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# Thread: This is the hardest trigonometry problem I've ever seen, reps available.

1. ## This is the hardest trigonometry problem I've ever seen, reps available.

Ok I have been studying for my trig exam for 4 hours strait, and I'm on my last problem I can't figure out.

It says simplify:

The back of the book says the answer is "1-sin x" but I cant get it simplified that far.

2. are those 2s?

3. Originally Posted by BTBAM
are those 2s?
correct, dividing the X by 2 but not the whole problem

4. Prob goes to Cos^2(x) + Sin^2(X) which goes to 1 then blah blah blah

I dunno

5. expand what's inside: [cos(x/2) + sin(x/2)^2] - 2sin(x/2)cos(x/2)

recognize that you have cos(y)^2 + sin(y)^2

= 1 - 2sin(x/2)cos(x/2)

recognize that you have a double angle identity on the right

= 1-sin(x)

6. Originally Posted by SlammaJamma
expand what's inside: [cos(x/2) + sin(x/2)^2] - 2sin(x/2)cos(x/2)

recognize that you have cos(y)^2 + sin(y)^2

= 1 - 2sin(x/2)cos(x/2)

recognize that you have a double angle identity on the right

= 1-sin(x)
how did you expand that in the first step?

7. Originally Posted by Ultimus
how did you expand that in the first step?
Just expand it like you would anything else. foil I guess

I put brackets just to show that what's inside reduces to 1

8. Originally Posted by Ultimus
Ok I have been studying for my trig exam for 4 hours strait, and I'm on my last problem I can't figure out.

It says simplify:

The back of the book says the answer is "1-sin x" but I cant get it simplified that far.
This is basically like proving the identity...except without the right hand side given.

Sin(x/2) = +- sq root( 1-cosx/2)
Cos(x/2) = +- sq root (1+cosx/2)

Since its squared the roots cancel out and all you have to do is combine after getting the lcd and you wind up with 1-sinx.

9. The answer is one my son.

well in short its 1 sin...but its really

1.000000000000000000000000000000000000000000000000 0000000000000000000000001

10. Originally Posted by FltMech
rofl what's that from?

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