My Math Forum Help Solving/Verifying Trig Identities

 Trigonometry Trigonometry Math Forum

 November 1st, 2017, 05:25 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, 06:19 PM   #2
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,767
Thanks: 1422

Quote:
 Originally Posted by Bluebird8 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 ...

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

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

$(\tan^2{\theta} + 1) + (1 + \cot^2{\theta}) = ?$

 November 2nd, 2017, 08:22 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, 11:20 AM #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, 01:34 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,521 Thanks: 1747 (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, 03:41 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, 04:06 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,856 Thanks: 1078 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, 05:07 AM   #8
Newbie

Joined: Nov 2017
From: Pennsylvania

Posts: 7
Thanks: 0

Quote:
 Originally Posted by greg1313 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.

 November 16th, 2017, 10:41 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,856 Thanks: 1078 Math Focus: Elementary mathematics and beyond Go beyond the homework. Lots of practice helps.
 November 17th, 2017, 06:36 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 half-angle identity.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ehh Trigonometry 7 June 20th, 2012 07:50 PM omo5031 Algebra 3 February 19th, 2012 07:45 PM Danicus Trigonometry 6 February 4th, 2012 01:19 PM bignasty36 Algebra 5 March 3rd, 2009 07:08 AM sultanofsurreal Algebra 2 May 4th, 2007 05:37 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top