Trigonometric problem

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sunrise

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Nov 11, 2012
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Hi,
I have some problems and couldnt find.
May you help me?

If cos(2x)*cos(x)=1/16*Sin(x) then sin(4x)=? I ve tried to

sin(4x)=2*Sin(2x)*Cos(2x)
cos(x)*Cos(y)=1/2[(Cos(3x)+Cos(x)]
1/16*Sin(x)=1/2*[(Cos(3x)+Cos(x)]
1/8*Sin(x)=[(Cos(3x)+Cos(x)] but then i couldnt thought how i will solve.

Second one is,

If cos(x)+5Sec(x)=1.1 then Cos(x)=? I know sec=1\cosx but i couldnt find?
 
sunrise said:
Hi,
I have some problems and couldnt find.
May you help me?

If cos(2x)*cos(x)=1/16*Sin(x) then sin(4x)=? I ve tried to

sin(4x)

= 2*Sin(2x)*Cos(2x)

= 2 * 2 * sin(x) * cos(x) * cos(2x)

= 4* sin(x) * 1/16 *sin(x)

= 1/4 * sin2(x)

cos(x)*Cos(y)=1/2[(Cos(3x)+Cos(x)]
1/16*Sin(x)=1/2*[(Cos(3x)+Cos(x)]
1/8*Sin(x)=[(Cos(3x)+Cos(x)] but then i couldnt thought how i will solve.

Second one is,

If cos(x)+5Sec(x)=1.1 then Cos(x)=? I know sec=1\cosx but i couldnt find?
Click to expand...

cos(x)+5Sec(x)=1.1

cos(x) + 5/cos(x) = 1.1

cos2(x) + 5 = 1.1 * cos(x) ....................square term was displayed as 2x before.

Now you have a quadratic equation - solve it using your favorite method
 
Last edited by a moderator:
You are great:)

Thank you for ansvers, I learned sin 4(x) = 2 Sin 2(x) *Cos 2(x)

But how you did to this equation?

= 2 * 2 * sin(x) * cos(x) * cos(2x)

Thanks

 
sunrise said:
You are great:)

Thank you for ansvers, I learned sin 4(x) = 2 Sin 2(x) *Cos 2(x)

But how you did to this equation?

= 2 * 2 * sin(x) * cos(x) * cos(2x)

Thanks
Click to expand...

What is the difference between

2 sin (2x) *cos (2x) and

2 * 2 * sin(x) * cos(x) * cos(2x)

(some terms got replaced by some other terms)

Which term got replaced by which terms?
 
sunrise said:
You are great:)

Thank you for ansvers, I learned sin 4(x) = 2 Sin 2(x) *Cos 2(x)
Click to expand...
It would be better to write "sin(2x)" and "cos(2x)" But yes, sin(2a)= 2sin(a)cos(a) for any angle a

But how you did to this equation?

= 2 * 2 * sin(x) * cos(x) * cos(2x)

Thanks
Click to expand...
Doing the same thing again: sin(2x)= 2sin(x)cos(x) so 2 sin(2x)cos(2x)= 2(2 sin(x)cos(x))cos(2x)
 
2Cos^2(x)-1.1-Cos(x)-5=0 I couldnt find to the rootsI found delta=-4.09 and X1=2.19 and x2=0.9but answer is 0.5 where did i mistake?Thanks
 
sunrise said:
2Cos^2(x)-1.1-Cos(x)-5=0 I couldnt find to the rootsI found delta=-4.09 and X1=2.19 and x2=0.9but answer is 0.5 where did i mistake?Thanks
Click to expand...

Where did that equation come from???
 
Trigonometric question

Hi, may you help me?
I had a question yesterday. I solved as

If cos(x)+5Sec(x)=1.1 then Cos(x)=?

2Cos(x)+5Sec(x) =11/10

2Cos(x)+5/Cos(x)=11/10

2Cos(x)*Cos(x)+5=11/10*Cos(x)

2Cos^2(x)+5=11/10*Cos(x)

2Cos^2(x)-11/10-Cos(x)-5=0 then i used delta=b^2-4ac

then i found delta=-4.79

X1=-b+sqrt(delta)/2a
then i found x1=2.19 and x2=-b-sqrt(delta)/2a but as i learnd answer is 0.5 where is the my mistake?
 
sunrise said:
Hi, may you help me?
I had a question yesterday. I solved as

If cos(x)+5Sec(x)=1.1 then Cos(x)=?

2Cos(x)+5Sec(x) =11/10

2Cos(x)+5/Cos(x)=11/10

2Cos(x)*Cos(x)+5=11/10*Cos(x)

2Cos^2(x)+5=11/10*Cos(x)

2Cos^2(x)-11/10-Cos(x)-5=0 then i used delta=b^2-4ac

then i found delta=-4.79

X1=-b+sqrt(delta)/2a
then i found x1=2.19 and x2=-b-sqrt(delta)/2a but as i learnd answer is 0.5 where is the my mistake?
Click to expand...
First, always start a new question with a new thread. That being said, where did the 2 come from in your second step??

Anyway, you should have \(\displaystyle \cos^2(x)-1.1\cos(x)+5=0\), but this will lead to imaginary answers when using the quadratic formula. Are you sure you copied the original problem down correctly?
 
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