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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 ali_zulfikar@yahoo.com 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


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Sin 43 degrees 30'


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HOW TO CALUCULATE SIN 30DEGREES 45MINUTES?


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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 ticalc.org web site to look for such a program.

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k24674_54.jpg

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.

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1 Answer

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


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

Is there an angle greater than 360???
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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))'
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)]

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Hii..this can be done as follow

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2cos^2 x - sin x -1=0
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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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