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Quote:
Originally Posted by greg1313 View Post
1. $2\cdot\frac12\sin(A)\cos(A)=\sin(2A)$

2. $\frac{\sin(2x)}{\sin^2(x)-1}+3\tan(x)=-\frac{2\sin(x)\cos(x)}{\cos^2(x)}=-2\tan(x)+3\tan(x)=\tan(x)$
Thanks a lot!

Is there a particular reason someone might struggle with verifying and simplifying trig expressions? I am working on some others and can't get them either. It's depressing.
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November 16th, 2017, 11:41 AM   #9
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Go beyond the homework. Lots of practice helps.
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November 17th, 2017, 07:36 AM   #10
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I think you're right. Thankfully, I was able to verify the others. That said, I do have a simplification problem that I'd like some perspective on:

4sin^2 22.5 degrees * cos^2 22.5 degrees. I decided to plug this one into an online calculator and got 1/2 as the answer. I found one way to get 1/2 also, BUT that's without using that "4." Can you or anyone else kindly explain? The calculator simply used the sin^2 half-angle identity.
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