Question

about trigonometric identities . answer this pls (sinx+cosx)^4 = (1+2sinxcosx)?

Answer 1

have you tried to expand the LHS?

Answer 2

not yet

Answer 3

pls help me

Answer 4

where did you get this question from? i think there's something missing.

Answer 5

in my book .

Answer 6

RHS should be (1+2sinxcosx)^2

Answer 7

oh sorry typos .

Answer 8

(sinx+cosx)^4 = (1+2sinxcosx)^2

Answer 9

that's what I thought.
so if you expand the LHS you get (sinx +cosx)^2 x (sinx +cosx)^2

Answer 10

then we have (sin^2x + 2sinxcosx + cos^2x)(sin^2x + 2sinxcosx + cos^2x)
see if you can do the rest?

Answer 11

oh pls continue

Answer 12

you used distributive ?

Answer 13

\[[(sinx +cosx)^2]^2\]
\[[\sin^2x+\cos^x + 2sinx cosx]^2\]
\[(1+2sinx cosx)^2\]

Answer 14

then ?

Answer 15

LHS = RHS , Hence Proved.

Answer 16

ah what identity did u used ?

Answer 17

\[\sin^2x + \cos^x = 1\]

Answer 18

pythagorean . can you pls answer this too 1+sinx/cos x= cos x/1+sinx

Answer 19

As in \[\frac{1 + \sin x}{\cos x} = \frac{\cos x}{1 + \sin x}\]

Answer 20

yes

Answer 21

please check your question i think it should be 1-sinx

Answer 22

okay wait

Answer 23

\[\frac{(1+\sin x)(1-sinx) }{ cosx(1-sinx) }\] multiplying and dividing by (1-sinx)

Answer 24

Too many typos?

Answer 25

\[\frac{ 1-\sin^2x }{ cosx(1-sinx) }\]
\[\frac{ \cos^2x }{ cosx(1-sinx) }\]
\[\frac{ cosx }{ 1-sinx }\]

Answer 26

Well done cp