Prove Pythag Trig Identities

[ThineMadCatter]

I have been doing these trig identities in class and I am having a hard time knowing how to do this problem. Any help is appreciated

 

[blamocur]

Hint: express all terms through [imath]\sin[/imath] and [imath]\cos[/imath].

 

[Deleted member 4993]

I have been doing these trig identities in class and I am having a hard time knowing how to do this problem. Any help is appreciated

start with converting the LeftHandSide of the equation to sin(x) and cos(x) using:

tan(x) = \(\displaystyle \frac{sin(x)}{cos(x)} \)..........&........ sec(x) = \(\displaystyle \frac{1}{cos(x)} \) ...... and continue after that using algebra

There are of course other ways to solve the problem.

Please show us what you have tried and exactly where you are stuck.

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[Dr.Peterson]

I have been doing these trig identities in class and I am having a hard time knowing how to do this problem. Any help is appreciated

The suggestion others have made is a good one, based on the fact that the RHS contains cosines, so it is appropriate to change the LHS in that direction. This is a good general principle. Another is that changing to sines and cosines often makes it easier to see what to do next, by reducing the complexity of a problem and using the most familiar forms.

Another approach you could take is to recall that tangent and secant are related by a Pythagorean identity of their own, and write that out as a hint to yourself. Then you might either intentionally aim in that direction; or immediately use it to rewrite the numerator of the LHS (and then think about factors); or multiply top and bottom by the conjugate of the denominator (which is often a helpful idea when you see something of the form 1+a or 1-a, and know that 1-a^2 would be of use).

I suppose you could even start with the RHS, though I can't quite imagine doing that, because it's usually safest to start on the "more complicated" side, which includes "the side that has something more than sines and cosines".

 

[JeffM]

I’d add only one thing to the excellent advice already given. Once you have an expression in sine and cosine, convert to cosine as soon as possible because that is where you want to end up. The fact that the LHS involves a square suggests that [imath]sin^2(x) = 1 - cos^2(x)[/imath] is going to come in very handy.