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)
Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).
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