cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
Try to write everything in terms of sines and cosines:1 / cos B - cos B = (sin B / cos B) sin B1 / cos B - cos B = sin2B / cos BMultiply by the common denominator, cos B:1 - cos2B = sin2BUse the pithagorean identity on the left side:sin2B + cos2B - cos2B = sin2Bsin2B = sin2B
All others can be derived from these and a little calculus: sin2x+cos2x=1 sec2x-tan2x=1 sin(a+b)=sin(a)cos(b)+sin(b)sin(a) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) eix=cos(x)+i*sin(x)
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
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)
Thanks to the pre-existing addition and subtraction theorums, we can establish the identity:sin(a+b) = sin(a)cos(b)+sin(a)cos(b)Then, solving this, we getsin(a+b) = 2(sin(a)cos(b))sin(a)cos(b) = sin(a+b)/2a=b, sosin(a)cos(a) = sin(a+a)/2sin(a)cos(a) = sin(2a)/2Therefore, the answer is sin(2a)/2.
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
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.
The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).
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
.385
We can use the law of cosines here. ( remember, side b is opposite angle B) DEGREE MODE! b^2 = a^2 + c^2 - 2ac cos(B) 16^2 = 10^2 + 12^2 - 2(10)(12) cos(B) 256 = 244 - 240(cos B ) 12 = -240(cos B ) -0.05 = cosB arcos(-0.05) = B B = 93 degrees