How do you evaluate #tan(arctan(10))#? - Socratic
2017年1月25日 · tan(arctan(10))=10 Because arctan(x)=tan^(-1)(x), we can see that tan(x) and arctan(x) are inverses of each other. Remember that f(f^(-1)(x))=x In this case, replace f with tan and x with 10. Then we get tan(tan^(-1)(10))=tan(arctan(10))=10
What is tan(arctan 10)? - Socratic
2015年2月5日 · tan and arctan are two opposite operations. They cancel each other out. Your answer is 10. Your formula in words would be: "Take the tangent of an angle. This angle has a size that 'belongs' to a tangent of 10" arctan 10=84.289^0 and tan 84.289^0=10 (but you don't have to do all this) It's a bit like first multiplying by 5 and then dividing by 5. Or taking the square root of a number and then ...
How do you show arctan (x) ± arctan (y) = arctan [ (x ± y ... - Socratic
2016年5月25日 · How do you show #arctan (x) ± arctan (y) = arctan [(x ± y) / (1 ± xy)] #? Trigonometry Trigonometric Identities and Equations Sum and Difference Identities 1 Answer
How do you simplify the expression #sin(arctan x)#? - Socratic
2016年7月31日 · sin(arctan x)=x/(+-sqrt(x^2+1)) The solution: arctan x is an angle whose tangent function =x/1 Considering the sides of the right triangle.
How do you find the exact value of arctan 1? | Socratic
2017年7月13日 · arctan 1 = pi/4 Based on the definition of arctan, you know that tan theta = 1; therefore, arctan 1 = theta. The only values of theta that make the equation tan theta = 1 true are pi/4 and (5pi)/4. However, we must consider the range of the arctan function, which is (-pi/2, pi/2). Only pi/4 falls into this interval. Thus, arctan 1 = pi/4.
How do you find the exact value of #tan(arctan(10))#? - Socratic
Answer is 10 Suppose arc tan 10 = x, then tan x= 10. Now, this what the problem is about. Thus tan (arctan10)= 10
How do you find the exact value of #arctan(sqrt(3))#? - Socratic
2015年2月13日 · arctan( sqrt (3) ) = 60^@ or pi/3 radians If theta = arctan ( sqrt (3)) then tan (theta) = sqrt (3) by definition of tan-arctan relationship
How do you evaluate #arctan(-1/sqrt3)#? - Socratic
2016年7月11日 · How do you use inverse trigonometric functions to find the solutions of the equation that are in
How do you simplify #cos(arctan(x)) - Socratic
2016年7月26日 · Let a = arctan ( x ). The principal value of #a in ( − π/2, π/2 )#. Then #tan a = x# and #0 <= cos a in [ 0, 1 )# ( wrongly marked as [ 0, - 1 ], in my . previous answer, two years ago). Now, the given expression is. #cos a = 1/sqrt ( 1 + x^2 ), x in ( - pi/2, pi/2 )#. It is important that #cos a >= 0#, for #a in Q_1# or #Q_4#.
How do you calculate the #arctan(0)#? - Socratic
2015年11月21日 · arctan(0) = 0 arctan(0) is at which at which x tan(x) = 0 and -pi/2 < x < pi/2 (or -90 < x < 90 if you use degrees) tan(x) = sin(x)/cos(x) tan(x) = 0 when sin(x) = 0 Use the attached unit circle from wikipedia: x = 0+n*2*pi or x = pi+n*2*pi so in this case x= n*pi where n is an integer Only x = 0 satisfies -pi/2 < x < pi/2 sin(0)=0 tan(0)=0 arctan(0)=0
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