There is no figure given!!!!
However,
In a triangle the three angles are A B & 90 degrees.
The sides opposite to A,B, & 90 are 'a' , 'b' & 'h' respectively.
Hence Sin A = a/h
SinB = b/h
CosA = a/h ( Check ; 'a/h')
Csc B ( CosecantB) = 1/ (a/h) = h/a
CotB = CosB / SinB = (a/h) / (b/h) = a/b
2sin(x), or 5.4cos(y), tan(z)
sin4x=(4sinxcosx)(1-2sin^2x)
2 pi
Thanks to the pre-existing addition and subtraction theorums, we can establish the identity:sin(a+b) = sin(a)cos(b)+sin(a)cos(b)Then, solving this, we getsin(a+b) = 2(sin(a)cos(b))sin(a)cos(b) = sin(a+b)/2a=b, sosin(a)cos(a) = sin(a+a)/2sin(a)cos(a) = sin(2a)/2Therefore, the answer is sin(2a)/2.
(/) = theta sin 2(/) = 2sin(/)cos(/)
4Sin(x)Cos(x) = 2(2Sin(x)Cos(x)) = 2Sin(2x) ( A Trig. identity.
y = 2sin(x)? If that's your function, well we know that sin(x) oscillates between y = 1 and y = -1, but in our case we have double that from 2sin(x), so our range is -2 to 2.
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
R^2sin(theta)d(theta)d(phi)(r-hat)
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
3
3