Question 1

why do ppl invent triangles and calculus T.T

Accepted Answer

People don't invent mathematics, they discover it.

Question 2

How do you prove the half-angle identities?

Accepted Answer

The easiest way is to see that cos 2φ = cos²φ - sin²φ = 2 cos²φ - 1 or 1 - 2sin²φ by the cosine double angle formula and the Pythagorean identity.  Now substitute 2φ = θ into those last two equations and solve for sin θ/2 and cos θ/2.  Then the tangent identity just follow from those two and the quotient identity for tangent.

Question 3

I am kind of struggling on solving sinusoidal equations (advanced) since I don't do all the identities. I don't check all of the solutions. 
Here is some that I know: 
sin(θ)=(θ+360)
sin(θ)=pi-θ
sin(θ)=θ+2pi 
cos(θ)=2pi-θ
cos(θ)=θ+2pi 
is there any others missing? am I doing anything wrong?

Accepted Answer

First of all, we should probably make the notation a bit more rigorous, because the way you've phrased it isn't quite correct.  Instead, write:        
sin(θ)=sin(θ+360)=sin(θ+2pi)
sin(θ)=sin(pi-θ)
sin(θ)=sin(θ+2pi) see above
cos(θ)=cos(2pi-θ)
cos(θ)=cos(θ+2pi)
... and yes, there are lots of others - technically, an infinite number of others since sin and cos are periodic and repeat every 2pi, positive or negative. So, for example, sin(θ)=sin(θ+2npi), where n is any integer.

I'd probably add to the list:
sin(-θ)=-sin(θ)
cos(-θ)=cos(θ),
if we're sticking to sin(a)=sin(b) and cos(a)=cos(b) style identities.

Question 4

In the Angle sum explanation diagram, why is the bottom triangle's adjacent side to angle theta (cos theta)(cos phi)?

Accepted Answer

The bottom triangle is a right triangle with hypotenuse length h = cos phi.    So if x were your unknown side, doing normal trig on it gives cos theta = x/h = x / (cos phi), or in other words x = (cos theta)(cos phi).  All of the sides in that diagram are defined in the same way, relative to the one side that was defined to be of length 1.

Question 5

This isn't exactly related to this, but I don't know where else to ask. I was doing problems related to this on KA, and needed to find the tangent of 15 degrees. I used tan(45-30) in order to find it, which gave me (3-sqrt3)/(3+sqrt3). In the KA "hints," they used tan(60-45) and got (sqrt(3)-1)/(sqrt(3)+1) ... These seem to be two ways of expressing the same value, as putting both into a calculator returns the same result. But for the life of me, I cannot seem to algebraically manipulate my answer to get KA's answer. If I start with tan(60-45), I get that form easily, but how can I prove (3-sqrt3)/(3+sqrt3) equals (sqrt(3)-1)/(sqrt(3)+1) ? I want to be able to more easily choose the right answer in the future, without having to evaluate all of the multiple choices with a calculator and compare them to the evaluation of my own expression.

Accepted Answer

(3-√3)/(3+√3)
Multiply numerator and denominator by 3-√3 and multiply out to get:
(12-6√3)/6
Cancel the 6s to get:
(2-√3)/1

Multiply and divide by √3 +1 to get:
(2-√3)*(√3+1)/(√3+1) =(2√3+2-3-√3)/(√3+1) =(√3-1)/(√3+1)

Question 6

I still don't know how to get the half angle identities.Who can help me?

Accepted Answer

I won't do all three, but you can get the idea from the cosine case.
I assume you're comfortable with the double angle formula: cos(2θ) = cos²θ - sin²θ

Make the substitution φ = 2θ or θ = φ/2

cos(φ) = cos²(φ/2) - sin²(φ/2)
Using the Pythagorean identity we get
cos(φ) = cos²(φ/2) - (1 - cos²(φ/2))
       = 2cos²(φ/2) -1

Solving for cos(φ/2) gives
cos²(φ/2) = (cos(φ) + 1)/2
or 
cos(φ/2) = ±√((cos(φ) + 1)/2)
Which is the result we wanted.

Now once you have that, you can get the sine case by substituting for sin(φ/2) in terms of cosines

ie √(1 - sin²(φ/2)) = √((cos(φ) + 1)/2)

or (1 - sin²(φ/2)) = (cos(φ) + 1)/2
or sin²(φ/2) = 1 - (cos(φ) + 1)/2

or sin²(φ/2) = 2/2 - (cos(φ) + 1)/2
             = (1 - cos(φ))/2

Hence sin(φ/2) = ±√((1 - cos(φ))/2)
Note that the half angle formula for sine gives a result in terms of cosine.

I'll leave the case of tan(φ/2) to you.

Question 7

I have a table of trig identities in my Calculus textbook that has the double cosine identity as:
`cos 2x = cos^2 x - sin^2 x`
Makes sense, because that's the way you would get it if you applied the rule of adding 2 different angles.  How do you get from there to what they have here:
`cos 2x = 2cos^2 x - 1`?

Accepted Answer

The Pythagorean identity states:
sin²𝑥 + cos²𝑥 = 1
This means:
sin²𝑥 = 1 - cos²𝑥
The standard double cosine identity is:
cos 2𝑥 = cos²𝑥 - sin²𝑥
Substituting for sin²𝑥:
cos 2𝑥 = cos²𝑥 - (1 - cos²𝑥)
cos 2𝑥 = 2cos²𝑥 - 1
Comment if you have questions.

Question 8

Do you need to remember all these identities or at least know how to derive them using the unit circle.

Accepted Answer

It's okay just know how to derive them using the unit circle but if you remember all of them, it'll be faster when you solve questions. And I recommend you to remember it because when you are taking a test, you don't have time to derive using the unit circle.
+ It's not that hard to remember though. There is a pattern. And once you proved why it is then it'll be way easier to memorize it.

Question 9

Does one have to remember the angle difference identities if the angle sum identities are already known? If we already knew that `sin(θ+ϕ)=sinθcosϕ+cosθsinϕ`, for instance, could I just use `sin(θ+(-ϕ))=sinθcos(-ϕ)+cosθsin(-ϕ)` instead of having to memorise the sine angle difference identity?

Accepted Answer

Certainly. The symmetric properties of sine and cosine are easier to remember or re-derive anyway.

Question 10

My teacher, as well as textbook, say that the cosine symmetry identity is cos(−θ)=-cos(θ)
but this says that it's cos(−θ)=+cos(θ) which one is correct?

Accepted Answer

The correct identity is:
cos(−θ)=+cos(θ)

Your teacher and the book probably mixed up sine and cosine. The sine identity is:
sin(−θ)=-sin(θ)

Precalculus

Course: Precalculus > Unit 2

Trig identity reference

Reciprocal and quotient identities

Pythagorean identities

Identities that come from sums, differences, multiples, and fractions of angles

Symmetry and periodicity identities

Cofunction identities

Appendix: All trig ratios in the unit circle

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