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Anyone that can please help, can you please? Been stuck for hours on these. suppose to prove that each Identity are true.

Cos^4 θ - Sin^4 θ = Cos^2 θ- Sin^2 θ



Csc^4 θ - cot^4 θ = 2cot^2 θ +1


(csc θ -cot θ)^2 = (1- cos θ) / (1+ cos θ)


(csc^2 θ + 2csc θ -3) / (csc^2 θ -1) = (csc θ +3) / ( csc θ+1)

Well, the first one is really easy.
a^2 - b^2 = (a-b)(a+b), remember?

All you have to do is:

cos^4 (x) - sin^4 (x) = (cos^2 (x) - sin^(x)) * (cos^2(x) + sin^2(x)), and the right bracket is, of course, equal to 1.

As for the other ones, I don't really know. Look into the the properties of the csc function, and use calculation formulas like the one I used above. It shouldn't be too hard.

Trig identity that will be useful:


Rewrite the original as:


Substitute in the trig identity from above:


Foil:


Simplify:


Tada!


Used mathematica to write the equations for me, much easier to read I think.

EDIT: I quoted the wrong problem. Fixed.

2. Csc^4 θ - cot^4 θ = 2cot^2 θ +1

Working with the left side... Change it so that the powers of the left and right side are the same

(Csc^2 θ - cot^2 θ)(csc^2 θ + cot^2 θ)

Then use 1 + cot^2 θ = csc^2 θ (which is the same as sin^2 θ + cos^2 θ = 1) to replace all the csc's

(1 + cot^2 θ - cot^2 θ)(1 + cot^2 θ + cot^2 θ)

And simplify

2cot^2 θ + 1


3. (csc θ -cot θ)^2 = (1- cos θ) / (1+ cos θ)

Working with the right side, first multiply by the conjugate as one way of doing this is to get rid of the denominator since the left side doesnt have a demoninator.

[(1 - cos θ)(1 - cos θ)]/[(1+ cos θ) (1 - cosθ)]

Distribute the top and bottom

(1 - 2cos θ + cos^2 θ)/(1 - cos^2 θ)

Using sin^2 θ + cos^2 θ = 1, the bottom is sin^2 θ... and 1/sin^2 θ is also equal to csc^2 θ (which removes the denominator)... So when you distribute that, you end up with

csc^2 - 2 csc θ cot θ + cot^2

Which can be factored to

(csc θ - cot θ)^2

4. (csc^2 θ + 2csc θ -3) / (csc^2 θ -1) = (csc θ +3) / ( csc θ+1)

Working with the right side. Well this one is really easy... Just try to get the denominators the same, and you can easily do that by multiplying the conjugate.

[(csc θ +3)(csc θ - 1)] / [( csc θ+1)(csc θ - 1)]

And you end up with the left side after distributing.

I feel myself very un-educated after reading this topic.
Thank you SRF.

F**K TRIG IDENTITIES! they are the downfall of my precalc grade

Lmao it was so cake the surprise test was sweeeet.

Did my 'images' help? lmao... I failed. Apparently wolphram doesn't allow linking after a certain time period... the-more-you-know.jpg