Question

hey CAN U PLSS HELP ME WITH TRIGONOMETRY

Answer 1

rewrite \[\sin(2x)\] using the "double angle" formula
do you know it?

Answer 2

ya \[\sin (sx)=2\sin x \cos x=2\tan x \div 1+\tan ^{2}x\]

Answer 3

use the first one

Answer 4

sorry i am slow at typing

Answer 5

\[2\cos(x)+2\cos(x)\sin(x)=0\] factor etc

Answer 6

how would i solve that

Answer 7

factor out the common factor of \(2\cos(x)\) then set each factor equal to zero and solve

Answer 8

can u show me one plsss

Answer 9

\(\color{#000000 }{ \displaystyle a+ab~~\Longrightarrow ~~a\cdot 1+a\cdot b~~\Longrightarrow ~~a(1+b) }\)

Answer 10

Same there, bu instead of «a» and «b», you have cosine and sine (respectively).

Answer 11

knowing that
\(\color{#000000 }{ \displaystyle 2a+2ab~~\Longrightarrow ~~2a\cdot 1+2a\cdot b~~\Longrightarrow ~~2a(1+b) }\)

\(\color{#000000 }{ \displaystyle 2\cos(x)+2\cos(x)\sin(x) ~~\Longrightarrow ~~2\cos(x)\cdot 1+2\cos(x)\cdot \sin(x)~~\Longrightarrow ~~? }\)

Answer 12

thnx guys