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Mathematics 23 Online
OpenStudy (anonymous):

sin2x/tanx ??

OpenStudy (anonymous):

2sinxcosx/tan .... what do i do after that?

OpenStudy (anonymous):

@freckles

OpenStudy (anonymous):

@satellite73 @bohotness

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle \frac{\sin^2x }{\tan x} }\) like this?

OpenStudy (anonymous):

no sin2x/tanx

OpenStudy (solomonzelman):

If so, use the fact that \(\large\color{black}{ \displaystyle \tan x = \frac{\sin x }{\cos x} }\). \(\normalsize\color{black}{ \displaystyle \frac{\sin^2x }{\tan x}~\color{blue}{\Rightarrow}~(\sin^2 x)\div(\tan x)~\color{blue}{\Rightarrow}~(\sin^2 x)\div\left(\frac{\sin x }{\cos x}\right)~\color{blue}{\Rightarrow}~(\sin^2 x)\times\left(\frac{\cos x }{\sin x} \right). }\)

OpenStudy (solomonzelman):

then the sines cancel, and you are multiplying the sine function times the cosine.

OpenStudy (anonymous):

2sinxcosx/sinx/cosx

OpenStudy (solomonzelman):

oh \(\large\color{black}{ \displaystyle (\sin {\tiny ~}2x)\div(\frac{\sin x }{\cos x})}\) like this ?

OpenStudy (anonymous):

what is the next step?

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle\color{blue}{ (\sin {\tiny ~}2x)}=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x) \\ =\color{blue}{2\sin(x)\cos(x)}}\)

OpenStudy (solomonzelman):

replace the sin(2x) by the expression in terms of an angle of (a single) x. And then simplify by canceling things out

OpenStudy (anonymous):

2cos^2x ?

OpenStudy (solomonzelman):

yup

OpenStudy (solomonzelman):

well done

OpenStudy (solomonzelman):

By the way, do you see me viewing the question? I don't see that, that is why I am asking you..... (The site is really bad for me after the recent update)

OpenStudy (anonymous):

THANK YOU SO MUCH! Can you help me with another question please?

OpenStudy (anonymous):

Which of the following are solutions for this equation: 2cos x sin x - cos x = 0 Answers are in degrees a)30 b)90 c) both 30 and 90 d) neither 30 nor 90

OpenStudy (anonymous):

@freckles @Nnesha the second question ^^

OpenStudy (freckles):

you can factor a cos(x) from the left hand expression

OpenStudy (freckles):

cos(x)(2sin(x)-1)=0 set both factors equal to 0 cos(x)=0 or 2sin(x)=1

OpenStudy (freckles):

or if you want a cheaper less expensive way just plug in 30 and 90 and see which work

OpenStudy (anonymous):

the answer will be both 30 and 90.... cosx=0 is the same as cosx=90 degrees and sinx=1/2 is 30 degrees

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