=
Note: Conversion is based on the latest values and formulas.
What is the answer for sin theta cos theta? - Socratic 25 Feb 2018 · #Sin thetacos theta# probably is the simplest form of trigonometric expression so it may not have any answer . but it can be written as #tantheta*cos^2theta# or #cotthetasin^2theta# or #1/(secthetacsctheta)#.
cos^2(2x)+sin^2(2x)=? what value is this equal to? - Socratic 12 Jun 2018 · Well the x refers to any number so if your number is 2x, then cos^2 2x+sin^2 2x=1 You can also prove this by using the double angle formula cos^2(2x)+sin^2(2x) =(cos^2x-sin^2x)^2+(2sinxcosx)^2 =cos^4x-2sin^2xcos^2x+sin^4x+4sin^2xcos^2x =cos^4x+2sin^2xcos^2x+sin^4x =(cos^2x+sin^2x)^2 =1^2 =1
How do you simplify #sin(arccos(x))#? - Socratic 21 Oct 2016 · From Pythagoras, we have: #sin^2 theta + cos^2 theta = 1# If #x in [-1, 1]# and #theta = arccos(x)# then:. #theta in [0, pi]#
Fundamental Identities - Trigonometry - Socratic The best videos and questions to learn about Fundamental Identities. Get smarter on Socratic.
What is sin(x) times sin(x)? + Example - Socratic 16 Apr 2015 · sin(x)xxsin(x) = sin^2(x) There are other answers, for example, since sin^2(x)+cos^2(x) = 1 you could write sin(x)xxsin(x) = 1-cos^2(x) (but that's not much of a simplification)
How do you simplify #Sin(Cos^-1 x)#? - Socratic 9 May 2016 · sin(cos^(-1)(x)) = sqrt(1-x^2) Let's draw a right triangle with an angle of a = cos^(-1)(x). As we know cos(a) = x = x/1 we can label the adjacent leg as x and the hypotenuse as 1. The Pythagorean theorem then allows us to solve for the second leg as sqrt(1-x^2). With this, we can now find sin(cos^(-1)(x)) as the quotient of the opposite leg and the hypotenuse. sin(cos^( …
The value of sin(x)tan(x) is equal to what? - Socratic 27 Feb 2018 · Remember how #tan(x)=sin(x)/cos(x)#? If you substitute that in the expression above, you will get: #sin(x)*sin(x)/cos(x)# . Now it is just a matter of multiplying: #sin^2(x)/cos(x)#
What is sin (x)+cos (x) in terms of sine? - Socratic 15 Apr 2015 · Please see two possibilities below and another in a separate answer. Explanation: Using Pythagorean Identity. #sin^2x+cos^2x=1#, so #cos^2x = 1-sin^2x#
What does cosx sinx equal? - Socratic 7 Mar 2016 · cos(x)sin(x) = sin(2x)/2 So we have cos(x)sin(x) If we multiply it by two we have 2cos(x)sin(x) Which we can say it's a sum cos(x)sin(x)+sin(x)cos(x) Which is the double angle formula of the sine cos(x)sin(x)+sin(x)cos(x)=sin(2x) But since we multiplied by 2 early on to get to that, we need to divide by two to make the equality, so cos(x)sin(x) = sin(2x)/2
How do you integrate #sin(x)cos(x)#? - Socratic 2 Aug 2016 · Depending on the route you take, valid results include: sin^2(x)/2+C -cos^2(x)/2+C -1/4cos(2x)+C There are a variety of methods we can take: Substitution with sine: Let u=sin(x).