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November 1st, 2017, 06:25 PM  #1 
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0  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!!! 
November 1st, 2017, 07:19 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,778 Thanks: 1430  Quote:
$\dfrac{\cos{\theta}}{\sin^3{\theta} + \sin{\theta}\cos^2{\theta}}$ 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:
$\dfrac{\sec^2{\theta}}{\csc{\theta}} = \dfrac{\sin{\theta}}{\cos^2{\theta}} = \dfrac{1}{\cos{\theta}} \cdot \dfrac{\sin{\theta}}{\cos{\theta}}$ ... finish it Quote:
$\tan^2{\theta} + 2\tan{\theta}\cot{\theta} + \cot^2{\theta}$ $\tan^2{\theta} + 2 + \cot^2{\theta}$ $(\tan^2{\theta} + 1) + (1 + \cot^2{\theta}) = ?$  
November 2nd, 2017, 09:22 AM  #3 
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0 
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! 
November 14th, 2017, 12:20 PM  #4 
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0 
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. 
November 14th, 2017, 02:34 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,951 Thanks: 1842 
(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. 
November 14th, 2017, 04:41 PM  #6 
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0 
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. 
November 14th, 2017, 05:06 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,884 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
1. $2\cdot\frac12\sin(A)\cos(A)=\sin(2A)$ 2. $\frac{\sin(2x)}{\sin^2(x)1}+3\tan(x)=\frac{2\sin(x)\cos(x)}{\cos^2(x)}=2\tan(x)+3\tan(x)=\tan(x)$ 
November 16th, 2017, 06:07 AM  #8  
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0  Quote:
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.  
November 16th, 2017, 11:41 AM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,884 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
Go beyond the homework. Lots of practice helps.

November 17th, 2017, 07:36 AM  #10 
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0 
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 halfangle identity. 

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