#Pre-calc: equivalent representations
76 messages · Page 1 of 1 (latest)
The first one you can use 2sin(x)cos(x) = sin(2x)
Where does that come from, is it just a rule?
The second one is cos²(x) - sin²(x) = cos(2x)
Those are identities
You said you couldnt find resources or some
Are you allowed to use trigonometric table?
So what are all the trig identities
I think so
There are plenty, but you need those two
If so then 5π/4 and 7π/6 should be on the table
well then
if you still don't know the values tho you can make use of sine and cosine addition theorem
adonhs
For example 5π/4 = 4π/4 + π/4 = π + π/4
π and π/4 are very easy angles to memorize.
adonhs
The other two identities I just provided above as well.
I don't know a specific name for that
You can prove it though using sine addition theorem
You may try it out yourself first 😄
||sin(2x) = sin(x+x) = sin(x)cos(x) + sin(x)cos(x) = 2sin(x)cos(x)||
Would the exact value be -√2/2?
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it's good
I think 1 and 2 makes more sense, what about the verify questions how would u do those
You mean proving the identities?
Try to make use of what I provided
2sin(x)cos(x) = sin(2x)
cos²(x) - sin²(x) = cos(2x)
cot(x) = cos(x)/sin(x)
I’m not sure maybe we’re doing different processes to solve😭 cause our teacher taught us to change the trig identities. Like for 3 we would change the equation on the right of the equal sign to look like the one on the left
yea
that's how you approach it
Those trig identities might just help you
Ok I think I did it right
show
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You made use of 1 + cot² = csc²
and also
csc = 1/sin
But question did you use the internet, or did your teacher tell you some identites by themself?
A little bit of both honestly
that's good
I know the csc = 1/sin stuff already, but had to search up the 1 + cot part
you're doing great so far
there are people who are very clueless
but you know what you are doing, you just need the right puzzles to begin with
I’d say I’m close to clueless but yeah!
you are rather indecisive😎
Pfffft I guess
How would I do 4, I’m not sure since it’s 1 - sin(2x) instead of the trig identities that I went off of in the previous question which was just sin(2x)
honestly i would try (a-b)² = a² - 2ab + b²
😄
I am sure you know
1 = sin² + cos²
that should work 😄
👏🏻 🔥