Question

f(x)=cosx+5cos3x+cos5x/cos6x+6cos4x+15cos2x+10 then f(0)+f'(0)+f''(0)=

Answer 1

Hello @samigupta8 ! This is INeedHelpPlease assisting you with your problem today! :)
May I know what is the problem that you are encountering in this particular question?

Answer 2

may i know what easy thing u find in dis ques

Answer 3

Can you rewrite the equation involved in the question in proper equation form using LaTex?That would be pretty helpful.

Answer 4

\[\frac{ cosx+5\cos3x+\cos5x }{ \cos6x+6\cos4x+15\cos2x+10}\]

Answer 5

Okay,that would be very helpful thanks!
So we want to find the derivative of this function,maximum twice.
We surely won't apply blind quotient rule directly,our aim would be to simplify what is given to us with the help of trigonometry formulaes,do you get it?

Answer 6

From the numerator,I would chose the first and the last term i.e
cos x and cos 5x,use the formula for cos C+cos D here,
We know that,
\[\cos C +\cos D = 2 \cos (\frac{C +D}{2}) \cos (\frac{C -D}{2})\]
\[\cos x+\cos 5x= 2 \cos 3x+ \cos 2x\]
Is that okay?

Answer 7

Now the question becomes..
\[\frac{2 \cos 3x+\cos 2x + 5 \cos 3x}{\cos6x+6\cos4x+15\cos2x+10}=\frac{7 \cos 3x + \cos 2x}{\cos6x+6\cos4x+15\cos2x+10}\]

Answer 8

You can consider doing the same thing in the denominator,apply the formula and reduce it,do it as much as you can then ping me,after you think the function is reduced to maximum and no more simplification is possible you can differentiate it easily :)

Answer 9

Then just compute the following:
f(0)
f'(0)
f''()
and add them.
For f(0)
f(x)=cosx+5cos3x+cos5x/cos6x+6cos4x+15cos2x+10
Put x=0 in this.
similarly do for others.

Answer 10

another way to simplify this fraction is using power reduction formulas
http://en.wikipedia.org/wiki/Trig_Identity#Power-reduction_formula
notice
\[\cos^{5} x = \frac{\cos(5x) +5\cos(3x)+10\cos x}{16}\]
\[\cos^{6} x = \frac{\cos(6x)+6\cos(4x)+15\cos(2x)+10}{32}\]
solve each for cos(5x) and cos(6x) respectively and sub into fraction
you will end up with
\[\frac{16\cos^{5} x -9\cos x}{32\cos^{6}}\]
\[= \frac{\sec x}{2} - \frac{9}{32}\sec^{5} x\]
then taking derivative is more straightforward

Answer 11

@samigupta8 Now you have 2 wonderful solutions to your problem.
Are you alive?

Answer 12

@samigupta8
The answer is\[-\frac{11}{16} \]Will present a Mathematica solution in a few minutes.

Answer 13

ans is 1

Answer 14

@INeedHelpPlease? u hve ur second step wrongly written

Answer 15

why?

Answer 16

it's 2cos3xcos2x+5cos3x

Answer 17

Yup,that was intentional to check if you are following me or not :)

Answer 18

k....then we willl simply get co3x(2cos2x+5)

Answer 19

wt happnd is it not ryt clubbing will not give us ans or will it give

Answer 20

\[\frac{2\cos 3x \cos 2x+5\cos 3x}{\cos6x+6\cos4x+15\cos2x+10}=\frac{2\cos 3x \cos 2x+5\cos 3x}{\cos2(3x)+6 \cos2(2x)+15 \cos(2x)+10}\]
\[\frac{2\cos 3x \cos 2x+5\cos 3x}{2 \cos^2(3x)-1+6( 2\cos^2(2x)-1)+15( 2\cos^(2x)-1)+10}=\]
\[\frac{2\cos 3x \cos 2x+5\cos 3x}{2 \cos^2(3x)-1+12\cos^2(2x)-6+30\cos^2(2x)-15+10}=\]
\[\frac{2\cos 3x \cos 2x+5\cos 3x}{2 \cos^2(3x)+42\cos^2(2x)+18}=\]
\[\frac{2\cos 3x \cos 2x+5\cos 3x}{\cos(6x)+\cos(4x)+40}=\]
\[\frac{2\cos 3x \cos 2x+5\cos 3x}{2\cos(5x)\cos(x)+40}=\]
don't think clubbing will help :P

Answer 21

Refer to the Mathematica solution attached.