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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.

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Q: How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?

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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)

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).

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No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a

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)

sin2A + cos2A

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.

Probably you should start by looking up the double-angle formulas, reducing the "4a" to some combination of "2a".

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).

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.

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=

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).

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)

Firing from a hill Firing a projectile from an elevated position increases its range. If you know the initial velocity, you should be able to use the usual formulas for displacement (distance) in the horizontal and vertical directions to determine the initial vertical position.1 When you say the initial velocity is known, I assume that includes magnitude and direction. Since velocity is a vector, you should be able to calculate the vertical and horizontal components.2 If you know the horizontal velocity and the horizontal displacement (distance traveled), you should be able to calculate the time in flight. Once you determine the time in flight, you should be able to use that value in the formula for vertical displacement to determine the initial vertical displacement. Hint: The vertical displacement of the projectile when it hits the ground is zero (assuming you have selected the origin -- the axes of the plane in which the projectile is moving -- properly). ----------- 1. d = d0 + V0t + [1/2]at2, where d0 is the initial displacement, v0 is the initial velocity, and a is acceleration. For motion in the vertical direction, a = -g. For motion in the horizontal direction, a = 0 (for projectile problems). 2. Vx = Vcos(theta); Vy = Vsin(theta), where theta is the angle of elevation. Maximum range is achieved when theta = 45 degrees. At that angle, Vx = Vy.*********PLEASE NOTE: formatting has been messed up in this so things that are supposed to be raised to a power, the number is not a superscript. This needs to be relooked at.Maximum range is achieved when theta=45o only if the vertical displacement is zero (i.e. the projectile begins and ends at the same elevation). If launched from a certain height h, the angle for maximum range is given byanglemax = 1/2 cos-1 [(gh)/(v^2 + gh)]Returning to the problem, let h = launch height, R = horizontal distance from base of launch site to landing spot, V = launch speed, A = launch angle and T = time in air. The horizontal component of the launch velocity is constant since there is no acceleration in that direction. Therefore:Vx = R/TV cosA = R/TSolving for T:T = R/[V cosA]Consider the vertical part of the problem. This solution is given for a projectile launched from an angle above the horizontal so that the initial vertical component of the velocity is positive when the acceleration due to gravity (g) is negative. Also assumed is that the launch position is above the landing position. Let the initial position be the origin.d = do + viT + 1/2aT^2-h = (V sinA)T - 1/2gT^2Substituting for T:(equation A) -h = [VR sinA ]/[V cosA] - [gR^2]/[2V^2 cos2A](equation B) h = -R [tanA] + [gR^2]/[2V^2cos2A]******If you want to find the launch angle for a given height and launch speed that gives the maximum range, multiply both sides of equation A by 2cos2(theta) and rearrange to get:(g/v^2)R^2 - (2sinAcosA)R - 2hcos2A=0Using these trig identities:2sinAcosA = sin2Acos2A = 1/2 [1 + cos2A]the equation becomes:(g/v^2)R^2 - (sin2A)R - h[1+cos2A] = 0Solving for R using the quadratic formula:R(A) = v^2/(2g)[sin2A + (sin22A + (8gh/v^2)cos2A)1/2]Find the derivative ofR(A): R'(A)=v^2/(2g)[2cos2A+1/2(sin22A+(8gh/v^2)cos2A)-1/2(4sin2Acos2A+(8gh/v^2)(-2sinAcosA))]Set this equal to zero to find angle (A) for maximum range (R).Rearrange and use some trig identities to get:1/(cos22A) - v^2/(gh)(1/cos2A) - (1+v^2/(gh))=0Use the quadratic formula to solve for 1/(cos2A):1/(cos2A) = (1/2)[v^2/(gh) + (v^4/(g^2h^2)+4(1+v^2/(gh)))1/2] 1/(cos2A) = v^2/(gh) + 1A = 1/2cos-1[(gh)/(v^2+gh)]

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