You are watching: If tan x° = and cos x° = , what is the value of sin x°?
We understand the signs.
We have actually $$sin x = -2sqrt2 cos x$$ and
$$sin^2 x+ cos^2 x = 1$$
Substitute the an initial equation into the second, and you deserve to solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should have the ability to recover $sin x$ making use of the first equation.
Take $sin x=y$ which means that $cos x= pmsqrt1-y^2$ wherein the sign depends on the quadrant the interest. Currently equate their ratio with $ an x$ and also solve. Again, you select the proper value of $y$ depending upon your name: coordinates of interest.
The given suggestions are great and simpler, together an alternate for a direct calculation recall the for $ hetain(-pi/2,pi/2)$
$$y= an heta iff heta=arctan y$$
and therefore since in that instance $xin(-pi/2,0)$ through $y=-2sqrt2$ we can use that by composition formulas$sin x= sin (arctan y)=fracysqrt1+y^2$$cos x= cos (arctan y)=frac1sqrt1+y^2$
Use the simple relations$$cos^2x=frac11+ an^2x,enspace extwhence quad sin^2x= an^2xcos^2x=frac an^2x1+ an^2x.$$So $sin x$ is recognized up to its sign. ~ above $igl<-fracpi2,0igr>$, that is an unfavorable or $0$.
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how to uncover $ an left(fracx2 ight) $ learning that $cos left(x ight)+sin left(x ight)=frac75 $
Verify the identification $frac an(a+b) an(a-b)$ = $fracsin(a)cos(a)+sin(b)cos(b)sin(a)cos(a)-sin(b)cos(b)$
reflecting that, if $ aneta=fracsinalpha-cosalphasinalpha+cosalpha$, then $sqrt2sineta=sinalpha-cosalpha$
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