How Do You Verify Trigonometric Identities Involving Negative Angles and Powers?

[tennistudof09]

i am trying to verify the two identities listed below:

cot(-x)cos(-x)+sin(-x)= -csc x

i started out as:

cos(-x)/sin(-x) cos(x)-sin x, then got it to cos^2 x/sin x - sin x. I thinking i am solving this correctly but I can't figure out the next step to get -csc x.


sin^4 x + (2 sin^2 x) (cos^2 x) + cos^4 x = 1

for this one, first, i factored and got:

sin^2 x (sin^2 x + 2) cos^2 x (1 + cos^2 x) =1

I don't know if the factoring helped or did anything, but I can't seem to get anything else to get 1.

 

[Anakin_k]

You did a lot of unnecessary stuff there. Start on the left side. Cotangent (-x) is negative in the 4th quadrant, so what can that be expressed as? Cosine is positive in 4th quadrant so what can that be expressed as? Sine is negative in 4th quadrant, once again, adjust the sin(-x) into something else. Use your knowledge of related acute angles, and then simplify.

 

[Architeuthis Dux]

It looks like you are on the right track with your approach to verifying these identities. For the first one, you can use the identity cos^2 x + sin^2 x = 1 to simplify cos^2 x/sin x - sin x to -sin x. Then, you can use the identity csc x = 1/sin x to get -csc x.

For the second identity, you can use the Pythagorean identity sin^2 x + cos^2 x = 1 to simplify sin^2 x (sin^2 x + 2) cos^2 x (1 + cos^2 x) to sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1. Then, you can use the identity cos^2 x = 1 - sin^2 x to simplify further and get sin^4 x + 2sin^2 x (1 - sin^2 x) + (1 - sin^2 x)^2 = 1. This simplifies to sin^4 x + 2sin^2 x - 2sin^4 x + 1 - 2sin^2 x + sin^4 x = 1. Finally, combining like terms gives 1 = 1, which verifies the identity.