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give the formula for cos (2x) in terms of cos x and sin x

using the trig identities, I just can't remember how this is done

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Give the formula of cos(2x) in terms of cos(x) and sin(x)


Give the formula of cos(2x) in terms of cos(x) and sin(x)
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How can I simplify cos2x cosx + sin2x sinx to a single ...


... where A=2x and B=x, so we can apply the formula: Cos ... How can I simplify cos2x cosx + sin2x sinx to a single primary trigonometric function? ... #sin^2(x)+cos ...
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Give The Formula For Cos(2x) In Terms Of Cos X And... | Chegg.com


Answer to Give the Formula for Cos(2x) in terms of cos x and sin x please show stops so i ... Give the Formula for Cos(2x) in terms of cos x and sin x ... Give Us ...
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The Sine and Cosine Functions - Properties - Math2.org


Formula: sin(x + y) = sin(x)cos(y) + cos(x)sin(y) ... we can express the sine function in terms of cosine and vice versa: sin(x) ... , sin 2 (x) + cos 2 (x) ...
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1 Pretest Review x - BYU Math Department


Give the formula for cos(x y) in terms of cosx, ... Give the formula for sin(2x) in terms of cosx, sinx: Law of cosines and sines 13. In the picture, = 1 4 ˇ; = 2 3
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Double and half-angle formulas - Windstream


Doubling the sin x will not give you the value of sin 2x. ... sin 2 x + 2sin x cos ... We need to match this with the double angle formula for sin (1/2)(2sin 22.5 cos ...
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Trigonometric Identities and Formulas - analyzemath.com


sin 2 X + cos 2 X = 1 1 + tan 2 X = sec 2 X 1 + cot 2 X = csc 2 X Negative Angle Identities sin ... Difference of Squares Formulas sin 2 X - sin 2 Y = sin(X + Y)sin(X ...
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Trigonometric Identities | Purplemath


Lists the basic trigonometric identities, ... sin(2x) = 2 sin(x) cos(x) cos(2x) = cos 2 ... Terms of Use; Privacy; Contact;
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Suggested Questions And Answer :


if n is a +ive Z, and..... (Calculus)

The first part is covered by Proof by Induction, and gives us the iteration formula I_(n-1) – I_n = 2π For brevity and clarity I am going to use the notation int[f(x)] to represent the integral, from 0 to 2π, of f(x) with respect to x. To prove: int[sin^2(nx/2) / sin^2(x/2)] = 2nπ sin^2(nx/2) / sin^2(x/2) = 2(1 – cos(nx)) / ({2(1 – cos(x))} = (1 – cos(nx)) / (1 – cos(x)) Therefore, int[sin^2(nx/2) / sin^2(x/2)] = int[(1 – cos(nx)) / (1 – cos(x))] int[] = int[(1) / (1 – cos(x))] –  int[(cos(nx)) / (1 – cos(x))] But, from part 1,  int[(cos(nx)) / (1 – cos(x))] = I_n, therefore, int[] = int[(1) / (1 – cos(x))] – I_n int[] = int[(1) / (1 – cos(x))] – I_(n-1) + 2π int[] = int[(1) / (1 – cos(x))] – int[(cos((n-1)x)) / (1 – cos(x))]  + 2π int[] = int[(1 – cos((n-1)x)) / (1 – cos(x))]  + 2π, i.e. int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-1)x)) / (1 – cos(x))]  + 2π The above is an iteration formula, with the nth term on the lhs and the (n-1)th term on the rhs. We can expand this iteration sequence as follows, int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-2)x)) / (1 – cos(x))]  + 2(2π) int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-3)x)) / (1 – cos(x))]  + 3(2π) ... int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-(n-1))x)) / (1 – cos(x))]  + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos(x)) / (1 – cos(x))]  + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = int[ (1) ] + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = 2π + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = 2nπ int[sin^2(nx/2) / sin^2(x/2)] = 2nπ  
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what is cosh x in terms of i as we know that sinh x=1/i(sin ix)

QUESTION: what is cosh x in terms of i as we know that sinh x=1/i(sin ix). By definition, cosh(x) = (1/2)(e^x + e^(-x)) and e^(ix) = cos(x) + i.sin(x) and e^(-ix) = cos(x) - i.sin(x) so adding the above two equations gives us, e^(ix) + e^(-ix) = 2.cos(x) (1/2)(e^(ix) + e^(-ix)) = cos(x) i.e. cosh(ix) = cos(x) or, cosh(x) = cos(ix)
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given : sin (x+y) = sin x cos y + cos x sin y and cos (x+y) = cos x cos y-sin x sin y derive a formula for cot (x+y) and derive a formula for cos(2x)

cot(x+y)=cos(x+y)/sin(x+y)=(cosxcosy-sinxsiny)/(sinxcosy+cosxsiny). There are variations of this: divide top and bottom by cosx: (cosy-tanxsiny)/(tanxcosy+siny), then by cosy: (1-tanxtany)/(tanx+tany). Dividing by sinx and siny instead gives: (cotxcosy-siny)/(cosy+cotxsiny), then: (cotxcoty-1)/(coty+cotx). (Using these variations and putting y=x, you can work out cot(2x).) cos(2x)=cos^2(x)-sin^2(x), putting y=x. This can also be written: 1-2sin^2(x) or 2cos^2(x)-1 because sin^2(x)+cos^2(x).
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intergration

Question: intergrate the following question:  sin(3x+a).sin(x-a). There is a formula for the product of two sines. sinA.sinB = (1/2)(cos(A-B) - cos(A+B)) Using this formula gives us, I = int sin(3x+a).sin(x-a) dx = int (1/2)(cos((3x+a) - (x-a)) - cos((3x+a) + (x-a))) dx I = int (1/2)(cos(2x+2a) - cos(4x)) dx = (1/2)((1/2)sin(2x+2a) - (1/4)sin(4x)) Answer: I = (1/4).sin(2x+2a) - (1/8).sin(4x)  
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intergration

Question: intergrate the following question:  sin(3x+a).sin(x-a). There is a formula for the product of two sines. sinA.sinB = (1/2)(cos(A-B) - cos(A+B)) Using this formula gives us, I = int sin(3x+a).sin(x-a) dx = int (1/2)(cos((3x+a) - (x-a)) - cos((3x+a) + (x-a))) dx I = int (1/2)(cos(2x+2a) - cos(4x)) dx = (1/2)((1/2)sin(2x+2a) - (1/4)sin(4x)) Answer: I = (1/4).sin(2x+2a) - (1/8).sin(4x)
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give the formula for cos (2x) in terms of cos x and sin x

kosine(2x)=kosine^2 (x) -sine^2 (x) & then yu hav sine^2+kosine^2=1, so yukan get variashuns on this
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give the formula for cos (2x) in terms of cos x and sin x

kosine(2x)=kosine^2(x) - sine^2(x)
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Find a power series for pi.

Start with 1/(1+w) = 1 - w + w2 - w3 + ... Now subsitute x2 for w: 1/(1+x2) = 1 - x2 + x4 - x6 + ... Then integrate both sides (from x=0 to x=y): arctan y = y - y3/3 + y5/5 - y7/7 +... and plug in y=1, to get Pi/4 = 1 - 1/3 + 1/5 - 1/7 + ... Voila! There are other pi formulas that converge faster. Presentation Suggestions: An alternate way to present this is to start with the well-known formula for Pi, and then present this as a "justification". The Math Behind the Fact: Well, we glossed over the issue of why you can integrate the infinite series term by term, so if you wish to learn about this and more about Taylor series, this material is often covered in a fun course called real analysis.  
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inverse properties

All the trig functions have inverses: inverse of sin is arcsine or sin-1; cos is arccos or cos^-1; tan is arctan or tan^-1; and so on. The inverse property is sin(arcsin(x))=arcsin(sin(x)). This property is shared by all the inverses: if f is the trig function and f^-1 is the inverse, then f(f^-1(x))=f^-1(f(x)). The limits for the angles are between -90° and +90°, and limits for the inverses are between -1 and +1. However, some of the functions are not defined at the extremes: at 0, 90°, -90°, for example, tan 90° cannot be defined (infinity). There are also inverses to the reciprocal functions: cosecant or cosecant or csc (1/sin), secant or sec (1/cos), cotangent or cot (1/tan). These give rise to other properties: sin^-1(1/x)=cosec-1(x), cos^-1(1/x)=sec^-1(x), etc. It's useful to draw a right-angled triangle to help picture the relationship between sides and angles in terms of the trig functions, their reciprocals, their inverses, and the inverses of the reciprocals.
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What is the exact value (not decimal) of the sin of a 40 degree angle of a right triangle?

sin3A=sin(2A+A)=sin2AcosA+cos2AsinA=2sinAcos^2A+sinA(1-2sin^2A)=  2sinA(1-sin^2A)+sinA-2sin^3A=2sinA-2sin^3A+sinA-2sin^3A=3sinA-4sin^3A.  If A=40, sin120=3sin40-4sin^3(40)=sqrt(3)/2. Put x=sin40: 3x-4x^3-sqrt(3)/2=0 or 8x^3-6x+sqrt(3)=0, which reduces to: x^3-(3/4)x+sqrt(3)/8=0, so one solution of this cubic will be sin40. The cubic will also give sin20 as a solution, because sin120=sin60=sin(3*20)=sqrt(3)/2, and sin(-80) (=-sin80), because sin(-240)=sqrt(3)/2 (quadrant II), the same as sin120. sin40 is the most positive solution, of course, and it is an irrational number so cannot be represented as a fraction a/b where a and b are integers. The cubic factorises: (x-sin20)(x-sin40)(x+sin80)=0=x^3-x^2(sin20+sin40-sin80)+x(sin20sin40-sin20sin80-sin40sin80)+sin20sin40sin80. Comparing this with the cubic coefficients, sin20+sin40-sin80=0 because there is no x^2 coefficient, and sin40=sin80-sin20 (check: sin80-sin20=2cos((80+20)/2)sin((80-20)/2)=2cos50sin30=2sin40sin30=sin40, because sin30=1/2). Similarly, we can equate expressions for -3/4 and sqrt(3)/8, relating sin40 to sin20 and sin80. sin40=sin80-sin20 can be written sin(2*20)=sin(2*40)-sin20; 2sin20cos20=2sin40cos40-sin20; 2sin20cos20=4sin20cos20(2cos^2(20)-1)-sin20; 2cos20=4cos20(2cos^2(20)-1)-1; 8cos^3(20)-6cos20-1=0. Writing y=cos20: 8y^3-6y-1=0, which contains no irrational numbers, but is otherwise similar to the cubic obtained earlier. Although the solution includes cos20, sin40 can easily be calculated from it. Other solutions are cos100 (cos(180-80)) and cos140 (cos(180-40)). If the solution to the cubic equations can be expressed in terms of square roots, or other roots, I'll update this answer later.
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