0
Find each trigonometric ratio.Use the illustration on the right.1.sin A 2.sin B 3.cos A 4. csc B 5.cot B?
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
What else can I help you with?
Which trigonometric factor can be greater than 1?
2sin(x), or 5.4cos(y), tan(z)
Prove identity sin4x equals 4sinxcosx times 1 minus 2sin squared x?
sin4x=(4sinxcosx)(1-2sin^2x)
What is the period of the function y equals 2sin x?
2 pi
Use trigonometric identities to write each expression in terms of a single trigonometric function or a constant. simplify cos t sin t?
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.
How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
What is 4sin theta cos theta?
4Sin(x)Cos(x) = 2(2Sin(x)Cos(x)) = 2Sin(2x) ( A Trig. identity.
What is the range of the function y 2sin x?
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.
What does 2sin squared theta equal and why?
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
Where is the area vector for a sphere?
R^2sin(theta)d(theta)d(phi)(r-hat)
What is the derivative of sin under root theta?
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
Integral of sin 2x divided by 1 plus cosine squared x?
3
What is the second derivative of y equals 2sinx cosx?
y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)
