sin2x/tanx ??
2sinxcosx/tan .... what do i do after that?
@freckles
@satellite73 @bohotness
\(\large\color{black}{ \displaystyle \frac{\sin^2x }{\tan x} }\) like this?
no sin2x/tanx
If so, use the fact that \(\large\color{black}{ \displaystyle \tan x = \frac{\sin x }{\cos x} }\). \(\normalsize\color{black}{ \displaystyle \frac{\sin^2x }{\tan x}~\color{blue}{\Rightarrow}~(\sin^2 x)\div(\tan x)~\color{blue}{\Rightarrow}~(\sin^2 x)\div\left(\frac{\sin x }{\cos x}\right)~\color{blue}{\Rightarrow}~(\sin^2 x)\times\left(\frac{\cos x }{\sin x} \right). }\)
then the sines cancel, and you are multiplying the sine function times the cosine.
2sinxcosx/sinx/cosx
oh \(\large\color{black}{ \displaystyle (\sin {\tiny ~}2x)\div(\frac{\sin x }{\cos x})}\) like this ?
what is the next step?
\(\large\color{black}{ \displaystyle\color{blue}{ (\sin {\tiny ~}2x)}=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x) \\ =\color{blue}{2\sin(x)\cos(x)}}\)
replace the sin(2x) by the expression in terms of an angle of (a single) x. And then simplify by canceling things out
2cos^2x ?
yup
well done
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)
THANK YOU SO MUCH! Can you help me with another question please?
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
@freckles @Nnesha the second question ^^
you can factor a cos(x) from the left hand expression
cos(x)(2sin(x)-1)=0 set both factors equal to 0 cos(x)=0 or 2sin(x)=1
or if you want a cheaper less expensive way just plug in 30 and 90 and see which work
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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