Verifying Trig
Storyboard Text
- OMG, what did I just go through??
- TODAY'S LESSONcosx+sinxtanx=secxcos+sinxsinx/cos= cosx/1(cosx) + sin2x/coscos^2x+sin^2x/cosx = 1/cosx1/cosx=secx
- Yes please, that would be amazing. Can we meet in a classroom during lunch?
- Hey, I learned this before! I can help you if you need.
- That'll be good, I'll see you then.
- DURING LUNCH...
- THE PROBLEM: sinx/1+cosx + cosx/sinx = cscx
- Okay, so do we find a formula that equals sinx/cosx ?
- Okay, so when dealing with verifying trig identites, always start on the side that's the easiest to solve, for our case it'll be the left side.
- Actually we do not but that's a great start. Since the denominator has 1+cosx instead of cosx, we have to find the common denominator in both problems.
- THE PROBLEM:sinx/1+cosx + cosx/sinx = cscxsin^2x/(1+cosx)(sinx) + cosx+cos^2x/(1+cosx)(sinx)
- But why can't we cancel out the denominators with one another?
- We know that we have to find a common denominator so we can multiple it to the opposite side.
- Since both denominators are the same we can proceed combining the two fractions together.
- We can not do that because one has to be on the opposite side of the numerator and denominator.
- THE PROBLEM:sinx/1+cosx + cosx/sinx = cscxsin^2x/(+cosx)(sinx) + cosx+cos^2x/(1+cosx)(sinx)1+cosx/(1+cosx)(sinx)
- Doesn't six^2x+cos^2x=1?
- That's correct, so we can change the numerator from sin^2x+cos^2x to 1+cosx since we substituted the 1 here.
- Now that we have a new equation, we can go ahead and find out if we could use a trig formula here.
- THE PROBLEM:sinx/1+cosx + cosx/sinx = cscxsin^2x/(+cosx)(sinx) + cosx+cos^2x/(1+cosx)(sinx)1+cosx/(1+cosx)(sinx)
- Now we can cancel out the cosx since there's one in the numerator AND denominator!
- Now we are left with 1+cosx/(1+cosx) / (sinx). Do you see anything we can do to verify that it equals cscx?
- That is correct!
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