# Thread: differentiation of trigo functions

1. ## differentiation of trigo functions

1.
Given that y=2 $sin^3$ x - 3 sin x, find an expression for $\frac{dy}{dx}$ in terms of cos x and then find $\frac{d^2y}{dx^2}$.

2.
The diagram shows an isosceles triangle ABC inscribed in a circle of radius r cm and centre at O. Given that $\angle {BAO}$= $\theta$ radian and $\angle$ ADC = $\frac{\pi}{2}$ radians, show that area , S $cm^2$, of triangle ABC is given by S = $r^2$ sin 2 $\theta$(1+ cos 2 $\theta$).

for the first question, what i tried to do:
$\frac{dy}{dx}$
= 6( $sin^2$ x)( $cos^2$ x) - 3 cos x
=3 cos x [ 2 $sin^2$ x cos x -1]

then.....how do i continue to simplify it?the answer is 3 cos x(1- 2 $cos^2$ x)......

for question 2 ....

AB= $\frac{2r}{cos \theta}$

S
= ( $\frac{1}{2}$)( $\frac{2r}{cos \theta}$)( $\frac{2r}{cos \theta}$)sin 2
$\theta$
=( $\frac{2r^2}{cos^2 \theta}$) sin 2 $\theta$

and......then....how do i simplify it to $r^2$sin 2 $\theta$(1+ cos 2 $\theta$)....??? is my working correct in first place???

any help is much appreciated. thanks in advance.

2. 1. Redo your differentiation, cause it's false.
Then there's a simple way of relating $sin^2(x)$ to $cos^2(x)$.

2. The formula for AB is correct.
Start from it and compute AD and BD.

3. Originally Posted by wintersoltice
1.
Given that y=2 $sin^3$ x - 3 sin x, find an expression for $\frac{dy}{dx}$ in terms of cos x and then find $\frac{d^2y}{dx^2}$.

2.
The diagram shows an isosceles triangle ABC inscribed in a circle of radius r cm and centre at O. Given that $\angle {BAO}$= $\theta$ radian and $\angle$ ADC = $\frac{\pi}{2}$ radians, show that area , S $cm^2$, of triangle ABC is given by S = $r^2$ sin 2 $\theta$(1+ cos 2 $\theta$).

for the first question, what i tried to do:
$\frac{dy}{dx}$
= 6( $sin^2$ x)( $cos^2$ x) - 3 cos x
=3 cos x [ 2 $sin^2$ x cos x -1]

then.....how do i continue to simplify it?the answer is 3 cos x(1- 2 $cos^2$ x)......

for question 2 ....

AB= $\frac{2r}{cos \theta}$

S
= ( $\frac{1}{2}$)( $\frac{2r}{cos \theta}$)( $\frac{2r}{cos \theta}$)sin 2
$\theta$
=( $\frac{2r^2}{cos^2 \theta}$) sin 2 $\theta$

and......then....how do i simplify it to $r^2$sin 2 $\theta$(1+ cos 2 $\theta$)....??? is my working correct in first place???

any help is much appreciated. thanks in advance.
Let me solve both parts from the beginning:

$y=2(sinx)^3-3sinx$
$
\frac{dy}{dx}=6(sinx)^2*cosx-3cosx$

$=6(1-(cosx)^2)*cosx-3cosx$
$=cosx((6(1-(cosx)^2)-3)$

Now the second part:
$\angle {BOD}=2\theta$ (it's circle property i.e angle twice at centre than at the corner)

$AD(DO+OA)=rcos2\theta+r$
$BC=2rsin2\theta$
Area $= \frac{1}{2}(rcos2\theta+r)(2rsin2\theta)$
$=r^2 sin 2\theta(1+ cos 2\theta)$

Hope this helps

4. To take $\frac{d^2y}{dx^2}$ take derivative of the $\frac{dy}{dx}$ function i.e calculated in first part of the previous post.