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# Stuck on Differentiation Question- Trig watch

1. Hello!
The question is as follows-
Find the minimum value of y= tanx-8sinx which is given by a value of x between 0 and pi/2

So firstly I differentiated y=tanx-8sinx. Which gave me dy/dx= sec^2x-8cosx.
Then I made that = 0, and solved for x
0=1/cos^2x - 8cosx
0=cosx(1/cos^3x -8)
cosx=0
x=pi/2
But the limits do not include pi/2, so I discount this value for x.
1/cos^3x=8
cosx=(plusorminus)1/2
x= pi/3 or 2pi/3
But I managed to get 2 answers, both are wrong.

The answer is -3root3. So I'm stumped.
Would love it if you can spot where I went wrong!
Thanks XD
2. (Original post by Ruineth)
Hello!
The question is as follows-
Find the minimum value of y= tanx-8sinx which is given by a value of x between 0 and pi/2

So firstly I differentiated y=tanx-8sinx. Which gave me dy/dx= sec^2x-8cosx.
Then I made that = 0, and solved for x
0=1/cos^2x - 8cosx
0=cosx(1/cos^3x -8)
cosx=0
x=pi/2
But the limits do not include pi/2, so I discount this value for x.
1/cos^3x=8
cosx=(plusorminus)1/2
x= pi/3 or 2pi/3
But I managed to get 2 answers, both are wrong.

The answer is -3root3. So I'm stumped.
Would love it if you can spot where I went wrong!
Thanks XD
I put a line in bold; from there try putting the right hand side as one fraction.
3. (Original post by Ruineth)
Hello!
The question is as follows-
Find the minimum value of y= tanx-8sinx which is given by a value of x between 0 and pi/2

So firstly I differentiated y=tanx-8sinx. Which gave me dy/dx= sec^2x-8cosx.
Then I made that = 0, and solved for x
0=1/cos^2x - 8cosx
0=cosx(1/cos^3x -8)
cosx=0
x=pi/2
But the limits do not include pi/2, so I discount this value for x.
1/cos^3x=8
cosx=(plusorminus)1/2
x= pi/3 or 2pi/3
But I managed to get 2 answers, both are wrong.

The answer is -3root3. So I'm stumped.
Would love it if you can spot where I went wrong!
Thanks XD
It looks like you've done it correctly, it's just asking for the minimum value of y.. soo..
4. (Original post by Andy98)
I put a line in bold; from there try putting the right hand side as one fraction.
0= (1-8cos^3)/cos^2x
But if the top = 0, then I get the same answers as I got before?
5. (Original post by SeanFM)
It looks like you've done it correctly, it's just asking for the minimum value of y.. soo..
Ohhhh yeahhh - whoops, would help if I read the question properly

(Original post by Ruineth)
0= (1-8cos^3)/cos^2x
But if the top = 0, then I get the same answers as I got before?
Sean's right, you've got x - it wants y
6. (Original post by Andy98)
Ohhhh yeahhh - whoops, would help if I read the question properly

Sean's right, you've got x - it wants y
Omg.
I should not be allowed to sit exams XD
Thanks both of you! Hahaha
7. (Original post by Ruineth)
Omg.
I should not be allowed to sit exams XD
Thanks both of you! Hahaha
Hey, I read it wrong too

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Updated: November 10, 2015
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