# Trigonometric problem

1. May 26, 2008

### Anony-mouse

1. The problem statement, all variables and given/known data

1)
COS4$$\theta$$-COS2$$\theta$$$$/SIN4\theta-SIN2\theta$$=-TAN3$$\theta$$

2)sinx/cosx+1 + cosx-1/sinx = 0

2. Relevant equations
1) Verify
2)verify

3. The attempt at a solution
1)
cos(2$$\theta$$-2$$\theta$$)-cos2$$\theta$$ / sin(2$$\theta$$+sin2$$\theta$$)-sin2$$\theta$$

when simplified i get a large answer :S

2)
sinx/cosx+1 X cos-1/cos-1(reciprocal) + cos-1/sinx
= sinx cosx-1/ cos2 -1 + cosx-1/sinx
=sinx cosx-1/ Sin2x + cosx-1/sinx
=cosx-1/sinx + cosx-1/sinx
=2(cosx-1)/sinx :S

thats it i hope u can read it
formulas used
Trigonometric Identities
sum and difference Formulas of cosines and sines
and double angle formulas

my problem is that there is so many formulas and its hard to tell which one to use
they are all usable but not all give u the answer

Last edited: May 26, 2008
2. May 26, 2008

### rock.freak667

Find what

Sin(A+B)-Sin(A-B) and similar for cos ,for the first part.

$$\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}$$

Just bring them to the same denominator

3. May 26, 2008

### Anony-mouse

well how do u get them = to zero
this far, and i don't know if its right :P
cosx-1/sinx + cosx-1/sinx they gave the same denominator but they dotn' = zero

4. May 26, 2008

### rock.freak667

$$\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}$$

$$\frac{????}{(sinx)(cosx+1)}$$

bring them to a common denominator like that one.

5. May 26, 2008

### Anony-mouse

$$\frac{sin^{2}x+cos^{2}x-1}{(sinx)(cosx+1)}$$

?

6. May 26, 2008

### rock.freak667

correct.

What is $sin^2x+cos^2x$ equal to?

7. May 26, 2008

### Anony-mouse

1
thx man
too simple and i didn't look that :P

8. May 27, 2008

### BrendanH

Are you still looking for help on this one?
Hint: this is a very straightforward case of sum-to-product substitution

Last edited: May 27, 2008
9. May 27, 2008

### Anony-mouse

2cos theta / 2sin theta

when i use the double angle formula I end up with squared cosines and sines :S

10. May 27, 2008

### rock.freak667

Don't use the double angle formula here,it'll get too tedious

Consider this
sin(A+B)=sinAcosB+sinBcosA
sin(A-B)=sinAcosB-sinBcosA

if we add those two we get

sin(A+B)+sin(A-B)=2sinAcosB

Let P=A+B and Q=A-B, you'd eventually get A=(P+Q)/2 and B=(P-Q)/2

hence then

SinP+SinQ=2sin[(P+Q)/2]cos[(P-Q)/2]

now do the same for

cos(A+B)-cos(A-B)

11. May 27, 2008

### Anony-mouse

thx that helps

Last edited: May 27, 2008