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(sin-cos)(sin+cos) help!!

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(sinX-cosX)(sinX+cosX) =(sin^2 - Cos^2X)
=Sin^2X - (1-Sin^2X)
=Sin^2X -1 + Sin^2X
=2Sin^2X -1

Zulfikar Ali
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Posted on Mar 15, 2009

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X=ln(seca+tana) find coshx


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.

Dec 07, 2013 | Casio ClassPad 300 Calculator

1 Answer

Find the trig function given its period cos 5 pi


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

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

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I need to rewrite Y=5(sqrt2)sin(x)-5(sqrt2)cos(x) as Y=Asin(Bx-c)


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).

1_22_2012_4_04_59_am.jpg

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

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Differentiate each of the following w.r.t.x; 29.sin2xsinx


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)]

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

1 Answer

If f(x) = sin x, what is the 99th derivative of f(x)?


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.

Oct 20, 2009 | Casio FX-300MS Calculator

1 Answer

Cos+tan(sin)=sec


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:

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

1 Answer

Help


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

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

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Solve it plz......


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

Aug 28, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

4 Answers

Trig Identities


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.

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

4 Answers

Cosh Squared or Cubed


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

(
sin
3
)
)
^
2

Sep 24, 2007 | Texas Instruments TI-89 Calculator

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