trigonometric identity question

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Locus

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Good morning/evening everyone. I have a doctrinal question here and I am not quite sure if the question is wrong or the answer I deduct from my own question is wrong.

I am at the trigonometric identities chapter, specifically secx = 1/cosx. Another identity is tanx = sinx/cosx.
Other rules/identities:
tanx=opposite/adjacent; sinx=opposite/hypotenuse;
cosx=adjacent/hypotenuse.
Further on, the book goes on to say "this relationship between trigonometric ratios is called an identity because it is true for all values of x in a given domain".

Then here is a problem:

Show that tanxsinx + cosx = secx.

First trial was (opposite/adjacent)x(opposite/hypotenuse)+(adjacent/hypotenuse)=1/adjacenthypotenuse

That is not precisely hypotenuse/adjacent as i expect secx to be.

On a second trial, where i use (sinx/cosx)sinx + cosx= (sinsquaredx + cossquaredx)/cosx that brings me to the expected result of course.

My question is, am I missing a step where 1/adjacenthypotenuse becomes hypotenuse/adjacent or is any of the trigonometric identities not much of an identity?

Thank you in advance.
 
tanxsinx + cosx = (sinx/cosx) + cosx = sinx/cosx + cos^2x/cosx which is NOT sec x

I suspect that the problem was or should have been tanxsin^2x + cosx = secx
 
Hi Steven. tanx multiplied by sinx does indeed become sinsquaredx/cosx. Add a cosx to it (which should become cossquaredx/cosx in order to be added to the fraction) and the result will be 1/cosx as the problem requires. however this does not hold true if, instead of the cos and sines, we use their original form, which is adjacent/hypotenuse and opposite/hypotenuse respectively. so my question is, what kind of identity are we talking about if they do not hold to be true in every possible scenario? either that, or I am doing something very wrong, but i honestly don't see it.
 
Locus said:
Hi Steven. tanx multiplied by sinx does indeed become sinsquaredx/cosx. Add a cosx to it (which should become cossquaredx/cosx in order to be added to the fraction) and the result will be 1/cosx as the problem requires. however this does not hold true if, instead of the cos and sines, we use their original form, which is adjacent/hypotenuse and opposite/hypotenuse respectively. so my question is, what kind of identity are we talking about if they do not hold to be true in every possible scenario? either that, or I am doing something very wrong, but i honestly don't see it.
Click to expand...
Steven G's correction is wrong. The original expression is correct:

[imath]tan(x)~sin(x) + cos(x) = sec(x)[/imath]

[imath]\dfrac{o}{a} \cdot \dfrac{o}{h} + \dfrac{a}{h}[/imath]

[imath]= \dfrac{o^2}{ah} + \dfrac{a}{h}[/imath]

[imath]= \dfrac{o^2}{ah} + \dfrac{a^2}{ah}[/imath]

[imath]= \dfrac{o^2+a^2}{ah}[/imath] <-- [imath]o^2 + a^2 = h^2[/imath]

[imath]= \dfrac{h^2}{ah}[/imath]

[imath]= \dfrac{h}{a}[/imath]

[imath]=sec(x)[/imath]

I urge you to take a look at the Trigonometric version, though, as Trig functions are not always simple to express in terms of triangles. (In that some of the o's and a's are negative.)
[imath]tan(x)~sin(x) + cos(x) = sec(x)[/imath] <-- [imath]tan(x) = \dfrac{sin(x)}{cos(x)}[/imath]

[imath]= \dfrac{sin^2(x)}{cos(x)} + cos(x)[/imath]

[imath]= \dfrac{sin^2(x)}{cos(x)} + \dfrac{cos^2(x)}{cos(x)}[/imath]

[imath]= \dfrac{sin^2(x)+cos^2(x)}{cos(x)}[/imath] <--[[imath]sin^2(x) + cos^2(x) = 1[/imath]

[imath]= \dfrac{1}{cos(x)}[/imath]

[imath]=sec(x)[/imath]

-Dan
 
Thank you Dan. I just failed to account for the obvious, the part where the hypotenuse squared is equal to 1squared. So the trigonometric identity is indeed correct and the universe is safe.
 
topsquark said:
Steven G's correction is wrong. The original expression is correct:

[imath]tan(x)~sin(x) + cos(x) = sec(x)[/imath]

[imath]\dfrac{o}{a} \cdot \dfrac{o}{h} + \dfrac{a}{h}[/imath]

[imath]= \dfrac{o^2}{ah} + \dfrac{a}{h}[/imath]

[imath]= \dfrac{o^2}{ah} + \dfrac{a^2}{ah}[/imath]

[imath]= \dfrac{o^2+a^2}{ah}[/imath] <-- [imath]o^2 + a^2 = h^2[/imath]

[imath]= \dfrac{h^2}{ah}[/imath]

[imath]= \dfrac{h}{a}[/imath]

[imath]=sec(x)[/imath]

I urge you to take a look at the Trigonometric version, though, as Trig functions are not always simple to express in terms of triangles. (In that some of the o's and a's are negative.)
[imath]tan(x)~sin(x) + cos(x) = sec(x)[/imath] <-- [imath]tan(x) = \dfrac{sin(x)}{cos(x)}[/imath]

[imath]= \dfrac{sin^2(x)}{cos(x)} + cos(x)[/imath]

[imath]= \dfrac{sin^2(x)}{cos(x)} + \dfrac{cos^2(x)}{cos(x)}[/imath]

[imath]= \dfrac{sin^2(x)+cos^2(x)}{cos(x)}[/imath] <--[[imath]sin^2(x) + cos^2(x) = 1[/imath]

[imath]= \dfrac{1}{cos(x)}[/imath]

[imath]=sec(x)[/imath]

-Dan
Click to expand...
I'm wrong! What am I going to do?!
 
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