Proving Trig Identities

Question

So, I've been learning proofs since I was in about 8th grade math. But I can't grasp it no matter how many times its explained to me. Its one of those things that makes sense when someone else does it, but as soon as I go to do a problem on my own, I'm totally lost.

So, can you please explain each step on how to prove secx-tanxsinx=(1/secx), and why you did each step you did? Thanks ^^

Answer 1

secx - tanx(sinx) = 1/secx

one method is convert everything to sines & cosines, if you don't see anything better

secx = 1/cosx, tanx = sinx/cosx, 1/secx = cosx

=1/cosx - (sinx/cosx)(sinx) = cosx

= 1/cosx - (sin^2(x)/cosx = cosx

multiply by cosx to eliminate fractions

1 -sin^2(x) = cos^2(x)

add sin^2(x) to both sides

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

the right side is the most common trig identity = 1

1 = 1

QED

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

is a version of the pythagorean theorem

in polar coordinates x=rcosA, y= rsinA

y^2 + x^2 = 1 which is the pythagorean theoem applied to the point (x,y) on a unit circle

Answer 2

There are certain "tricks" to use that sometimes help. Almost always, it's good to start with the more complicated side and "morph" it into the less complicated side using known identities. In this case, we'll manipulate the left side to make it become the right side.

sec(x) - tan(x)sin(x) = 1/sec(x)

Using the Reciprocal Identity sec(x) = 1/cos(x) and the identity tan(x) = sin(x)/cos(x), we can make those substitutions on the left side:

1/cos(x) - [sin(x)/cos(x)]sin(x) = 1/sec(x)

1/cos(x) - sin²(x)/cos(x) = 1/sec(x)

The two expressions on the left now have a common denominator so they can be combined together:

[1 - sin²(x)]/cos(x) = 1/sec(x)

Using the Pythagorean Identity sin²(x) + cos²(x) = 1 we can subtract sin²(x) from both sides to get:

cos²(x) = 1 - sin²(x)

Now we can substitute cos²(x) for "1 - sin²(x)" in our equation:

cos²(x)/cos(x) = 1/sec(x)

cos(x) = 1/sec(x)

Now, we can use the Reciprocal Identity cos(x) = 1/sec(x) we can substitute on the left side:

1/sec(x) = 1/sec(x)

And we are finished. We have proven this identity.

Answer 3

Hi Scarlett,

I know these identities can be confusing and a little hard to remember.

In this case it's important to remember the following:

For any x, sec(x) = 1/cos(x), sin²(x) + cos²(x) = 1, and tan(x) = sin(x)/cos(x)

These are the identities that will help us prove your question;

So we want to prove that: secx-tanxsinx=(1/secx)

Let's start with the first term using the identities above:

sec(x)-tan(x)sin(x) = 1/cos(x) - [(sin(x)/cos(x)) . sin(x)]

= 1/cos(x) - [sin²(x)/cos(x)]

We now have a common denominator (cos(x))

So 1/cos(x) - [sin²(x)/cos(x)] = (1- sin²(x)) / cos(x)

we can use the identity sin²(x) + cos²(x) = 1

So (1- sin²(x)) / cos(x) = [sin²(x) + cos²(x) - sin²(x)] / cos(x)

= cos²(x) / cos(x) (because the sin²(x) cancels)

= cos (x)

and cos (x) = 1/ sec(x)

So secx-tanxsinx=(1/secx)

I hope that helps