if sinA= 4/5 and angle A is in the 1st quadrant find the value of sin2A

the expression sin2A/2cos(squared)A equals?

I have lots more too ne help would b great

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Posts: 1422

if sinA= 4/5 and angle A is in the 1st quadrant find the value of sin2A

the expression sin2A/2cos(squared)A equals?

I have lots more too ne help would b great

the expression sin2A/2cos(squared)A equals?

I have lots more too ne help would b great

Posts: 8233

maybe you want one for english as well..

Posts: 1422

yea b/c using proper english on a forum of a bunch of teenagers really impresses ne1

Posts: 3179

you have to use an identity for sin2A, i really dont feel like looking for it but i think theres a few of them

Posts: 1256

yeah i might have it in my bag if u dont no what the identity is.

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Posts: 1626

"if sinA= 4/5 and angle A is in the 1st quadrant find the value of sin2A"

so sin2A = 2sinAcosA, so sin2A = 2 (.8) (.6) = .96

someone correct me if i am wrong

so sin2A = 2sinAcosA, so sin2A = 2 (.8) (.6) = .96

someone correct me if i am wrong

Ηα†⊂h ς |< ì

Posts: 3179

hat doesnt look right, .8^2 is .64, i dont know where you got .96

Posts: 3831

there are a few ways to do this. the easiest way is to go to ur calculator, plug in sin^-1(4/5) and take the result and multiply it by two and then take the sign of that reslut. so its sin(2sin^-1(4/5))

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Posts: 10270

is that 4 or 536? cause im on 436 and we didnt do sin 2a?

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Posts: 1626

why would you square .8?

i got .96 by multiplying 2x.8x.6

.8 being the sin of A, which is approximately 53.13 degrees

.6 being the cos of A

i got .96 by multiplying 2x.8x.6

.8 being the sin of A, which is approximately 53.13 degrees

.6 being the cos of A

Ηα†⊂h ς |< ì

Posts: 3179

well i dont know im just real confused right now, in less than a month i forgot how to solve identities, and now i cant do this

Posts: 3574

ijust had a test on this today. think i did pretty well. dont feel like anymore work

use your trig properties/identities and trig values

use your trig properties/identities and trig values

yadadadamean?

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Posts: 1626

"the expression sin2A/2cos(squared)A equals?"

sin2A = 2(sinA)(cosA), so substitute it in and you get 2sinAcosA/2(cosA)(cosA), cancel a cosA in the numerator and denominator and you're left with 2sinA/2cosA which equals sinA/cosA which equals tanA

someone correct me if i am wrong again

sin2A = 2(sinA)(cosA), so substitute it in and you get 2sinAcosA/2(cosA)(cosA), cancel a cosA in the numerator and denominator and you're left with 2sinA/2cosA which equals sinA/cosA which equals tanA

someone correct me if i am wrong again

Ηα†⊂h ς |< ì

Posts: 1626

haha yeah i don't remember much of identities either, i am just getting all the stuff for this straight from my book

Ηα†⊂h ς |< ì

Posts: 4478

I'm pretty sure you're right. I didn't check the numbers but you definitely have the right method.

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"some were liek how many sides does a rhombus was i had no idea ont he cups in 5.5 gallons and what time period we are in but besides those i think im smarter thena 5th grader"

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Posts: 1422

id ont need your help ne more im going to just copy off someone at school or have them do it for me. I have like 16 problems left

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Posts: 1422

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