## Trigonometry Identities

There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve. One of the identities is: **sin ^{2} + cos^{2} = 1. ** To prove this consider a right angled triangle with side a,b and c as shown below

From this we can use Pythagoras theorem to say: a^{2}+b^{2}=c^{2 now we know sin t = b/c so b = csin t cos t = a/c so a = ccos t} substituting these values in the above equation we get (csint)^{2} +(ccost)^{2} = c^{2} canceling the c^{2} we get **sint ^{2} + cost^{2} = 1**

There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over.

Using these a cos

^{2}+ sin

^{2}= 1 we can calculate other identities

**tan**We can obtain this by dividing through by cos

^{2}t + 1 = sec^{2}t^{2}as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec. Other similar identities can be obtained for cosec and cot.

Assume that the Pythagoras theorem until now not exist, how to prove the trig identity. Thx.

Rohedi, let taking the absolute square of Euler formula e^(ix)=cosx+isinx,

where i=(-1)^(1/2). Then we will find…

[e^(ix)][e^(-ix)] = [cosx+isinx][cosx-isinx]

1 = (cosx)^2 + (sinx)^2

A simple proof, isn’t it.

Thanks Nadya for your hint, Next I apply square operation for the Euler Formula

[e^(ix)][e^(ix)] = [cosx+isinx][cosx+isinx]

e^(i2x) =[ (cosx)^2 – (sinx)^2] + i[2cos(x)sin(x)] , (1)

we know from the definition your Euler Formula

e^(i2x) =cos(2x) + i sin(2x) , (2)

ho ho ho…then from the equality real part of eq.(1) and eq.(2) we find

cos(2x)=(cosx)^2 – (sinx)^2

wihile from the equality imaginary part of eq.(1) and eq.(2) we find

sin(2x)=2sin(x)cos(x)

Okay Rohedi, this is also simple derivation, isn’t it.

Nadya and Denaya, please stay on the topic. Back to Rohedi’s question, but let assume that the Euler formula also not exist up today, how to prove the trig identity?

Hi Juliani, Nadya, and Denaya, the common way to get the Euler formula e^(ix)=cosx+isinx was performed by using Argand Diagram and Maclaurin Series. Let assume the two tools until now do not exist. My question, how to get the Euler Formula?

Ho ho ho ho ho, Mr.Rohedi, I believe that the new breakthrough for your problem in creating the Euler Formula is only by using Rohedi’s Formula for solution the Bernoulli differential equation of constant coefficients : dy/dx + Py = Qy^n, for n is not equal 1, that is of form:

y(x) = 1/[( 1/y0^(n-1 )-Q/P )e^{P(n-1)(x-x0)} + Q/P]^(n-1)

with both initial values of x0 and y0. Let’s inspire to get it.

O yeah, I also ever read the nice explanation about the above Rohedi’s Formula when visiting to this address :

http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34&st=0&sk=t&sd=a&start=20.

Are you Maduranish from Indonesian coutry???

Sorry, which one of Rohedi that your purpose who create the BDE’s Formula. Coz after searching at google there many rohedi’s name, such as nikita rohedi, cucun rohedi, yoedi rohedi, DR.Yossi Rohedi, etc, and finally of course the rohedi’s name on ROHEDI LAboratory. I think all rohedi’s name originate from Indonesia.

Of course the rohedi’s name that purposed by Nadya Fermega. He is the Head of ROHEDI LAboratory who has created the second form ol solving the Bernoulli DE :

y(x)=[(P/2Q)^1/(n-1)][1+tanh(P(n-1)(x-x0)+atanh( (2Q/P)y0^(n-1) -1) ) ]^1/(n-1)

Oh my God, indeed the above formulation is very nice. For detail explanation please visit to this address : http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34 on the topic of Stable Modulation Technique (SMT).

Apologize, there is small correction. Rohedi’s formula for BDE’s solution dy/dx+Py=Qy^n is of form

y(x)=[(P/2Q)^1/(n-1)][1+tanh(-P(n-1)(x-x0)/2 +atanh( (2Q/P)y0^(n-1) -1) ) ]^1/(n-1)

Oh my GOD, I am also wrong in writing Rohedi’s formula for dy/dx+Py=Qy^n. This is the true solution:

y(x) = 1/[( 1/y0^(n-1 )-Q/P )e^{P(n-1)(x-x0)} + Q/P]^(1/(n-1))

Thx Mr.Pramono.