(sinX-cosX)(sinX+cosX)
=(sin^2 - Cos^2X)

=Sin^2X - (1-Sin^2X)

=Sin^2X -1 + Sin^2X

=2Sin^2X -1

Zulfikar Ali

ali_zulfikar@yahoo.com

9899780221

Posted on Mar 15, 2009

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Posted on Jan 02, 2017

Sorry I do not like to work with secant and cosecant.

sec(a)+tan(a)=(1+sin(a))/cos(a)

ln(sec(a)+tan(a))=** ln( (1+sin(a))/cos(a))=X**

2*cosh(X)= e^(X)+e^(-X)

**e^(X)=(1+sin(a))/cos(a)**

**e^(-X)= cos(a)/(1+sin(a))**

2cosh(X)=(1+sin(a))/cos(a) +cos(a)/(1+sin(a))= 2/cos(a)

**cosh(X)=1/cos(a)=sec(a)**

Now that you see how you can do it, I trust you will discover any mistake I might have made.

If you want to use the classPad function sequence**Action>Transformation>simplify(,** do it step by step as I have detailed above.

Good Luck.

sec(a)+tan(a)=(1+sin(a))/cos(a)

ln(sec(a)+tan(a))=

2*cosh(X)= e^(X)+e^(-X)

2cosh(X)=(1+sin(a))/cos(a) +cos(a)/(1+sin(a))= 2/cos(a)

Now that you see how you can do it, I trust you will discover any mistake I might have made.

If you want to use the classPad function sequence

Good Luck.

Dec 07, 2013 | Casio ClassPad 300 Calculator

cos(5PI)=cos(4PI+PI)=cos(PI)=-1

sin(19PI/6)=sin(18PI/6+ PI/6)=sin(3PI +Pi/6)=sin(2PI+PI+PI/6)=sin(PI+PI/6)=sin(-PI/6)=-sin(PI/6)=-1/2

sin(19PI/6)=sin(18PI/6+ PI/6)=sin(3PI +Pi/6)=sin(2PI+PI+PI/6)=sin(PI+PI/6)=sin(-PI/6)=-sin(PI/6)=-1/2

Dec 12, 2011 | Super Tutor Trigonometry (ESDTRIG) for PC

Use the fact that cos(pi/4)=sin(pi/4)= 1/square root(2). Trigonometric identity cos(a+b)=cos(a)cos(b)-sin(a)sin(b).

Nov 07, 2010 | SoftMath Algebrator - Algebra Homework...

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...

Hello,

Frankly my dear, who wants to compute the 99th derivative of sin(x)?

d/dx(sin(x))= cos(x)

d^2/dx^2 (sin(x))= d/dx( cos(x))=-sin(x)

d^3/dx^3 (sin(x))= -cos(x)

d^4/dx^4 (sin(x))=d/dx(-cos(x))=sinx.

In 100 successive differentiations you perform the foregoing sequence 100/4=25 times, and you end up with sin(x). Hence in the 99th step the derivative was -cos(x).

Hope it helps.

Frankly my dear, who wants to compute the 99th derivative of sin(x)?

d/dx(sin(x))= cos(x)

d^2/dx^2 (sin(x))= d/dx( cos(x))=-sin(x)

d^3/dx^3 (sin(x))= -cos(x)

d^4/dx^4 (sin(x))=d/dx(-cos(x))=sinx.

In 100 successive differentiations you perform the foregoing sequence 100/4=25 times, and you end up with sin(x). Hence in the 99th step the derivative was -cos(x).

Hope it helps.

Oct 20, 2009 | Casio FX-300MS 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

cos x + root 3 sin x =root 2

cos x + ö3 * Sin x = ö2

squaring both the side

(cos x + ö3 * Sin x)2 = (ö2)2

Cos2 x + 3 * Sin2 x = 2

Cos2 x + Sin2 x + 2 * Sin2 x = 2

1 + 2 * Sin2 x= 2

2 * Sin2 x = 2-1

2 * Sin2 x = 1

Sin2 x = ½

Sin x = ö½

Sin x = 1/V2= Sin 45

X = 450

cos x + ö3 * Sin x = ö2

squaring both the side

(cos x + ö3 * Sin x)2 = (ö2)2

Cos2 x + 3 * Sin2 x = 2

Cos2 x + Sin2 x + 2 * Sin2 x = 2

1 + 2 * Sin2 x= 2

2 * Sin2 x = 2-1

2 * Sin2 x = 1

Sin2 x = ½

Sin x = ö½

Sin x = 1/V2= Sin 45

X = 450

Aug 28, 2008 | 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

if you want to square sin(3), press:

(

sin

3

)

)

^

2

(

sin

3

)

)

^

2

Sep 24, 2007 | Texas Instruments TI-89 Calculator

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