can anyone do the proof of (cal 1) cos x= -sin x ? i will appreciated :]?

Question

i already seen the proof for sin x = cos x. I just would like to see the proof for cos x = -sin x. i will consider anything that could help me like a website or something. by the way its a calculus 1 problem.

Answer 1

From first principle

Right hand limit
Rf'(x) = Lim h->0+ [{f(x+h) - f(x)}/h]

Left hand limit
Lf'(x) = Lim h->0- [{f(x+h) - f(x)}/h]

Here, f(x) = cos x
So, f(x+h) = cos (x+h)

So, Right hand limit
Rf'(x) = Lim h->0+ [{cos(x+h) - cos(x)}/h]

= Lim h->0+ [{2*sin(x+(h/2))*sin(-h/2)}/h]

[since cosA - cosB = 2*sin{(B-A)/2}*sin{(B+A)/2}]

= Lim h->0+ [{-2*sin(x+(h/2))*sin(h/2)}/h]

[since sin(-x) = -sin(x)]

= Lim h->0+ [-sin(x+(h/2))] * Lim h->0+ [2*sin(h/2)}/h]

[since limit of products = product of limits]

= -sin(x) * Lim (h/2)-->0+ [sin(h/2)}/(h/2)]

[since as h tends to 0+, h/2 also tends to 0+]

= -sin(x)*1 = -sin(x)

[since Lim z->0 [sin(z)/z] = 1]


Similarly, show Lf'(x) = -sin(x)

Hence, Lf'(x) = Rf'(x). Thus the derivative exists for all 'x' and its value is -sin(x).

Answer 2

There are a couple of ways to do it
I think you want d/dx(cos x)

let d/dx(cos x) = t

we know

sin ^ 2 x + cos ^2 x = 1
differentiate both sides

2sinx d/dx(sinx) + (2 cos x ) t = 0
or 2 sinx cos x + 2t cos x =0

so t = - 2sinx cosx /(2cosx) = - sin x


2nd method
let d/dx(cos x) = t
we know
exp(ix) = cos x + i sin x
differentiate both sides
i exp(ix) = t + i cos x

devide both sides by i
exp(ix) = -it + cos x = cos x -it
equating imaginary part t = -sin x

Answer 3

There are a couple of ways to do it
I think you want d/dx(cos x)

let d/dx(cos x) = t

we know

sin ^ 2 x + cos ^2 x = 1
differentiate both sides

2sinx d/dx(sinx) + (2 cos x ) t = 0
or 2 sinx cos x + 2t cos x =0

so t = - 2sinx cosx /(2cosx) = - sin x


2nd method
let d/dx(cos x) = t
we know
exp(ix) = cos x + i sin x
differentiate both sides
i exp(ix) = t + i cos x

devide both sides by i
exp(ix) = -it + cos x = cos x -it
equating imaginary part t = -sin x

Answer 4

If you already know that d/dx sin(x) = cos(x), then you can use a simple trig identity to give you d/dx cos(x) = -sin(x).

sin(x)^2 + cos(x)^2 = 1

Now bring sin(x)^2 to the other side:

cos(x)^2 = 1 - sin(x)^2

Now take the derivative of both sides, keeping in mind the chain rule:

2*cos(x)*[ d/dx cos(x) ] = -2*sin(x)*cos(x)

You can cancel the 2*cos(x) from both sides, and you get the desired result:

d/dx cos(x) = -sin(x)