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Question 1 of 3
Given that x=π12, find the slope (m) of the normal line
y=sin3x
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Derivatives of Trigonometric Functions
First, find the derivative of the equation
y |
= |
sin 3x |
y’ |
= |
cos 3x⋅3 |
Derive sin 3x |
|
= |
3 cos 3x |
Next, substitute the value of x and get the slope of the tangent
m |
= |
3 cos 3x |
|
|
= |
3 cos 3⋅(π12) |
Substitute x=π12 |
|
|
= |
3 cos π4 |
Simplify |
|
|
= |
3⋅1√2 |
cos π4=1√2 |
|
|
= |
3√2 |
Finally, to find the normal line, get the negative reciprocal of the tangent, which is 3√2
|
|
3√2 |
|
|
= |
-√23 |
Get the negative reciprocal |
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Question 2 of 3
Given that x=π2, find the slope (m) of the tangent line
y=tan(x2)
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Derivatives of Trigonometric Functions
First, find the derivative of the equation
y |
= |
tan x2 |
y’ |
= |
(sec x2)2⋅12 |
Derive tan x2 |
|
= |
12(sec x2)2 |
Finally, substitute the value of x and get the slope of the tangent
m |
= |
12(sec x2)2 |
|
|
= |
12(sec π22)2 |
Substitute x=π1 |
|
|
= |
12(sec π4)2 |
Simplify |
|
|
= |
12(√2)2 |
sec π4=√2 |
|
|
= |
12⋅2 |
|
|
= |
1 |
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Question 3 of 3
Find the equation of the tangent to the curve y=cos2x at the point x=π4
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Derivatives of Trigonometric Functions
Point Gradient Formula
y-y1=m(x-x1)
First, find the derivative of the equation
y |
= |
cos 2x |
y’ |
= |
-sin 2x2⋅2 |
Derive cos 2x |
|
= |
-2 sin 2x |
Next, substitute the value of x to the derived function and get the slope of the tangent
-2 sin 2x |
= |
-2 sin 2(π4) |
Substitute x=π4 |
|
|
= |
-2 sin π2 |
Simplify |
|
|
= |
-2⋅1 |
sin π2=1 |
|
|
= |
-2 |
Next, substitute the value of x to the original expression and get value of y
y |
= |
cos 2x |
|
y |
= |
cos 2(π4) |
Substitute x=π4 |
|
|
= |
cos π2 |
Simplify |
|
|
= |
0 |
cos π2=0 |
Finally, substitute the known values to the slope-gradient formula to get the equation
y-y1 |
= |
m(x-x1) |
|
y-0 |
= |
−2(x−π4) |
Substitute known values |
|
y |
= |
-2x+π2 |
Simplify |