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trig inequality (1 Viewer)
- Thread starter kractus
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Average Boreduser
Rising Renewal
You factorise brother. Take 6 to the other side, maybe let tank=u or smn initially, then solve for tanx
kractus
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ye, where do u go from there tho to isolate x or do u keep it like that?(tan x - 2) < 0 \\ \\ \therefore -3 < tan x < 2)
Average Boreduser
Rising Renewal
Keep it like that
Average Boreduser
Rising Renewal
I’m in the car so I can’t be really that accurate, but from the looks of it, most likely you take reciprocal of cosec and sinx/cosx of tan and try making into form of the rhs
You want make use of the following theorems:
f^-1 (f(x)) = x
f(f^-1(x)) = x
# The theorem are fairly easy to prove if (x, y) is point of "f" than (y, x) is a point of f^-1 // By definition of inverse function
# Similarly if (x, y) is point of "f^-1" than (y, x) is a point of f
You will also need to consider the unit circle definition of tan(theta) i.e. tan(theta) = y/x as well as ASTC.
kractus
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oh bruh thats such an easy proof i just think of compound angle straight away cause like tanpi/4 is just 1 its so tempting lol}+\frac{\cos(4x)}{\sin(4x)}\\\\=\frac{\cos(4x)+1}{sin(4x)}\\\\=\frac{2\cos^2(2x)-1+1}{2\sin(2x)\cos(2x)}\\\\=\frac{(\cos^2(x)-\sin^2(x))^2}{2\sin(x)\cos(x)(\cos^2(x)-\sin^2(x))}\\\\=\frac{\cos^2(x)-\sin^2(x)}{2\sin(x)\cos(x)}\\\\=\frac{1}{2}(\cot(x)-\tan(x))\\\\=RHS)
kractus
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wait i thought it was pi/4 I'm blind ahhahayeah but then you have to find out=?????)
You want to isolate the x. But because it's an inequality, it's a little tricky because you can't just use the inverse tangent function without accounting for what your domain is. Remember that the inverse tangent function takes the range (-pi/2,pi/2)
One way to do it is to draw a graph. So you want to graph y = tan(x), y=3 and y=-2, and then find the parts of the tangent graph that are 'between the two lines'. Depending on your domain, you might get multiple segments.
For example, if your domain is