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Trigonometry C3 Maths A Level Question

Solve
sin (Θ+40°)+ sin(Θ+50°)=0
for 0≤ Θ≤ 360

Please help me with this, thank you
Original post by Ken Poison
Solve
sin (Θ+40°)+ sin(Θ+50°)=0
for 0≤ Θ≤ 360

Please help me with this, thank you


Expand using compound angle formulae to

sinθcos40+cosθsin40+sinθcos50+cosθsin50=0\sin \theta \cos 40 + \cos \theta \sin 40 + \sin \theta \cos 50 + \cos \theta \sin 50 = 0

Divide through by cosθ\cos \theta and proceed.



Alternatively, note that -sin(x) = sin(-x) hence the eqn is the same as

sin(θ+40)=sin(θ50)\sin(\theta+40) = \sin(-\theta -50)

So just inverse both sides and obtain all the solutions you need.
(edited 4 years ago)
Reply 2
Another method you can use is to use the sum-product formula: sin(A)+sin(B)=2sin(A+B2)cos(AB2). \sin (A) + \sin (B) = 2 \sin \big( \frac{A+B}{2} \big) \cos \big( \frac{A-B}{2} \big) .


Here we have: sin(θ+40)+sin(θ+50)=0    2sin(θ+45)cos(5)=0. \sin ( \theta + 40^{\circ} ) + \sin ( \theta + 50 ^{\circ} ) = 0 \, \iff \, 2 \sin( \theta + 45 ^{\circ} ) \cos ( 5 ^{\circ} ) = 0.
...and yet another approach would be to reason that, for the equation to sum to zero, (theta + 40) and (theta + 50) must be arranged symmetrically either side of 180n.

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