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Posted on Jan 02, 2017

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Let use an example to convert from polar to rectangular. There may be other ways to do it on this particular calculator, but this should work on all calculators.

Let's use 20 degrees and 5 units for our polar.

Sin 20 degrees = y/5, cross multiplying we get y = 5 sin 20 degrees.

Similarly, Cos 20 degrees = x/5. Cross multiplying we get x = 5 cos 20 degrees.

Thus, I get (1.71, 4.70) as the rectangular coordinates of the point.

Good luck.

Paul

Let's use 20 degrees and 5 units for our polar.

Sin 20 degrees = y/5, cross multiplying we get y = 5 sin 20 degrees.

Similarly, Cos 20 degrees = x/5. Cross multiplying we get x = 5 cos 20 degrees.

Thus, I get (1.71, 4.70) as the rectangular coordinates of the point.

Good luck.

Paul

Nov 06, 2015 | Canon F-603 Calculator

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

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

try change the phone or try reflashing it

Aug 22, 2008 | Nokia 6233 Cellular Phone

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

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