Question

Simplify
\[\frac{asinx-bcosx}{acosx+bsinx}\]

Answer 1

\[ \frac{asinx}{acosx+bsinx} -\frac{bcosx}{acosx+bsinx} \]

Answer 2

hm..its not working; anyone able to give me a hint?

Answer 3

Is this allowed \[\frac{asinx*sinx}{acosx+bsinx*sinx} -\frac{bcosx*cosx}{acosx+bsinx*cosx} \]?
I havent simplified things in a while lol

Answer 4

are you allowed to multiply by the conjugate

Answer 5

well i dont know

Answer 6

do you still need help @Omniscience ?

Answer 7

yes i do lol

Answer 8

@lgbasallote

Answer 9

Find the first derivative
Remeber these rules:
The derivative of sinx = cosx
The derivative of cosx = -sinx

y' = A(-sinx) + B(cosx)

NOw find the second derivative

y'' = A(-cosx) + B(-sinx)

Plug everything into the equation: y'' - y = sinx

-Acosx - Bsinx - Acosx - Bsinx = sinx

Simplify

-2Acosx - 2Bsinx = sinx

Since we don't need cosx at all, we can make A=0

-2Bsinx = sinx

Now this is a simple algebra problem, just sovle for B

-2B = 1

B = -1/2

A = 0
B = -1/2

Answer 10

what is this? this is not a calculus problem..

Answer 11

Im trying to simplify the expression; not solving for a and b

Answer 12

Multiply and divide by \(acos(x) - bsin(x)\)

Answer 13

No I think:

put a = cos(y) and b = sin(y)

Answer 14

hmm..i dont think a substitution is required..and multipying and dividing by acosx-bsinx will look ugly..

Answer 15

Then don't you think this is the best simplified form ??

Answer 16

what do you mean its already simplfied?

Answer 17

Try that substitution..

Answer 18

well i havent learnt about substitutions to simplify so i dont think that its required

Answer 19

\[\frac{\sin(x)\cos(y) - \cos(x)\sin(y)}{\cos(x)\cos(y) + \sin(x)\sin(y)} \implies \frac{\sin(x-y)}{\cos(x-y)} \implies \tan(x-y)\]

Answer 20

By that substitution:

b = cos(y)
a = sin(y)

so:

b/a = tan(y)

or y = tan^{-1}(b/a)

Answer 21

well thank you.

Answer 22

is there another method that i dont need a substitution?

Answer 23

Divide numerator and denominator by \(acos(x)\):

\[\frac{\tan(x) - \frac{b}{a}}{1 + \frac{b}{a}\tan(x)} \implies \frac{\tan(x) - \tan(\tan^{-1}(\frac{b}{a}))}{1 + \tan(x) \cdot \tan(\tan^{-1}(\frac{b}{a}))} \implies \tan(x - \tan^{-1}{\frac{b}{a}})\]

Note that :

\(tan^{-1}(\frac{b}{a})\) is nothing but \(y\) as I have done earlier by substitution..

This one is now without substitution..

Answer 24

thank youu.

infact, i have figured out another method as well; in which gives the same answer xD
\[\frac{\sqrt{a^{2}+b^{2}}\sin\left(x-\tan^{-1}\frac{b}{a}\right)}{\sqrt{a^{2}+b^{2}\cos\left(x-\tan^{-1}\frac{b}{a}\right)}} => \tan\left(x-\tan^{-1}\frac{b}{a}\right) \]

Answer 25

Here you have used the same substitution concept:

a = cos(y) and b = sin(y)

Square both and then add them:

\[a^2 + b^2 = \sin^2(y) + \cos^2(y)\]

\[a^2+ b^2 = 1\]

Answer 26

lol i didnt know that..

i used the auxilliary angle thingy