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January 3rd, 2017, 10:50 AM   #1
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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.
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January 3rd, 2017, 12:05 PM   #2
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(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.
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January 3rd, 2017, 02:12 PM   #3
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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.
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January 3rd, 2017, 03:01 PM   #4
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do you know the Pythagorean identity ...

$\cos^2{t} + \sin^2{t} = 1$

???
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January 3rd, 2017, 03:27 PM   #5
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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
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January 3rd, 2017, 03:37 PM   #6
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$\cos^2{t}+\sin^2{t} = 1 \implies \color{red}{\sin^2{t} = 1-\cos^2{t}}$

$\sqrt{\dfrac{\color{red}{1-\cos^2{t}}}{\cos^2{t}}} = \sqrt{\dfrac{\color{red}{\sin^2{t}}}{\cos^2{t}}}$

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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