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How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
How do you confirm this trig identity?
sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
Cot 70 plus 4cos70 equals?
cot 70 + 4 cos 70 = cos 70 / sin 70 + 4 cos 70 = cos 70 (1/sin 70 + 4) = cos 70 (csc 70 + 4) Numerical answer varies, depending on whether 70 is in degrees, radians, or grads.
Show that cos3t equals 4cos cubed t - 3cos t?
cos(3t) = cos(2t + t) = cos(2t)*cos(t) - sin(2t)*sin(t) = [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin(t)*sin(t) = [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin2(t) then, since sin2(t) = 1 - cos2(t) = [2*cos2(t) - 1]*cos(t) - 2*cos(t)*[1 - cos2(t)] = 2*cos3(t) - cos(t) - 2*cos(t) + 2*cos3(t) = 4*cos3(t) - 3*cos(t)
How do you express cosine in terms of cotangent?
cos(x)=sin(x-tau/4) tan(x)=sin(x)/cos(x) sin(x)=tan(x)*cos(x) cos(x)=tan(x-tau/4)*cos(x-tau/4) you can see that we have some circular reasoning going on, so the best we can do is express it as a combination of sines and cotangents: cos(x)=1/cot(x-tau/4)*sin(x-tau/2) tau=2*pi
What is the exact value using a sum or difference formula of the expression cos 11pi over 12?
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
What is the exact value of the expression cos 7pi over 12 cos pi over 6 -sin 7pi over 12 sin pi over 6?
cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2
What is integral tan3x sec2x?
-(4*log(2*cos(4*x)-4*cos(2*x)+3)-3*log(2*cos(4*x)+2)-2*log(2*cos(2*x)+2))/12
If sin B 1 over 2 show that 3 cos B - 4 cos cubed B equals 0?
All you need is knowledge of the sine and cosine ratios in a right angle triangle, and Pythagoras. Using Pythagoras you can work out the third side of the triangle which means you now know the value of cos B and so can substitute that into 3 cos B - 4 cos³ B and simplify it to see what the result is; if the result is 0, you have shown the required value. Try working it out before reading any further. -------------------------------------------------------------------------------- If you can't work it out by your self, read on: The sine ratio is opposite/hypotenuse The cosine ratio is adjacent/hypotenuse In a right angled triangle Pythagoras holds: hypotenuse² = adjacent² + opposite² → adjacent = √(hypotenuse² - opposite²) If sin B = 1/2 → opposite = 1, hypotenuse = 2 → adjacent = √(2² -1²) = √(4 - 1) = √3 → cos B = adjacent/hypotenuse = (√3)/2 → 3 cos B - 4 cos³ B = 3 × (√3)/2 - 4 × ((√3)/2)³ = 3(√3)/2 - 4((√3)²/2³) = 3(√3)/2 - 4(3(√3)/8) = 3(√3)/2 - 3(√3) × 4/8 = 3(√3)/2 - 3(√3)/2 = 0
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
Is four sixths greater than four thirds?
No, four sixths is less than four thirds. I like to think of it this way: 1) Reduce: 4/6 = 2/3 4/3 = 4/3 (cannot be reduced) 2) Think of the fraction as a pie. You divide the pie in 3 ways (the denominator when both fractions are reduced) For the first pie, only 2 out of the 3 slices are being eaten. For the second pie, 4 out of 3....Then you think: well, 3/3 is one whole pie, so 4/3 means there is no excess, there is not enough. 3) Also, you can divide the numerator (the top number) into the denominator (the bottom number) to get percentages, and see which one is greater. So: 4 divided by 6 = .666...66 % 4 divided by 3 = 1.3...over 100% I hope this helped!
What is the cos of angle A if the opposite is 3 and the adjacent is 4 and the hypotenuse is 5?
4/5
How do you confirm this trig identity?
sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
What is a faction less then one?
1/3 is a fraction less than one, because 3/3 is a whole. You only have 1 of the the 3 that is needed to make a whole. Think of pie (the kind you eat); you have 4 pieces of pie, then someone eats 2 pieces. How many pieces of pie do you have? 2 pieces! So you have 2 pieces of pie out of the 4 you originlly had. 2/4 of the pie is left. Does this help?
What are the trigonometric functions for 240?
It is easiest to find these using the unit circle. Assuming you want exact values for sin, cos, and tan.240 degrees is equal to 4[Pi]/3 radians.cos(4[Pi]/3) and sin(4[Pi]/3) are easy to find using the unit circle,cos(4[Pi]/3) = -1/2sin(4[Pi]/3) = -(Sqrt[3])/2To find tan, you will need to do a little arithmetic. We know that tan(x) = sin(x)/cos(x), so,tan(4[Pi]/3) = sin(4[Pi]/3)/cos(4[Pi]/3)tan(4[Pi]/3) = (-(Sqrt[3])/2) / (-1/2)tan(4[Pi]/3) = ((Sqrt[3])/2) * (2/1)tan(4[Pi]/3) = ((Sqrt[3])*2) / (2*1)tan(4[Pi]/3) = (Sqrt[3]) / 1tan(4[Pi]/3) = Sqrt[3]Find the other 3 using reciprocal. This is just a little arithmetic and up to you. You know that cossec(x) = 1/sin(x), sec(x) = 1/cos(x) and cot(x) = cos(x)/sin(x).
What is the value of cos pie4?
cos pi over four equals the square root of 2 over 2 This value can be found by looking at a unit circle. Cos indicates it is the x value of the point pi/4 which is (square root 2 over 2, square root 2 over 2)
Is it possible for a pie cut into three equal parts to be shared by four people with each of them getting one third of the pie?
No. It is not possible for 4 people to each get 1/3 of something. There is only 1 pie and therefore there are only 3/3 of a pie available. To be shared equally amongst 4 people each would get 1/4.
