Thread: Verify trig equation

Could someone please help me with this? I can't figure out what properties to use.

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

1 + sec2x sin2x = sec2x

I assume you mean:



Right?

Yes. Sorry.

Hey, no problem, just wanted to make sure, and just use the caret symbol "^" to denote exponentiation.

You should be familiar with a Pythagorean identity involving the secant function. That along with rewriting (can you think of a simpler way to write this term?) should help quite a bit.

The only Pythagorean identity in my book (that I can find at least) is sin^2u+cos^2u=1

You can derive two other useful forms of this identity by dividing through by either or .

So then I divide both sides by sin^2x?

You want the form involving the square of the secant function, and since the right side of the Pythagorean identity you cited is 1, you should divide through by , or equivalently, multiply through by . In fact, if you do the latter, you get the very equation you are given to verify!

OK I'm kind of lost now. I see that (and why) we want to use cos^2x, but I'm not sure how to divide that through.

I would begin with the Pythagorean identity:



Now, multiply through by , that is, multiply every term on both sides by this. What do you get. Recall:

OK so we can turn it into

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

I'm hopeful :P

I only wanted you to use the definition of the secant function for the first term on the left to get 1, as you did, but for the other terms, just use the square of the secant function, and you will find you will get the equation you are trying to verify.

Oh ok so its just 1 + (1/cos^2x) (sin^2x)^2 = sec^2x

Don't write (1/cos^2(x)) on the left side, write sec^2(x), and you then have the exact equation you are trying to verify.

So we are just back at 1 + sec^2x sin2x = sec^2x