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When abc are angles of a triangle prove sin2a sin2b sin2c4sinasinbsinc?
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If sin2a plus cos2a equals 1 does sin2a minus cos2a equals 1 also ... 2 means square and a means alpha?
No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a
What are the important questions in intermediate MATHS 1A?
sin2a=
Sin2A plus sin2B plus sin2C 4sinAsinBsinC proove it?
The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.
Show that sin ²A plus Cos ²A equals 1?
One definition of sine and cosine is with a unitary circle. In this case, the sine is simply equal to the y-coordinate, and the cosine, the x-coordinate. Since the hypothenuse is 1, the equation in the question follows directly from Pythagoras' Law: x2 + y2 = r2, x2 + y2 = 1, cos2A + sin2A = 1. You can also derive it from the alternative definition of sine and cosine (ratios in a right triangle).
Evaluate sin squared a and cos squared a given cos 2a equals 0.3?
This is geometry/trigonometry, not calculus. That said, cos 2a = .3 The property you need: cos 2u = 1 - 2sin2u = 2cos2u - 1 Using the first equality: cos(2a) = 1 - 2sin2a Let u = a (.3) = 1 - 2sin2a Substitute cos(2a) = .3 .7 = sin2a Subtract .3 from both sides, get rid of negative sign sin a = .71/2 Square root both sides a = sin-1 .71/2= .991156 Not a special angle, use calculator.
If sin2a plus cos2a equals 1 does sin2a minus cos2a equals 1 also ... 2 means square and a means alpha?
No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a
What is sin-cos?
Sin is sine. Cos is cosine. http://en.wikipedia.org/wiki/Sine_curve http://en.wikipedia.org/wiki/Cosine_curve In terms of trigonometric identities sin2A=2sinAcosA cos2A=cos2A-sin2A sin2A-cos2A=2sinAcosA-cos2A+sin2A === === sin(A) - cos(A) = sqrt(2)sin(A-45)
How do you learn do sums of trigonometry?
sin2A + cos2A
What are the important questions in intermediate MATHS 1A?
sin2a=
What is cos of 2 theta?
cos(A + B) = cosAcosB - sinAsinB, with A=B, cos(2A) = cos2A - sin2A, then you can use cos2A + sin2A = 1, to produce more, like: [2cos2A - 1] or [1 - 2sin2A], and others.
Sin2A plus sin2B plus sin2C 4sinAsinBsinC proove it?
The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.
How do you solve sin2a - sin4a equals cos2a - cos4a?
Probably you should start by looking up the double-angle formulas, reducing the "4a" to some combination of "2a".
Show that sin ²A plus Cos ²A equals 1?
One definition of sine and cosine is with a unitary circle. In this case, the sine is simply equal to the y-coordinate, and the cosine, the x-coordinate. Since the hypothenuse is 1, the equation in the question follows directly from Pythagoras' Law: x2 + y2 = r2, x2 + y2 = 1, cos2A + sin2A = 1. You can also derive it from the alternative definition of sine and cosine (ratios in a right triangle).
How do you differentiate 2sinAcosA?
First using the double angle formula; 2sinAcosA = sin(2A) Therefore d/dA (sin2A) = cos(2A) * 2 = 2cos(2A) We multiply by the two due to the chain rule, where we must multiply by the derivative of the angle.
What is the smallest possible value for a where y equals Sin2a reaches its maximum?
The maximum value of the sine function is 1 and that the smallest value for a would be pie divided by 2. So, if 2a= pie divided by 2, then the answer is a=pie divided by 4(simply divide both sides by 2).
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)
In maths does one minus two sine squared an equal cos squared a minus sine squared a 1-2sin2an equals sin2a-cos2a?
No.Remember: sin2 + cos2 = 1So, in place of (1 - 2 sin2) we can write (sin2 + cos2 - 2 sin2).Massage that around slightly: (sin2 + cos2 - 2 sin2) = cos2 - sin2That's not equal to (sin2 - cos2), which is the original question we were asked.
