Trig Applications of Differentiation

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Question 1 of 3

Given that

$x=(pi)/12$ , find the slope (

$m$ ) of the normal line

$y=sin3x$

1. $m=-(sqrt2)/3$
2. $m=(sqrt2)/3$
3. $m=(sqrt3)/2$
4. $m=-(sqrt3)/2$

Question 2 of 3

Given that

$x=(pi)/2$ , find the slope (

$m$ ) of the tangent line

$y=tan(x/2)$

$m=$ (1)

Question 3 of 3

Find the equation of the tangent to the curve

$y=cos2x$ at the point

$x=pi/4$

1. $y=2x-pi$
2. $y=-2x+pi$
3. $y=2x-pi/2$
4. $y=-2x+pi/2$

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Derivatives of Trigonometric Functions

$y'(\text(sin))=\text(cos)$

$y'(\text(cos))=-\text(sin)$

$y'(\text(tan))=\text(sec)^2$

Point Gradient Formula

$y-$ $y_1$

$=$ $m$

$(x-$ $x_1$

$)$

First, find the derivative of the equation

$y$	$=$	$\text(cos) 2x$
$y’$	$=$	$-\text(sin) 2x^2*2$	Derive $\text(cos) 2x$
	$=$	$-2 \text(sin) 2x$

Next, substitute the value of $x$ to the derived function and get the slope of the tangent

$-2 \text(sin) 2x$	$=$	$-2 \text(sin) 2(pi/4)$	Substitute $x=pi/4$

	$=$	$-2 \text(sin) pi/2$	Simplify

	$=$	$-2*1$	$\text(sin) (pi)/2=1$

	$=$	$-2$

Next, substitute the value of $x$ to the original expression and get value of $y$

$y$	$=$	$\text(cos) 2x$

$y$	$=$	$\text(cos) 2(pi/4)$	Substitute $x=pi/4$

	$=$	$\text(cos) pi/2$	Simplify

	$=$	$0$	$\text(cos) (pi)/2=0$

Finally, substitute the known values to the slope-gradient formula to get the equation

$x_1$	$=$	$pi/4$

$y_1$	$=$	$0$
$m$	$=$	$-2$

$y-$ $y_1$	$=$	$m$ $(x-$ $x_1$ $)$

$y-$ $0$	$=$	$\color{#007986}{-2}\left(x-\color{#e65021}{\frac{\pi}{4}}\right)$	Substitute known values

$y$	$=$	$-2x+pi/2$	Simplify

$y=-2x+pi/2$

	$3/(sqrt2)$

$=$	$-(sqrt2)/3$	Get the negative reciprocal

Trig Applications of Differentiation

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Quiz summary

Information

1. Question

Hint

Well Done!

Derivatives of Trigonometric Functions

2. Question

Hint

Nice Job!

Derivatives of Trigonometric Functions

3. Question

Hint

Excellent!

Derivatives of Trigonometric Functions

Point Gradient Formula

Quizzes

$y$	$=$	$\text(sin) 3x$
$y’$	$=$	$\text(cos) 3x*3$	Derive $\text(sin) 3x$
	$=$	$3 \text(cos) 3x$

$m$	$=$	$3 \text(cos) 3x$

	$=$	$3 \text(cos) 3*((pi)/12)$	Substitute $x=(pi)/12$

	$=$	$3 \text(cos) (pi)/4$	Simplify

	$=$	$3*1/(sqrt2)$	$\text(cos) (pi)/4=1/(sqrt2)$

	$=$	$3/(sqrt2)$

$y$	$=$	$\text(tan) x/2$
$y’$	$=$	$(\text(sec) x/2)^2*1/2$	Derive $\text(tan) x/2$
	$=$	$1/2(\text(sec) x/2)^2$

$m$	$=$	$1/2(\text(sec) x/2)^2$

	$=$	$1/2(\text(sec) ((pi)/2)/2)^2$	Substitute $x=(pi)/1$

	$=$	$1/2(\text(sec) (pi)/4)^2$	Simplify

	$=$	$1/2(sqrt2)^2$	$\text(sec) (pi)/4=sqrt2$

	$=$	$1/2*2$

	$=$	$1$