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Help 0.5(cos15 + (squer of 3).sin 15)

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Simple 0.5(cos15 + (squer of 3).sin 15) =0.5(cos15+(3x3).Sin15) =0.5(0.9659+9x0.2588) =0.5(0.9659+2.3292) =0.5x3.2951 =1.64755 Zulfikar Ali 989980221

Posted on Mar 15, 2009


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How can i find sin inverse of 54.4. As it shows that range of sin is 0 to 1. plz help asap

Could you give us the original question to see how you got 54.4? As you say, sine values can only be numbers between 0 and 1, and 0 and 1.

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Sin 43 degrees 30'

Press SIN 4 3 2ND [ANGLE] 1 3 0 2ND [ANGLE] 2 ENTER

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Note that this usually won't give you an EXACT value, but only a ten-digit approximation.

Dec 05, 2012 | Texas Instruments TI-84 Plus Calculator

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If necessary, press SHIFT MODE 3 to switch the calculator to degrees mode. Then press
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Jul 02, 2011 | Casio FX-115ES Scientific Calculator

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I typed in solve(sin(x^3-x)>0,x) and

You demand is to say the list unreasonanble. What you are trying to do is to find all the values of x that make sin(x^3-x) larger than or equal to 0. There is an infinite number of them: remember that the sine function is periodic. The algorithm of the solve( looks for the values of x where the expression inside the command nears zero.
What you need is a program for the TI that solves inequalities. You may want to go to the web site to look for such a program.

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Apr 27, 2011 | Texas Instruments TI-89 Calculator

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Where is the 2nd F key, to find the size of θ (to the nearest degree) where θ is acute. example question: sin θ = 0·259

The key is called SHIFT on Casio calculators. To get the arcsine of an angle you press [SHIFT][SIN], then enter the angle measure in the appropriate unit (degree in your case) and press ENTER. Your result should be 15.01073397 or 15, to nearest degree.

Sep 05, 2010 | Casio FX-115ES Scientific Calculator

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Why limit of sin x is 1, when x tends to 0 ?

Sin 0 = 0
Sin 90 = 1
Sin 180 = 0
Sin 270 = -1
Sin 360 = Sin 0 = 0

Is there an angle greater than 360???
Is there any greater value of Sine than 1 ???

Aug 11, 2010 | Pyramid Calculus WIZ 2.0 (cp11395) for PC,...

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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))'
  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
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)]

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1 + sinx = 2cos^2x

Hii..this can be done as follow

cos^2 x = 1 - sin^2 x--------(1)

2cos^2 x=1+ sin x
2cos^2 x - sin x -1=0
Substituting formula (1)
2(1 - sin^2 x) - sin x - 1 = 0
2sin^2 x + sin x - 1 = 0

Factor this
(2 sin x - 1)(sin x +1) = 0
2 sin x - 1 = 0 or sin x +1 = 0
sin x = 1, and sin x = -1

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What is the value of SIN15 degree and COS15 degree.....?

Hi rowanwah

The sine of an angle is only applicable is a right triangle. If you just want a number, ie, the actual value of the sine 15 degrees you can look it up on Google. Do a search for "sine and cosine functions"

If you want the mathematical description of the sine of an angle it is described as follows
In a triangle ABC, there are 3 angles angle A, angle B and angle C. There are also 3 sides, Side AB, Side AC and side BC. The sine of angle A is equal to the side opposite Angle A divided by the Hypotenuse (the longest side opposite the right angle)
The Cosine of angle A is equal to the side adjacent to Angle A divided by the hypotenuse

Hope this helps Loringh PS Please leave a rating for me Thanks

Nov 15, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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