Question

7) If sin A+sin^2 A=1,find cos^2 + cos^4 A

Answer 1

are you missing an A?

Answer 2

zarkon!

Answer 3

looks that way

Answer 4

yea! missed one A..

Answer 5

If sin A+sin^2 A=1,find cos^2 A + cos^4 A

Answer 6

\[\sin^2(A)+\sin(A)-1=0\]
solve
\[x^2+x-1=0\]
\[x=\frac{-1 \pm \sqrt{5}}{2}\]
--------------
if \[\sin(A)=\frac{-1+\sqrt{5}}{2}\]

Imagine we have a right triangle again
sin means opp/hyp=(-1+sqrt{5}}/2

\[adj=\sqrt{2^2-(-1+\sqrt{5})^2}=\sqrt{4-(1-2\sqrt{5}+5)}=\sqrt{-2+2\sqrt{5}}\]

so
\[\cos(A)=\frac{-2+2\sqrt{5}}{2}=>\cos^2(A)=\frac{(-2+2\sqrt{5})^2}{2^2}\]
and \[\cos^4(A)=\frac{(-2+2\sqrt{5})^2}{2^4}\]

Answer 7

gosh i did something wrong

Answer 8

sctrach everything i said after so cos(A)=

Answer 9

\[\cos(A)=\frac{\sqrt{-2+2\sqrt{5}}}{2} \]
there we go

Answer 10

\[\cos^4(A)+\cos^2(A)=\frac{(\sqrt{-2+2\sqrt{5}})^4}{2^4}+\frac{(\sqrt{-2+2\sqrt{5}})^2}{2^2}\]

Answer 11

\[=\frac{(-2+2\sqrt{5})^2+4(-2+2\sqrt{5})}{16}\]

Answer 12

\[=\frac{4-8\sqrt{5}+20-8+8\sqrt{5}}{16}=\frac{-4}{16}=\frac{-1}{4}\]

Answer 13

the answer turned out beautiful didnt expect that

Answer 14

wait made a mistake

Answer 15

you must have made a mistake
it can't be negative

Answer 16

\[=\frac{4+20-8}{16}=\frac{24-8}{16}=\frac{16}{16}=1\]

Answer 17

i forgot the +20 there

Answer 18

completely ignored him

Answer 19

correct !

Answer 20

but i probably did it the long way
and usually when you choose long way you will make mistakes

Answer 21

want to see another way ;)

Answer 22

lol yes shorter?

Answer 23

wait!

Answer 24

let me try something else

Answer 25

shorter...maybe not...though I don't need to solve for A

Answer 26

or the sin(A)

Answer 27

ok go ahead

Answer 28

ok...I'm going to write in my latex editor then paste it here

Answer 29

ok

Answer 30

$$\cos^2(A)+\cos^4(A)$$

$$=\cos^2+(1-\sin^2(A))^2$$

$$=1-\sin^2(A)+1-2\sin^2(A)$$

$$=2-3\sin^2(A)+\sin^4(A)$$

$$=2-3\sin^2(A)+(\sin^2(A))^2$$

$$=2-3(1-\sin(A))+(1-\sin(A))^2$$

$$=2-3+3\sin(A)+1-2\sin(A)+\sin^2(A)$$

$$=\sin(A)+\sin^2(A)=1$$

Answer 31

using the \[\sin^2(A)=1-\sin(A)\] which is given

Answer 32

I think I might be missing a piece

Answer 33

i see a mistake zarkon made a mistake :)
well its only a little typo
you are missing an A on 2nd line

Answer 34

$$\cos^2(A)+\cos^4(A)$$

$$=\cos^2+(1-\sin^2(A))^2$$

$$=1-\sin^2(A)+1-2\sin^2(A)+\sin^4(A)$$

$$=2-3\sin^2(A)+\sin^4(A)$$

$$=2-3\sin^2(A)+(\sin^2(A))^2$$

$$=2-3(1-\sin(A))+(1-\sin(A))^2$$

$$=2-3+3\sin(A)+1-2\sin(A)+\sin^2(A)$$

$$=\sin(A)+\sin^2(A)=1$$

Answer 35

lol that too...

Answer 36

$$\cos^2(A)+\cos^4(A)$$

$$=\cos^2(A)+(1-\sin^2(A))^2$$

$$=1-\sin^2(A)+1-2\sin^2(A)+\sin^4(A)$$

$$=2-3\sin^2(A)+\sin^4(A)$$

$$=2-3\sin^2(A)+(\sin^2(A))^2$$

$$=2-3(1-\sin(A))+(1-\sin(A))^2$$

$$=2-3+3\sin(A)+1-2\sin(A)+\sin^2(A)$$

$$=\sin(A)+\sin^2(A)=1$$

Answer 37

I believe I have it now

Answer 38

yes looks good
i see no mistakes

Answer 39

good :)

Answer 40

very nice zarkon

Answer 41

you too

Answer 42

wow! :D thanx! :D

Answer 43

i keep on doing his the long way
and you always find a shorter way
for some reason i'm stuck on imagining a right triangle

Answer 44

I first solved it on my calculator (doing it a similar way as you) ...so I knew the answer was 1 ;)

Answer 45

i bet i will not change
and i will keep doing it this way lol

Answer 46

how do you do it on a cal?

Answer 47

you said you have n-spire?

Answer 48

or was that someone else?

Answer 49

yes

I had the calculator solve for A in

\[sin(A)+\sin^2(A)=1\]

for A (restricting A between 0 and 2pi)
then evaluateed

\[\cos^2(A)+\cos^4(A)\]

with the A the calc found

Answer 50

i probably can't do that with ti83

Answer 51

well i guess i can graph the first one and approximate solution
nvm i can do it

Answer 52

you probably could...it just wouldn't be as nice...my nspire solved the first equation obtaining exact values...on the 83 you will get numerical approximations.

Answer 53

i got 1.0000000071

Answer 54

lol

Answer 55

that was with approximation though

Answer 56

I figured it would be a little off due to rounding

Answer 57

so the answer is an approximation

Answer 58

the nspire gave me the exact answer ... 1

Answer 59

we can use newton's method to get us closer to this irrational number
and then getting us closer to 1 lol

Answer 60

what bout this sum guys?! If {x+cot40/tan50} - {1/2(cos35/sin35)}=4;find x

Answer 61

just isolate x

Answer 62

\[x=4-\frac{\cot(40)}{\tan(50)}+\frac{1}{2}*\frac{\cos(35)}{\sin(35)}\]

Answer 63

just like solving
x+3-5=4 for x
x=4-3+5

Answer 64

trig(some number) is just a regular number like any other number

Answer 65

what bout this? P.T {sin theta*cos(90-theta)*cos theta/sin(90-theta)} + {cos theta*sin(90-theta)*sin theta/cos(90-theta)}