inverse trig (1 Viewer)

[kractus]

how do u know which triangle to construct?

 

[Masaken]

the inverse bit in the bracket. when you let the bit in the bracket equal to a variable. let's say a = arcsin(3/4), rearrange that to get sina = 3/4. you make a triangle based off of that, and when you combine everything together it becomes tan 2a. that's just double angle = 2tan(a)/(1-tan^2(a)), and since it just requires you to know the value of tan(a) to reach the answer, you can deduce that from the triangle you drew out with sina = 3/4

 

[ExtremelyBoredUser]

View attachment 38357
how do u know which triangle to construct?

tan(2A) = 2tan(A)/1-tan^2(A)

sin(A) = 3/4
make a triangle with opp = 3, hyp = 4 and find tan(A)

sub tan(A)

done

 

[kractus]

the inverse bit in the bracket. when you let the bit in the bracket equal to a variable. let's say a = arcsin(3/4), rearrange that to get sina = 3/4. you make a triangle based off of that, and when you combine everything together it becomes tan 2a. that's just double angle = 2tan(a)/(1-tan^2(a)), and since it just requires you to know the value of tan(a) to reach the answer, you can deduce that from the triangle you drew out with sina = 3/4

ye thx but when u get the triangle for sina = 3/4 it can be either Q1 or Q2 triangle.
i assume since arcsin only outputs from -pi/2 to pi/2 that it has to be a Q1 triangle right

 

[kractus]

Also idk how to do these, like i can only prove it if i do like tan both sides but idk if that's legal or not

 

[Average Boreduser]

ye thx but when u get the triangle for sina = 3/4 it can be either Q1 or Q2 triangle.
i assume since arcsin only outputs from -pi/2 to pi/2 that it has to be a Q1 triangle right

alpha is between -pi/2 to pi/2, and sinA is positive so Q1

 

[Average Boreduser]

View attachment 38362
Also idk how to do these, like i can only prove it if i do like tan both sides but idk if that's legal or not

just let some of the values on one side be a variable like alpha, beta, etc.

 

[synthesisFR]

View attachment 38362
Also idk how to do these, like i can only prove it if i do like tan both sides but idk if that's legal or not

u can do that or convert ratios

 

[unofficiallyred12]

View attachment 38357
how do u know which triangle to construct?

you don't actually need to construct a triangle although that's probably the most obvious method. You can also note tan2theta = +-sqrt(sec^22theta-1). since tan 2theta is in this case negative since 2arcsin3/4 is in the 2nd quad, tan2theta = -sqrt(sec^2(2theta)-1) = -sqrt(1/cos^2(2theta) -1 )) = -sqrt(1/(1-2sin^2theta)^2 -1) = - sqrt63

 

[=)(=]

That maths in focus screen shot has tainted this entire forum, that is illegal contraband by virtue of it being made by Margret Grove, it has been scientifically proven that people who use maths in focus are guaranteed to have trust in later life after being physically assaulted by the difficulty of the hsc

 

[nsw..wollongong]

That maths in focus screen shot has tainted this entire forum, that is illegal contraband by virtue of it being made by Margret Grove, it has been scientifically proven that people who use maths in focus are guaranteed to have trust in later life after being physically assaulted by the difficulty of the hsc

Cambridge textbook >>>

 

[kractus]

View attachment 38363

Here's the conditions for it to be "legal"

Ya ik that but I've always been told to never touch both sides in a proof question

 

[kractus]

u can do that or convert ratios

Thank u

 

[=)(=]

There's nothing wrong with math in focus... it's a great textbook for explaining the basics of a concept unlike Cambridge/NSM which overcomplicate things/go off on unnecessary random tangents

The issue with maths in focus is that its maths in focus

 

[ExtremelyBoredUser]

There's nothing wrong with math in focus... it's a great textbook for explaining the basics of a concept unlike Cambridge/NSM which overcomplicate things/go off on unnecessary random tangents

worst take ive ever seen

The issue with maths in focus is that its maths in focus

factual.

 

[5uckerberg]

View attachment 38362
Also idk how to do these, like i can only prove it if i do like tan both sides but idk if that's legal or not

From what I know I believe you would have to prove these identities. Here we start, . Here . Simplifying we will have This gives us .

View attachment 38362
Also idk how to do these, like i can only prove it if i do like tan both sides but idk if that's legal or not