Value of sinx/1+cosx
is:
Answers
a
Answer:
ans.
Step-by-step explanation:
dy/dx=sinx.(-sinx) +cosx.(1+cosx)
= -sin^2x+cosx+cos^2x
=-1+cos^2 x+cosx+cos^2x
=2cos^2x +cosx -1
=2cos^2x+2cosx-cosx-1
=2cosx(cosx+1)-1(cosx+1)
(cosx+1)(2cosx-1)
For maxima or minima put dy/dx=0
( cosx+1)(2cosx-1)=0
cosx+1=0 => x=180°
2cosx-1=0 =>cosx=1/2 or x=60°
dy/dx=-sin^2x+cosx+cos^2x=cosx+ cos^2x-sin^2x
dy/dx=cosx+cos2x
d2ydx^2= -sinx-2.sin2x
d2y/dx^2 (at x=180° )=0
d2y/dx^2 ( at x=60°) -ve ,there exist maxima
at x=60°.
Maximum value =sin60°(1+cos60°).
= (3^1/2)/2 (1+1/2)=(1/4).(3)^3/2 , Answer.
What is the maximum value of a function f(x)=sinx(1+cosx) ?
What is the maximum value of sin x + cos x?
What is the maximum and minimum value of sin x +cos x?
How can I find the minimum value of (1+sin x) (1+cos x)?
What is the maximum value of sin (cos x)?
What is the maximum and minimum value of sin x +cos x?
Using the Maxima and Minima.
Rewrite the equation as a function of x .
f(x)=sin(x)+cos(x)
Find the first derivative of the function.
f′(x)=cos(x)−sin(x)
Find the second derivative of the function.
f''(x)=−sin(x)−cos(x)
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
cos(x)−sin(x)=0
sin(x)=cos(x)
tan(x)=1
x=tan−1(1)
x=π4orx=5π4
Evaluate the second derivative at
x=π4
If the second derivative is positive, then this is a minimum. If it is negative, then this is a maximum.
−sin(π4)−cos(π4)
Evaluate the second de
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How do I find minimum and maximum value of sin (cos (sin(x)))?
Note that in the interval [0,π2] the sine function is increasing and the cosine function is decreasing.
Maximize f(x)=sin(cos(sin(x))
Maximize cos(sin(x))
Minimize sin(x)
Minimize x
In our interval, x=0 is the required minimum value.
max (f(x))=sin(1)
Minimize f(x)=sin(cos(sin(x))
Minimize cos(sin(x))
Maximize s i n(x)
Maximize x
x=π2
min (f(x))=sin(cos(1))
Therefore, the range of f(x) in the interval [0,π2] is [sin(cos(1)),sin(1)] .
By similar argument, we can extend this to the interval [0,2π] , just by keeping track of the change in signs. The answer, however, would remain the sam
If f(x)=sin(x)-cos(x), what is the maximum value?
f(x) = sin x-cos x
f'(x)= cos x+sin x.
For maximum or minimum putting f'(x)=0
0=cos x+ sin x
or. -sin x=cos x
or. tan x=-1=> x= 3π/4
f”(x)= -sin x+cos x = - sin 3π/4+cos 3π/4= -1/√2–1/√2=-2/√2= -√2 (negative)
f”(x) is negative. There exist maximum at x=3π/4.
Maximum value = sin3π/4-cos 3π/4= 1/√2+1/√2 =2/√2. =√2. units. Answer.