Differentiation of Trig Functions (1 Viewer)

[Avenger6]

Hi, a struggled with these questions in the excercise:

In 1v) the answer was given as cosx/sinx=cotx, im sure this is correct but I have no idea why, could someone please explain to me why that is the answer? In 3 and 4 I am stuck with finding the gradient, i can derive the y value easy enough but I get stuck when I have to sub the x value in to work out the gradient. I got y'=3(sec^2)3x in question 3 and y'=-cos(pi-x) in question 4. Like I said though, I get stuck when I have to sub the x value in to work out the gradient.

Any help is greatly appreciated.

 

[lyounamu]

Hi, a struggled with these questions in the excercise:

In 1v) the answer was given as cosx/sinx=cotx, im sure this is correct but I have no idea why, could someone please explain to me why that is the answer? In 3 and 4 I am stuck with finding the gradient, i can derive the y value easy enough but I get stuck when I have to sub the x value in to work out the gradient. I got y'=3(sec^2)3x in question 3 and y'=-cos(pi-x) in question 4. Like I said though, I get stuck when I have to sub the x value in to work out the gradient.

Any help is greatly appreciated.

1v) y = ln (sinx)

y' = 1/sinx . d/dx(sinx)
= 1/sinx . cosx
= cosx/sinx
= cotx

3) y = tan3x
y' = 3sec^2(3x)
When x=pi/9
y' = 3sec^2(3.pi/9)
= 3sec^2(pi/3)
= 3 . 1/(cos^(pi/3))
= 3 . 1/((1/2)^2)
= 3 . 4
= 12

4) y = sin(pi - x) = sinx
y'=cosx
When x=pi/6
= square root of 3/2
Therefore, m=square root of 3/2 and (pi/6, 1/2)

Using the formula: y-y1 = m(x-x1)
y-1/2 = square root of 3/2 . (x-pi/6)
2y-1 = (square root of 3)x - (square root of 3)pi/6
12y-6 = 6(square root of 3)x - (square root of 3)pi
6(square root of 3)x - 12y - (square root of 3)pi + 6 =0

I am sorry if I introduced any mistakes, I did this at quite a quick pace. However, I am quite confident that I got this right.

 

[Avenger6]

Thanks for the reply. Im still a little confused however. In question 3, how did you turn 3sec^2(pi/3) into 3 . 1/(cos^(pi/3)), i dont see how the cos comes into it??? And in question 4 how does sin(pi - x) = sinx?? Sorry if these are basic questions but i am really struggling trying to get my head around these equations.

 

[lyounamu]

Thanks for the reply. Im still a little confused however. In question 3, how did you turn 3sec^2(pi/3) into 3 . 1/(cos^(pi/3)), i dont see how the cos comes into it??? And in question 4 how does sin(pi - x) = sinx?? Sorry if these are basic questions but i am really struggling trying to get my head around these equations.

I am really sorry for the confusion.

Sec is the reciprocal of the cosine. Therefore, I turned 3sec^2(pi/3) into 3 . 1/cos^2(pi/3).


For question 4, sin(pi-x) = sin(180degrees - x) because pi = 180 degrees, right?

So, according to the ASTC rule, sin(180 - x) = sinx.

 

[Avenger6]

Okay thanks. Now i am lost as to how you made 3 . 1/(cos^(pi/3)) eqaul 3 . 1/((1/2)^2) in question 3. And in question 4 how did you know that when x=pi/6 = square root of 3/2??? I must be missing the point somewhere because even the example in the book confuses me...

 

[lyounamu]

Okay thanks. Now i am lost as to how you made 3 . 1/(cos^(pi/3)) eqaul 3 . 1/((1/2)^2) in question 3. And in question 4 how did you know that when x=pi/6
= square root of 3/2??? I must be missing the point somewhere because even the example in the book confuses me...

cos^2(pi/3) = (cos(pi/3))^2
= (1/2)^2
= 1/4
so, 1/(cos^2(pi/3)) = 1/(1/4) = 4

cos(pi/3) = cos 60 = 1/2

cos (pi/6) = cos30 = (square root of 3)/2

 

[Avenger6]

Thanks, makes much more sense now. Im still a little lost as to how sin(pi-x) = sinx, i understand that sin180=0 so shouldn't it be sin-x, since its 0-x???

 

[lyounamu]

Thanks, makes much more sense now. Im still a little lost as to how sin(pi-x) = sinx, i understand that sin180=0 so shouldn't it be sin-x, since its 0-x???

sin (180-x) =/= sin 180 - sin 0

According to the ASTC(All Stations to China), Sin x and sin(180-x) are same. Have a loook at Fitzpatrick or Cambridge. You will get better understanding.

Good luck.

 

[vds700]

u should revise angles of any magnitude and exact value triangles as they can test u on prelim content in the hsc

They are very important concepts

 

[r170491]

sin(pi-x) is compound angle = sinacosb-cosasinb sinpicosx-cospisinx which will be left for 3unit... that is assuming that pi is not a constant if following your logic

 

[tommykins]

sin(pi-x) = sin x.

 

[donetha]

sin(pi-x) is compound angle = sinacosb-cosasinb sinpicosx-cospisinx which will be left for 3unit... that is assuming that pi is not a constant if following your logic

Wow..your confusing yourself. This isn't a compound angle....just a complimentary ratio