Question

Verify:
1.(cscx/cosx)-tan=cotx

Solve:
2. cos^2x+sin(2x)-sinx-1=-sin^2x

Answer 1

1) Write the right hand side in terms of sin and cos, then simplify.

Answer 2

2) Use... half angle? or power reducing in reverse... hmm. I would start there at least. Don't see it all off hand.

Answer 3

for 1) im getting stuck at (1/sinxcosx)-(sinx/cosx)

Answer 4

So lets work 1 first. I am going to frget the x, we all know what sin/cos, etc is and there is only one var. \[\frac{1/sin}{cos}-\frac{sin}{cos}=cot\] Right? And we can drop the cot for now. As long as we end with it, we are fine.

Answer 5

That becomes: \[\frac{\frac{1}{sin}-sin}{cos}\] Now you ned to just look at the top for a bit. Put that into a comon denominator so you can combined it. In fact, it might look easier this way:\[\left(\frac{1}{sin}-sin\right)\frac{1}{cos}\] which means the same thing.

Answer 6

That help?

Answer 7

Looks like you are gone. I'll push one more step and I think you will get it then: \[\left(\frac{1}{sin}-\frac{sin^2}{sin}\right)\frac{1}{cos}\] That should really get you close. I was done in 3 more single operations.

Answer 8

Well, perhaos 4 or 5 depenidng on what you call one operation and the fact you would need to get it to cot.

Answer 9

On the other, it is a solve so that is easier than I thought. Set it to 0 and see. Well, setting it to 0 simplifies it. \[\cos ^2x+\sin (2x)-\sin x -1=-\sin ^2x \\ \Rightarrow \cos ^2x+\sin ^2x +\sin (2x)-\sin x -1=0 \\ \Rightarrow 1 +\sin (2x)-\sin x -1=0 \\ \Rightarrow \sin (2x)-\sin x =0 \] Now, use the double angle formula for the sin(2x), then factor and solve for 0. The big key here was getting it into a fom that is more understandable. That is why doing the proofs is important. They teach you how to simplify while it is still in the trig form. Also, get reaaaaaly good at trig manipulations if you plan to go to calculus. They become critical there.

Answer 10

1.from what e.mccormick said"={(1/sinx)-(sin^2x/sinx)}1/cosx"={(1-sin^2x)/sinx}1/cosx={cos^2x/sinx}1/cosx[since (1-sin^2x=cos^2x)]=cosx/sinx=cotx.

Answer 11

are you able to understand my reply.

Answer 12

Yes thanks