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How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?
You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function:
sin(A+B) = cos A sin B + sin A cos B
cos(A+B) = cos A cos B - sin A sin B
cos even fn → cos(-x) = cos(x)
To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A
The steps are to work with the left hand side, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function:
(cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A
= (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A)
= cos(A - 2A) + sin(A + 2A)
= cos(-A) + sin 3A
= cos A + sin 3A
which is the right hand side as required.
What else can I help you with?
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)
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).
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 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.
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).
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.
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.
What are the important questions in intermediate MATHS 1A?
sin2a=
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)
When abc are angles of a triangle prove sin2a sin2b sin2c4sinasinbsinc?
To prove that ( \sin^2 a \sin^2 b \sin^2 c = 4 \sin a \sin b \sin c ) for angles ( a, b, c ) of a triangle, we can use the identity ( a + b + c = 180^\circ ). The sine of angle ( c ) can be expressed as ( \sin c = \sin(180^\circ - (a + b)) = \sin(a + b) = \sin a \cos b + \cos a \sin b ). By substituting and manipulating these identities, we can derive the relationship, confirming the equality holds for the angles of a triangle.
