• θ occurs only once. Example Go to topic • θ occurs several times but always as the argument of the same trigonometric function. Example Go to topic • the equation has different trigonometric functions but they all have the same argument, θ. Example Go to topic • the equation contains trigonometric functions with different arguments. Example Go to topic
2 Q^{ 2} = Q.Solve this equation first for Q:
Q (2 Q − 1) = 0 → Q = 0 and Q = ½.Then substitute back sin (θ) for Q:
sin(θ) = 0 and sin(θ) = ½.Now solve these quations for θ as above.
5 cos(θ) = 3 cos(115° − θ).Use trigonometric identity (3b) with the lower sign to separate angle 115° from angle θ on the RHS of the equation.
5 cos(θ) = 3 cos(115°) · cos(θ) + 3 sin(115°) · sin(θ)Evaluate the sin and cos of 115°. Notice that now the only angle remaining is θ:
5 cos(θ) = −1.268 cos(θ) + 2.719 sin(θ)Collect terms:
6.268 cos(θ) = 2.719 sin(θ)Divide both sides by 2.719 cos(θ) and use the basic trigonometric identity to turn sin/cos into tan:
tan(θ) = 2.305From the diagram above we see that the angle we want is θ = 66.5°. The other solution corresponds to having the vectors rotated by 180° and being in quadrants 3 and 4.
θ = tan^{−1}(2.305) = 66.5° or 246.5°
y_{1} = 4 sin(θ) and y_{2} = 3 cos(θ + 40°).Find all angles θ between 0 and 2π radians for which their sum equals 2.
4 sin(θ) + 3 cos(θ + 40°) = 2.Because one angle is 40° more than the other, we will use the sum of angles identity (3b) in the form
cos(θ + 40°) = cos(θ) · cos(40°) − sin(θ) · sin(40°)to split up the two angles, θ and 40°. This gives
4 sin(θ) + 3 cos(θ) · cos(40°) − 3 sin(θ) · sin(40°) = 2, orThis looks similar to the previous example but we have a “2” on the right-hand-side so we must use a different method to solve it. In the section on trigonometric identities we proved the following identity, which states that a sine and a cosine waveform can always be added to give a sine waveform with a phase shift:
(4 − 3 sin(40°)) · sin(θ) + 3 cos(40°) · cos(θ) = 2, or
2.072 sin(θ) + 2.298 cos(θ) = 2.
, whereWe can use this identity to replace the two occurrences of θ in our equation by one. In our case A = 2.072 and B = 2.298, so we get C = 3.094 and φ = 47.97° = 0.8372 radians, and our equation becomes
3.094 sin(θ + 0.8372) = 2.Now θ occurs just once and we simply isolate it:
sin(θ + 0.8372) = 0.6464Of the three angles listed on the right-hand-side,
θ + 0.8372 = sin^{−1}(0.6464) = 0.7029, π − 0.7029 and 0.7029 + 2 π
θ = −0.134, 1.602 and 6.149
v = 100 sin(1 t) + 50 cos(2 t).Determine the times t when v = 0 in the cycle shown (indicated by the black dots).
cos(2 t) = 1 − 2 sin^{2}(t)to get rid of the angle 2 t in favor of the angle t:
0 = 100 sin(t) + 50·{1 − 2 sin^{2}(t)}Distribute, transpose everything to the left side of the equation and divide by 50:
100 sin^{2}(t) − 100 sin(t) − 50 = 0
2 sin^{2}(t) − 2 sin(t) − 1 = 0This is a quadratic equation in sin(t). (In fact replacing sin(t) with the alias Q would turn it into the quadratic equation 2 Q^{ 2} − 2 Q − 1 = 0.) Plugging a = 2, b = −2 and c = −1 into the quadratic formula we get:
sin(t) = −0.366
t = sin^{−1}(−0.366)
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