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Is sin 2x equals 2 sin x cos x an identity?
YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx
What are five trigonometric identities?
All others can be derived from these and a little calculus: sin2x+cos2x=1 sec2x-tan2x=1 sin(a+b)=sin(a)cos(b)+sin(b)sin(a) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) eix=cos(x)+i*sin(x)
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
In a triangle ABC b equals 15 cm and c equals 25 cm and also angle B equals 32'15'Find the side a and other angles?
By the sine rule, sin(C)/c = sin(B)/b so sin(C) = 25/15*sin(32d15m) = 0.8894 so C = 62.8 deg or 117.2 deg. Therefore, A = 180 - (B+C) = 85.0 deg or 30.5 deg and then, using the sine rule again, a/sin(A) = b/sin(B) so a = sin(A)*b/sin(B) = 28 or a = 14.3
Express cos4x sin3x in a series of sines of multiples of x?
The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).
What is the value of a Winchester 1894 Serial?
1894 winchestre rifle sin 8161 approximate value
What is the value of Winchester 1894 serial?
1894 winchestre rifle sin 8161 approximate value
Is sin b equals 2.4 possible?
No. The absolute value of the sin function cannot exceed 1.
How do you find an angle when you know one other angle and two of the sides of a triangle?
If these two sides are opposite to these angles, and you know one of the angles, you can use the Law of Sines to find the other angle. For example, in the triangle ABC the side a is opposite to the angle A, and the side b is opposite to the angle B. If you know the lengths of these sides, a and b, and you know the measure of the angle B, then sin A/a = sin B/b multiply by a to both sides; sin A = asin B Use your calculator to find the value of arcsin(value of asin b), which is the measure of the angle A. So, Press 2ND, sin, value of asin B, ).
Is sin 2x equals 2 sin x cos x an identity?
YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx
What are five trigonometric identities?
All others can be derived from these and a little calculus: sin2x+cos2x=1 sec2x-tan2x=1 sin(a+b)=sin(a)cos(b)+sin(b)sin(a) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) eix=cos(x)+i*sin(x)
Can you use the law of sines if 3 sides are given?
Yes, but you would need to know a degree measure too. [Sin(A)/a] = [Sin(B)/b] = [Sin(C)/c] [a/Sin(A)] = [b/Sin(B)] = [c/Sin(C)]
What is the approximate value of something?
The approximate value of something is an estimate.
What is the approximate value of 37?
The approximate value of 37 is 37.
How do you derived the sine law?
Consider any triangle ABC, and let AD be the altitude from A on to BC. Then sin(B) = AD/AB so that AD = AB*sin(B) and sin(C) = AD/AC so that AD = AC*sin(C) Therefore AB*sin(B) = AC*sin(C) or c*sin(B) = b*sin(C) where the lower case letter represents the side opposite the angle with the upper case name. Divide both sides by bc to give sin(B)/b = sin(C)/c. Similarly, using the altitude from B you can show that sin(A)/a = sin(C)/c. Combining with the previous result, sin(A)/a = sin(B)/b = sin(C)/c.
How do you proof the formula sin2A equals 2sinAcosA?
First, note that sin(a+b)=sin(a)cos(b)+sin(b)cos(a)[For a proof, see: www.mathsroom.co.uk/downloads/Compound_Angle_Proof.pptFor the case of b=a, we have:sin (a+a)=sin(a)cos(a)+sin(a)cos(a)sin (2a)=2*sin(a)cos(a)
How do you get the sides of the triangle when 3 angles are given and its perimeter?
You need to use the sine rule. If the three angles are A, B and C and the sides opposite them are named a, b and c then, by the sine rule, a/sin(A) = b/sin(b) = c/sin(C) Therefore b = a*sin(B)/sin(A) = a*y where y = sin(B)/sin(A) can be calculated and c = a*sin(C)/sin(A) = a*z where z = sin(C)/sin(A) can be calculated. then perimeter = p = a + b + c = a + ay + az = a*(1 + y + z) therefore a = p/(1 + y + z) or a = p/[1 + sin(B)/sin(A) + sin(C)/sin(A)]. Everything on the right hand side is known and so a can be calculated. Once that has been done, b = a*y and c = a*z.
