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Draw a right angled triangle OPQ with OP the base, PQ the altitude and OQ the hypotenuse.
Draw a second right angled triangle OQR with OQ the base, QR the altitude and OR the hypotenuse.
Drop a perpendicular from R to intercept OP at T. The line RT crosses OQ at U.
Draw a perpendicular from Q to intercept RT at S.
Let angle POQ be A and angle QOR be B.
Angle OUT = angle QUR therefore angle URQ = A
sin (A + B) = TR/OR = (TS + SR)/OR = (PQ + SR)/OR = (PQ/OQ x OQ/OR) + (SR/QR x QR/OR) = sin A cos B + cos A sin B.
What else can I help you with?
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 is the answer to a squared plus b squared and what's the proof?
Something's missing...
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
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)
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
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)
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.
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
Well, darling, if we square the first equation and the second equation, add them together, and do some algebraic magic, we can indeed show that a squared plus b squared equals 89. It's like a little math puzzle, but trust me, the answer is as sassy as I am.
What is the answer to a squared plus b squared and what's the proof?
Something's missing...
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
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)
What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?
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'.
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.
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
What are five trigonometric identities?
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)
Who originally wrote the proof of a squared plus b squared equals c squared?
Pascal
Can you use the law of sines if 3 sides are given?
Yes, but you would need to know a degree measure too. [Sin(A)/a] = [Sin(B)/b] = [Sin(C)/c] [a/Sin(A)] = [b/Sin(B)] = [c/Sin(C)]
