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Posted on Jan 02, 2017

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)

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)

Jul 12, 2014 | Super Tutor Trigonometry (ESDTRIG) for PC

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.

Dec 30, 2012 | Office Equipment & Supplies

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).

As regards the cotangent, use the equivalent definition cot(X)=1/tan(X).

Jul 26, 2011 | Casio FX-9750GPlus Calculator

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!

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!

Mar 06, 2011 | Texas Instruments TI-30 XIIS Calculator

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

(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)]

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

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

(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)]

Jun 21, 2010 | Vivendi Excel@ Mathematics Study Skills...

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)

so

5sin(x) = -1

sin(x) = -(1/5)

arcsin( sin(x) ) = arcsin( -(1/5) )

x = -.201 (radians)

x = -11.5369 (degrees)

Jun 04, 2010 | Texas Instruments TI-84 Plus Calculator

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:

{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

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

Note:

Aug 14, 2009 | Casio FX-115ES Scientific Calculator

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

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

Feb 02, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

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.

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.

May 22, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

Aug 23, 2014 | Nokia 6233 Cellular Phone

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