Solving Trigonometric Problems with Multiple Formulas

[Anony-mouse]

Homework Statement

1)
COS4[tex]\theta[/tex]-COS2[tex]\theta[/tex][tex]/SIN4\theta-SIN2\theta[/tex]=-TAN3[tex]\theta[/tex]

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

Homework Equations

1) Verify
2)verify

The Attempt at a Solution

1)
cos(2[tex]\theta[/tex]-2[tex]\theta[/tex])-cos2[tex]\theta[/tex] / sin(2[tex]\theta[/tex]+sin2[tex]\theta[/tex])-sin2[tex]\theta[/tex]

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

 

[rock.freak667]

Find what

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

[tex]\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}[/tex]

Just bring them to the same denominator

 

[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

 

[rock.freak667]

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

[tex]\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}[/tex]


[tex]\frac{?}{(sinx)(cosx+1)}[/tex]


bring them to a common denominator like that one.

 

[Anony-mouse]

[tex]\frac{sin^{2}x+cos^{2}x-1}{(sinx)(cosx+1)}[/tex]


?

 

[rock.freak667]

[tex]\frac{sin^{2}x+cos^{2}x-1}{(sinx)(cosx+1)}[/tex]


?

correct.

What is [itex]sin^2x+cos^2x[/itex] equal to?

 

[Anony-mouse]

correct.

What is [itex]sin^2x+cos^2x[/itex] equal to?

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

 

[BrendanH]

Homework Statement

1)
COS4[tex]\theta[/tex]-COS2[tex]\theta[/tex][tex]/SIN4\theta-SIN2\theta[/tex]=-TAN3[tex]\theta[/tex]

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

 

[Anony-mouse]

2cos theta / 2sin theta

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

 

[rock.freak667]

2cos theta / 2sin theta

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

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)

 

[Anony-mouse]

thx that helps