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Using trigonometric identities

Trigonometric identities like sin²θ+cos²θ=1 can be used to rewrite expressions in a different, more convenient way. For example, (1-sin²θ)(cos²θ) can be rewritten as (cos²θ)(cos²θ), and then as cos⁴θ. Created by Sal Khan.

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• how is tan squared less 1 = secant? Each question for this section uses this central calculation to simplify the calculations, but it makes no logical sense
We must simplify (tan^2 theta - 1) <<<< note the 1 within this argument, we're taking an angle, and deducting 1
Start by simplifying the tan^2 theta angle
tan^2 = sin^2+cos^2 = 1 << this we can agree on
the solutions tell us to divide both sides by cos^2.
so sin^2/cos^2 + cos^2/cos^2 = 1/cos^2 and 1/cos^2 is sec^2 << still following
then somehow it says therefore tan^2-1 = sec^2 so it replaces the entire first argument with sec^2, completely ignoring that 1 we were supposed to deduct from tan.
how is this possible? tan^2 is equal to sec^2 according to the calculations, they're just ignoring the one at the end of that original argument we're trying to simplify, like it wasn't there.

If sin^2 + cos^2 =tan^2 = 1
then tan^2 - 1 should theoretically be 0, I know this isn't the answer, but you can see that the 1 in tan^2 - 1 can't be ignored, it's not the 1 from the calculation of tan^2, so how can the simplification of tan^2 wipe out this 1?
(6 votes)
• tan²θ = sin²θ + cos²θ = 1
That is wrong. tan²θ = sin²θ/cos²θ. Secondly, the identity is tan²θ + 1 = sec²θ, not tan²θ - 1.
Maybe this proof will be easier to follow:
tan²θ + 1
= sin²θ/cos²θ + 1
= sin²θ/cos²θ + cos²θ/cos²θ
= (sin²θ + cos²θ)/cos²θ //sin²θ + cos²θ = 1, which we substitute in.
= 1/cos²θ
= sec²θ
Therefore, tan²θ + 1 = sec²θ.
(16 votes)
• how to find the value of sin18?
(3 votes)
• The process is somewhat confusing to find the exact value, but here it is:
Let x = 18° (therefore 5x = 90°)
sin(3x) = cos(90° - 3x) = cos(5x - 3x) = cos(2x)
sin(3x) = cos(2x) (Remember that x = 18°, so that is why this is true.)
3sin(x) - 4sin^3(x) = 1 - 2sin^2(x) (I expanded these.)

Let y = sin(x)
4y^3 - 2y^2 - 3y + 1 = 0
4y^3 - 2y^2 - 2y - (y - 1) = 0
(4y^3 - 2y^2 - 2y) - (y - 1) = 0
(y - 1)(4y^2 + 2y) - (y - 1) = 0
(y - 1)(4y^2 + 2y - 1) = 0
Remember that sin18° is a root, so you can use the quadratic equation to solve for it. Also remember that sin18° is positive, which will help you choose the right answer.

4y^2 + 2y - 1 = 0
y = (-2 ± √[4 - 4(4)(-1)]) / 2(4)
y = (-2 ± √20) / 8
y = (-2 ± 2√5) / 8
y = -1/4 ± √(5)/4
sin18° is positive, so it must be the positive root, not the negative root here. So, the exact value of sin18° is -1/4 + √(5)/4.
(12 votes)
• how to find 2sin^2x-sinx-1=0
(5 votes)
• You can make a substitution to make factoring a bit easier.
Let a = sin(x)
2a^2 - a - 1 = 0
Then you can solve for a, substitute in sin(x) for a, and solve for x.
(5 votes)
• at sal adds cos^2 with sin^2 to get 1 i don't understand that
(2 votes)
• That is a very important identity that comes directly from applying the Pythagorean theorem on the unit circle.

In the video, he used the Pythagorean theorem to say x²+y² = 1, but in the graph, x = cos ⊝ and y = sin ⊝. Thus (cos ⊝)²+(sin ⊝)² = 1 and this is often written as cos² ⊝+ sin² ⊝ = 1.
(11 votes)
• how they got Sin theta over cos theta 2 equals to tan 2 theta
(3 votes)
• How do we know which identity to use when finding the other answers. For example when do I use (2pi-x) or(pi-x), or even the negative version of those. I haven't found any examples explaining this.
(4 votes)
• I felt stumped by this too.

What helped me was practicing A LOT. I did all the problems in my text book, then compared my answers to the back of the book. Then I watched all the videos in this section at least a few times, and went back and did all the textbook problems again. At some point, it seemed easy instead of impossible and it felt like a miracle.

I think what happened is I got a feel for which identities to use just by working with them over and over again. Hope this helps!
(3 votes)
• How to verify the identity?

1+cos x/ sin x =cot x/2
(2 votes)
• cot(x/2)=cos(x/2)/sin(x/2)

=>when we multiply cos(x/2) in numerator and denominator,
cot(x/2)=cos^2(x/2)/sin(x/2)*cos(x/2)

By the formulas:
``cos(2x)=2cos^2(x)-1 ==>cos^2(x/2)=(1+cosx)/2sin(2x)=2sinxcosx ``

cot(x/2)=(1+cosx)/2/sinx/2
=>cot(x/2)=(1+cosx)/sinx
(3 votes)
• Can someone help me with establishing an identity? I'm having a bit of trouble with those types of problems.
(2 votes)
• Basically, If you want to simplify trig equations you want to simplify into the simplest way possible. for example you can use the identities -
cos^2 x + sin^2 x = 1
sin x/cos x = tan x
You want to simplify an equation down so you can use one of the trig identities to simplify your answer even more.
some other identities (you will learn later) include -
cos x/sin x = cot x
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
hope this helped!
(3 votes)
• Find the value of cot25+cot55/tan25+tan55 + cot55+cot100/tan55+tan100 + cot100+cot25/tan100+tan25
(2 votes)
• i'm too lazy to work this out, but here:
https://www.wolframalpha.com/input/?i=cot25%2Bcot55%2Ftan25%2Btan55+%2B+cot55%2Bcot100%2Ftan55%2Btan100+%2B+cot100%2Bcot25%2Ftan100%2Btan25
hope this helps
(2 votes)
• Please answer
If α,β,γ ∈ [0,π], then prove that sin(α+β+γ)/sinα+sinβ+sinγ < 1
(2 votes)

Video transcript

Let's do some examples simplifying trigonometric expressions. So let's say that I have 1 minus sine squared theta, and this whole thing times cosine squared theta. So how could I simplify this? Well the one thing that we do know-- and this is the most fundamental trig identity, this comes straight out of the unit circle-- is that cosine squared theta plus sine squared theta is equal to 1. And then, if we subtract sine squared theta from both sides, we get cosine squared theta is equal to 1 minus sine squared theta. So we have two options. We could either replace this 1 minus sine squared theta with the cosine squared theta, or we could replace this cosine squared theta with the 1 minus sine squared theta. Well I'd prefer to do the former because this is a more complicated expression. So if I can replace this with the cosine squared theta, then I think I'm simplifying this. So let's see. This will be cosine squared theta times another cosine squared theta. And so all of this is going to simplify to cosine theta times cosine theta times cosine theta times cosine theta, well, that's just going to be cosine to the fourth of theta. Let's do another example. Let's say that we have sine squared theta, all of that over 1 minus sine squared theta. What is this going to be equal to? Well we already know that 1 minus sine squared theta is the same thing as cosine squared theta. So it's going to be sine squared theta over-- this thing is the same thing as cosine squared theta, we just saw that-- over cosine squared theta, which is going to be equal to-- you could view this as sine theta over cosine theta whole quantity squared. Well what's sine over cosine? That's tangent. So this is equal to tangent squared theta. Let's do one more example. Let's say that we have cosine squared theta plus 1 minus-- actually, let's make it this way-- plus 1 plus sine squared theta. What is this going to be? Well you might be tempted, especially with the way I wrote the colors, to think, hey, is there some identity for 1 plus sine squared theta? But this is really all about rearranging it to realize that, gee, by the unit circle definition, I know what cosine squared theta plus sine squared theta is. Cosine squared theta plus sine squared theta, for any given theta, is going to be equal to 1. So this is going to be equal to 1 plus this 1 right over here, which is equal to 2.