# Trigonometric problem

• Anony-mouse
In summary: A+B)+cos(A-B)=2cos(A-B)What is cos^2x+cos^2x equal to?correct.What is sin^2x+cos^2x equal to?correct.In summary, this conversation is about solving equations involving sines and cosines. The person is looking for help solving equations involving sines and cosines, and they mention that it can be difficult to tell which formula to use. There are a few formulas that can be used, and the person ends up getting the answer correct.

## Homework Statement

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

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

1) Verify
2)verify

## 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

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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

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

Anony-mouse said:
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

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

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

bring them to a common denominator like that one.

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

?

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

?

correct.

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

rock.freak667 said:
correct.

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

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

Anony-mouse said:

## Homework Statement

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

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

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2cos theta / 2sin theta

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

Anony-mouse said:
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)

thx that helps

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