Range of a function involving trigonometric functions

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Homework Statement
Range of ##\displaystyle \left| \dfrac{(√(cosx)-√(sinx))(√(cosx)+√(sinx))} {3(cosx+sinx)} \right|##
Relevant Equations
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I resolved the numerator to ## cosx-sinx##
We get $$mod\frac{cosx-sinx} {3(cosx+sinx)} $$
If we divide the numerator and denominator by cosx we get
$$mod\frac{1-tanx} {3(1+tanx)}$$(eq1)
We know that tan(π/4-x) is same as ##\frac{1-tanx} {1+tanx}##
So re writing eq1 we get
$$mod\frac{tan(π/4-x} {3}$$
As we know tangent function can take any value from -∞ to+∞
Considering the modulus function we can conclude that the range is 0 to ∞

However thats not the case ofcourse. I graphed it on desmos and while the original question lies on the graph of this simplified tan function, its range is bounded

Please tell me where I went wrong?
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Aren't the graph and range correct though?
 
Aurelius120 said:
Aren't the graph and range correct though?
They aren't. The original expression's graph is only the red part.

The expression I simplified and got (in terms of tan) is the whole blue part.
 
Ibix said:
What's ##\sqrt{\cos(\pi)}##?
Oh. So ill have to restrict the domain π/4 - x should not be equal to places where sin and cos get negative
 
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tellmesomething said:
They aren't. The original expression's graph is only the red part.

The expression I simplified and got (in terms of tan) is the whole blue part.
I see.
It could be because the square root function restricts the value of ##\cos x ## and ##\sin x##
 
Aurelius120 said:
I see.
It could be because the square root function restricts the value of ##\cos x ## and ##\sin x##
I am sorry I dont know how I overlooked that. Thankyou for your Help.
 
Ibix said:
What's ##\sqrt{\cos(\pi)}##?
But doesnt that just restrict the domain from 0 to π/2 ?
 
tellmesomething said:
@fresh_42 Please taje a look at this if you have some time.
I see a pattern. You do the right algebra on the formulas but forget what has already been hidden in the original expressions. The last time you solved ##ae^{2x} + be^x+c=0## and investigated the discriminant. However, you forgot that ##e^x>0## regardless of the solution of the quadratic. Now, you did the correct algebra again, but by using ##(\sqrt{a}+\sqrt{b})\cdot (\sqrt{a}-\sqrt{b})=a-b## you forgot that you lost ##a,b\geqq 0.## I'm not sure which kind of practice is appropriate in such a situation. My guess is, that your mathematical enthusiasm leads you directly into algebra, e.g. being happy that you correctly identified the formula that has to be used. I understand this, I have the same tendency to jump into the problem and calculate. That often leads to situations where I do not see the obvious. Hence, the only advice I can give both of us is to wait a moment and inspect the problem before doing any algebra. The only problem: who reminds us of this advice?
 
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fresh_42 said:
I see a pattern. You do the right algebra on the formulas but forget what has already been hidden in the original expressions. The last time you solved ##ae^{2x} + be^x+c=0## and investigated the discriminant. However, you forgot that ##e^x>0## regardless of the solution of the quadratic. Now, you did the correct algebra again, but by using ##(\sqrt{a}+\sqrt{b})\cdot (\sqrt{a}-\sqrt{b})=a-b## you forgot that you lost ##a,b\geqq 0.## I'm not sure which kind of practice is appropriate in such a situation. My guess is, that your mathematical enthusiasm leads you directly into algebra, e.g. being happy that you correctly identified the formula that has to be used. I understand this, I have the same tendency to jump into the problem and calculate. That often leads to situations where I do not see the obvious. Hence, the only advice I can give both of us is to wait a moment and inspect the problem before doing any algebra. The only problem: who reminds us of this advice?
Have to remind our own self
 
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