Calculation: We have trigonometric expression as follows: sin (tan-a) First solve tan-x Let e = tanx We have tan 0=x sin 0 = Thus, 0 = sin VI Now, we have sin (tan-x) = sin (sin Also, we know sin (sin x) = x %3D Hence, we have sin sin
Calculation: We have trigonometric expression as follows: sin (tan-a) First solve tan-x Let e = tanx We have tan 0=x sin 0 = Thus, 0 = sin VI Now, we have sin (tan-x) = sin (sin Also, we know sin (sin x) = x %3D Hence, we have sin sin
Calculation: We have trigonometric expression as follows: sin (tan-a) First solve tan-x Let e = tanx We have tan 0=x sin 0 = Thus, 0 = sin VI Now, we have sin (tan-x) = sin (sin Also, we know sin (sin x) = x %3D Hence, we have sin sin
Hello, I don't underdstand chapter 4.7 problem 87 even with the instructions on this site. Please help?
I've attached an image of the steps. I'm lost from sin theta = x/(sqrt1 + x^2).
Where is that coming from? My class skipped a few chapters so maybe that was explained before? I was trying to visualize and understand by drawing a triangle on the unit circle but that didn't help.
Transcribed Image Text:Calculation:
We have trigonometric expression as follows:
sin (tan-lx)
First solve
tan
Let 0 = tan-x
We have
tan 0=x
sin 0 =
Thus,
0= sin
Now, we have
sin (tan-x) = sin ( sin
Also, we know
sin (sin-!x) = x
Hence, we have
sin ( sin
sin (tan 3) = T
Conclusion:
Hence, we have sin (tan-x)
Two-dimensional figure measured in terms of radius. It is formed by a set of points that are at a constant or fixed distance from a fixed point in the center of the plane. The parts of the circle are circumference, radius, diameter, chord, tangent, secant, arc of a circle, and segment in a circle.
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