0
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.
Still curious? Ask our experts.
Chat with our AI personalities
Lao
Beau
CoachAdd your answer:
Earn +20 pts
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
