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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).
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.
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.
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.
