January 3rd, 2017, 10:50 AM  #1 
Newbie Joined: Jan 2017 From: Huddersfield Posts: 8 Thanks: 0  trignometric identities
Can anybody help me with the following? I cannot find any examples even remotely similar to this and I do not understand how to tackle them. Prove the following trigonometric identities: A) √(1cos^2 θ)/(cos^2 θ) = tanθ B) (3π/2) + ϕ = sin ϕ Last edited by skipjack; January 3rd, 2017 at 03:13 PM. 
January 3rd, 2017, 12:05 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,302 Thanks: 1131 
(A) $\sqrt{\dfrac{1\cos^2{t}}{\cos^2{t}}} = $ $\sqrt{\dfrac{\sin^2{t}}{\cos^2{t}}} = $ $\sqrt{\tan^2{t}} = \tan{t}$ recheck (B) ... something is missing. could it be $\cos\left(\dfrac{3\pi}{2}+t\right) = \sin{t}$ ? Last edited by skeeter; January 3rd, 2017 at 12:32 PM. 
January 3rd, 2017, 02:12 PM  #3 
Newbie Joined: Jan 2017 From: Huddersfield Posts: 8 Thanks: 0 
Yes, that is correct for B. Can you please show me your working for equation A? I do not understand how you have got to that point.
Last edited by skipjack; January 3rd, 2017 at 03:16 PM. 
January 3rd, 2017, 03:01 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,302 Thanks: 1131 
do you know the Pythagorean identity ... $\cos^2{t} + \sin^2{t} = 1$ ??? 
January 3rd, 2017, 03:27 PM  #5 
Newbie Joined: Jan 2017 From: Huddersfield Posts: 8 Thanks: 0 
Yes I'm aware of that, I'm just a bit unsure of how to transpose it into this equation as In the stages of changing the identity

January 3rd, 2017, 03:37 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,302 Thanks: 1131 
$\cos^2{t}+\sin^2{t} = 1 \implies \color{red}{\sin^2{t} = 1\cos^2{t}}$ $\sqrt{\dfrac{\color{red}{1\cos^2{t}}}{\cos^2{t}}} = \sqrt{\dfrac{\color{red}{\sin^2{t}}}{\cos^2{t}}}$ you should know the basic ratio identity, $\dfrac{\sin{t}}{\cos{t}} = \tan{t}$ ... $\dfrac{\sin^2{t}}{\cos^2{t}} = \left(\dfrac{\sin{t}}{\cos{t}}\right)^2 = (\tan{t})^2 = \tan^2{t}$ finally ... $\sqrt{\tan^2{t}} = \tan{t}$ as far as your part (B), use the sum identity for cosine ... $\cos(a+b) = \cos{a}\cos{b}  \sin{a}\sin{b}$ 

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