Question

Can we differentiate trigonometric functions to solve an equation?

Like : \(\sin^2 \theta - \cos^2 \theta = -1\)

Well, can I differentiate both sides and simplify it further to find solutions for theta?

Answer 1

No, you can't. Taking a derivative eliminates a constant, so the fact that the derivatives of the two sides is equal does not immediately imply that the the two sides are equal.

To make it slightly more obvious, if that was opaque, you'd get the same result if you were trying to solve
\[ \sin^2(\theta) - \cos^2(\theta) = 1/2 \]

Answer 2

Oh, I see.. Can this relate to the concept of Differentiation also? Like... differentiating means that we're finding the slope, right?

Answer 3

Yes, but clearly the fact that the slope of two functions is equal at a point says nothing whatsoever about the actual value of the functions at that point.

Answer 4

Well... what if I don't differentiate the RHS ? :P

\(4\sin \theta \cos \theta = -1\)
\(\cos (2\theta) = -1\)

?

Answer 5

Or is it stupidity to do so?

Answer 6

To expand on Jemurray's suggestion, \(f'(x) = g'(x)\) for all \(x\) when \(f(x) = g(x)\) is an identity, meaning that it is true for all \(x\). Here, it is clearly an equation, not an identity.

That is why if we're given \(f(x) = \rm blah\) for all \(x\), we can exploit that condition by differentiating both sides and finding conditions for the rate of change at any point.

Answer 7

That's nicely explained Parth....!

Answer 8

So, basically, I can not do like what I did above ? (not differentiating RHS and only finding the derivative of LHS ):P

Answer 9

As ... If \(f(x) = c\) then .. \(f'(x) \ne c\)

Right? or Wrong?

Answer 10

wrong. consider c = 0

f(x) = 0, but f'(x) = 0

Answer 11

Hmm... so, can we just add a condition that :

\(If ~ f(x) = c ~ \text{then} ~~ f'(x) \ne c ~~~~~~~~~~~~~~ c \ne 0 \)

Answer 12

you can prove by contradiction

Answer 13

Keep in mind the following:

\(\sin^2 \theta - \cos^2 \theta =1\) is an equation which has to be solved. Only some and not all values of \(\theta\) will satisfy it.

If you differentiate both sides, \(2(\sin\theta + \cos \theta) = 0\) is what you get. But is it necessary that for the value of \(\theta\) satisfying the equation, the slope of the tangent has to be zero? Think about it for a little.

Answer 14

@ParthKohli
When you will differentiate that you will get :

\(2\sin \theta \cos \theta + 2\cos \theta \sin \theta = 4\sin \theta \cos \theta = 0\)

May be you did a mistake there.

Answer 15

And yes, the idea for differentiating only LHS and not RHS , is absolutely wrong. Integrating it again will give a constant and for that, we don't what that constant is. To avoid the constant, we will have to do definite integration, and we don't the 2 values required for doing definite integration.

So, yeah, that's totally a stupid idea to differentiate LHS and equate it to the original RHS (constant).

Answer 16

Fair point. I messed up my chain rule a little.

Answer 17

As,

i) It will not give all the solutions.
ii) It will not ALWAYS be true. Even most of the times, it will give wrong solutions.

Answer 18

:-) I think, my doubt is resolved now.

Thanks @ParthKohli and @Jemurray3

Answer 19

It's also not necessary that the slope of the tangent of the function \(\sin^2 \theta - \cos^2 \theta\) would be zero.

Answer 20

Corrrrrrrrrrrrrrect! :)