Quardartic equation x2 sin2α cos2θ + 4xy sinα sin θ + y2 [4cosα ? (1+ cosα)2] = 0

csc(x)=1/sin(x)
sec(x)=1/cos(x)
csc(x)/sec(x)=(1/sin(x))*(cos(x)=cot(x)
csc(x)*cot(x)/sec(x)=(cot(x))^2=(tan(x))^(-2)

4/sqroot2 is the same value as 2sqroot2, about 2.828 . It's just a matter of how the calculator and the textbook chooses to represent the value.

To enter powers of trigonometric functions you must enclose the functions in parentheses and then apply the exponent to the whole. For example X1t =(sin T)^3 , Y1t=(cos T)^3 will give you a shape similar to a rhombus with concave sides. The symbol ^syands for the operation of raising to a power. The key is the one with the caret symbol ^ , and it is wedged between the xsqure and EXIT keys (third row of keys).
As regards the cotangent, use the equivalent definition cot(X)=1/tan(X).

Make sure your calculator is in degree mode.
It is probably in radian mode so it can't calculate the inverse sin.
To change this, go to Mode -> Deg.
Hope this helps, cheers!

Use the rule for differentiating products of functions: ()' signifies derivative
(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'
But

1. (29)'=0 derivative of a constant is zero
2. (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
3. (sin(X))'=cos(X)

Result is
(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form
sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]
then calculated the derivative of
29/2*[cos(X)-cos(3X)]
which is
29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent
29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

5sin(x)+1 = 0 is the equation you want to solve?
so
5sin(x) = -1
sin(x) = -(1/5)
arcsin( sin(x) ) = arcsin( -(1/5) )
x = -.201 (radians)
x = -11.5369 (degrees)

This is a trigonometry problem not a calculator's.
cos(x) +tan(x).sin(x) = ( cos(x) + (sin(x)/cos(x)).sin(x). After reduction to the same denominator (which is cos(x)) you obtain

{cos(x).cos(x) + sin(x).sin(x)} divided by cos(x).
The content of the bracket above is just 1.
Your fraction will have 1 as numerator and cos(x) as denomitor. That is exactly the definition of the secant function i.e. Function sec is the reciprocal (not the inverse) of the cos function, while the arccos is the inverse of cos.

A mild advice: Avoid writing function without specifying the argument (the variable on which a function acts).

Note:

sec^4X- sec^2X = 1/cot^4X + 1/cot^2X
RHS
1/cot^4X + 1/cot^2X
=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)
=Sin^4X/Cos^4X + Sin^2X/Cos^2X
=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X
=Sin^2X/Cos^4(Sin^2X + Cos^2X)
=Sin^2X/Cos^4X
=(1-Cos^2X)/Cos^4X
=1/Cos^4X - Cos^2X/Cos^4X
=1/Cos^4X - 1/Cos^2X
=Sec^4X - Sec^2X
=LHS

Change csc to 1/sin. Find a common denominator and add the two left terms.
1/sin - sin = (1 -sin^2)/sin. Rewrite formula
(1 - sin^2)/sin = cos^2/sin Divide out the /sin.
1 - sin^2 = cos^2 Rearange.
1 = cos^2 + sin^2 Yes, that's true. It's like the Pythagorean formula.

Mollweide's equations?


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