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 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) √(1-cos^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: 2,201 Thanks: 1049 (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: 2,201 Thanks: 1049 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: 2,201 Thanks: 1049 $\cos^2{t}+\sin^2{t} = 1 \implies \color{red}{\sin^2{t} = 1-\cos^2{t}}$ $\sqrt{\dfrac{\color{red}{1-\cos^2{t}}}{\cos^2{t}}} = \sqrt{\dfrac{\color{red}{\sin^2{t}}}{\cos^2{t}}}$ 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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