Question

help please @thesmartone.

Answer 1

@TheSmartOne here it is

Answer 2

i need help with f and g

Answer 3

@sshayer

Answer 4

\[\cos \left( c+d \right)=\cos c \cos d-\sin c \sin d\]
put d=c
\[\cos 2c=\cos c \cos c-\sin c \sin c=\cos ^2c-\sin ^2c\]
\[=\cos ^2c-(1-\cos ^2c)=2\cos ^2c-1\]
\[2 \cos ^2 c=1+\cos 2c\]
put 2c=x
\[c=\frac{ x }{ 2 }\]
\[2 \cos ^2 \frac{ x }{ 2 }=1+\cos x,\cos \frac{ x }{ 2 }=\pm \sqrt{\frac{ 1+\cos x }{ 2 }}\]
\[or~\left| \cos \frac{ x }{ 2} \right|=\sqrt{\frac{ 1+\cos x }{ 2 }}\]
similarly you can prove for sin

Answer 5

oh daaamnnnn. thank uuuuuuuuu soooooo much @sshayer i would give u like 10000 medals if i coudlve

Answer 6

yw

Answer 7

now i got this one can u help me with g plzzzzz

Answer 8

earlier we have proved
\[\cos 2c=\cos ^2c-\sin ^2c=1-\sin ^2c-\sin ^2c=1-2\sin ^2c,\]
\[2\sin ^2c=1-\cos 2c\]
put 2c=x and complete as above.

Answer 9

\[2\sin^2 (x)=1-\cos^2 (x)\]

Answer 10

like that? @sshayer

Answer 11

\[2\sin ^2\frac{ x }{ 2 }=1-\cos x\]

Answer 12

divde by2

Answer 13

n then do square root to have sin by itself

Answer 14

n then u have sin (x/2) = radical 1-........

Answer 15

correct

Answer 16

thanksssssssss sooo much

Answer 17

but add plus minus in front.

Answer 18

yw

Answer 19

oh yeah

Answer 20

i have soo much hw n I NEEEDDD HELP WITHTHEM ALL DO U MIND HELPING ME AGAIN

Answer 21

no problem,but i am leaving for dinner.

Answer 22

have to find the roots to these

Answer 23

\[f(x)=(x^2+2x-3+2x+2)e^x=(x^2+4x-1)e^x\]
\[e^x>0~ for ~all ~real~ values.\]
Hence we have to find the roots of \[x^2+4x-1=0,x=\frac{ -4 \pm \sqrt{4^2-4*1*-1} }{ 2 },\]
\[x=\frac{ -4\pm \sqrt{20} }{ 2 }=-2\pm \sqrt{5}\]

Answer 24

wow. May I ask u a sort of persoanl question? How are you soo smart?

Answer 25

@sshayer

Answer 26

@jigglypuff314

Answer 27

@.Sam.

Answer 28

@zzr0ck3r