if sin(x+y)= sinxcosy + cosxsiny, find sin(π/2 + α) Please explain answer
well set pi/2 = x and α = y in sin(x+y)
sin(π/2 + α)= sin(π/2)cosα + cos(π/2)sinα
the answer is cosα, but I don't know how to get there
sin(pi/2) = 1 cos(pi/2) = 0 thus we have sin(π/2 + α) = cos(α)(1) + sin(α)(0)
@ajdirig I think you forgot the unit circle |dw:1338512719727:dw|
where sin(x) = y cos(x) = x at pi/2 we have the point (0,1)
Indeed, you are correct. Thank you for pointing the obvious out to me. lol
you should just memorize it really easy to do
Just remember How to memorize Radians the top numbers going counter clockwise (dont forget to add the pi next to them) Read->2 3 5 4 5 7<-Read Read->5 7 11 its pretty easy to remember these sequences of numbers they form a pattern 2 3 5 4 5 7 5 7 11 5 diagonal line 7 diagonal line, 2 3 4 top corner, and 11 bottom corner they are all divided in order at the top 3, 4, 6 and in order on the bottom 3, 4, 6 then for the coordinates you can remember pi/2 and 3pi/2 for the other coordinates just go write at the top of the fraction the x,y x,y 1, 3 2, 2 3, 1 square these numbers and divide by 2 At the bottom 3. 1 2, 2 1, 1 square these numbers and divide by 2 and you have the unit circle This makes trig really easy if you memorize this :) here is the unit circle with these instructions you should be able to memorize it easy, if you want to
wow, that was way more help than I expected! Thanks alot, this should make things a lot easier!
i made a mistake it is 1, 2, 3 for the bottom half but seriously man the unit circle is super easy to memorize
not 1, 2, 1
Thanks, I'm just old & trying to relearn Trig has been kinda trying
yeah the identities can get ugly, good thing is you only have to memorize the main ones also to remember what csc(x), sec(x) and cot(x) is If you can remember SOH CAH TOA sin(x), cos(x) tan(x) csc(x), sec(x), cot(x) I just write down the letters sct csc <- super easy to remember then it is as easy as 1/sin(x) = csc(x) 1/cos(x) = sec(x) 1/tan(x) = cos(x)/sin(x) = tan(x)
sorry not I meant 1/tan(x) = cos(x)/sin(x) = cot(x)
am I missing some way to relate sin to sec? As in if sinx= 5/13, and tan x<0, find secx?
I have no idea what you are asking
read what I just posted above and disregard that I put 1/tan(x) = cos(x)/sin(x) = tan(x) I meant 1/tan(x) = cos(x)/sin(x) = cot(x) sin(x) and sec(x) are unrelated trig functions
I know they are not related, however there must be some way to answer this question. the anser is -13/12, but I have no idea how to go about getting there
can you write the question out how it is stated?
if sinx= 5/13 and tanx<0 then secx=?
I vaguely remember these questions I would square both sides of the equation which is perfectly allowed sin^(2)(x) = (5/13)^(2) by the identity sin^(2)(x) + cos^(2)(x) = 1 we get 1 - cos^(2)(x) = (5/13)^(2) (5/13)^(2) + 1 = cos^(2)(x) take square root of both sides ((5/13)^(2) + 1)^(1/2) = cos(x) thus we remember sec(x) as 1/cos(x) thus 1/((5/13)^(2) + 1)^(1/2)
If you plan on going further in Math Memorize the three main identities, google them
but if you dont want to memorize all three just remember this one it is vital sin^(2)(x) + cos^(2)(x) = 1
I just figured it out, so simple. Pythagorean to find adjacent side, then sec= h/a
it never goes away im in calc 2 and it still appears
what did you get as an answer I'm just wondering if my thinking was correct
13/12
sorry, -13/12 because the tan<0
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