3x + 2x = 5xis an identity that is always true, no matter what the value of x, whereas
3x = 15is an equation (or more precisely, a conditional equation) that is only true if x = 5.
1a. 1b. 1c.
2a. cos (−θ ) = cos ( θ )
2b. sin (−θ ) = − sin ( θ )
2c. tan (−θ ) = − tan ( θ )
3a.3b. 3c.
4. sin^{2}(θ) + cos^{2}(θ) = 1
1 − sin^{2}(θ) = cos^{2}(θ)or as
1 − cos^{2}(θ) = sin^{2}(θ)and these are also considered to be basic simplifications when they are used to replace two terms on the left with the one term on the right.
1 + tan^{2}(θ) = sec^{2}(θ)Alternatively, if we divide both sides of identity (4) by sin^{2}(θ) and simplify then it reads
1 + cot^{2}(θ) = csc^{2}(θ)
The identities that we have discussed so far (identities 1 to 4) are really just simplifications that are always applied from left to right (i.e. by replacing the lefthandside expression anywhere it appears by the righthandside expression.) The rest of the identities (identities 5 to 9, discussed next) are stated so that the lefthandside is the combined form of some expression and the righthandside is the broken apart form. They can be applied in either direction depending on what we want to accomplish:

5.
6a.
sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)
6b. 6c.
sin(α + β) = sin(α)+ sin(β) WRONG!This is not correct! It could only be correct if the sin curve was a straight line through the origin, instead of a wavy curve.
sin(α + β) = sin(α)·cos(β) + cos(α)·sin(β)which we just proved, is probably the most important identity of the group. Every other identity in the group (and in fact almost any trigonometric identity) can be derived from it. For example:
sin((α + 90°) + β),then applying identity (6a) and then using (3a) again to clean up.
7a.
sin(2α) = 2·sin(α)·cos(α)
7b.
cos(2α) = cos^{2}(α) − sin^{2}(α)
= 1 − 2 sin^{2}(α)
= 2 cos^{2}(α) − 1
7c.
8a. 8b.
9a.
sin(α)·sin(β) = ½[cos(α − β) − cos(α + β)]
9b.
sin(α)·cos(β) = ½[sin(α + β) + sin(α − β)]
9c. cos(α)·cos(β) = ½[cos(α + β) + cos(α − β)]
How to prove a trigonometric identity:

y_{1} = 3 sin(θ) and y_{2} = 4 cos(θ),as a single sin wave with a phase shift. Then plot all three waveforms in a graph.
A = C cos(φ) and B = C sin(φ).Here are the steps in the proof:
y = 5 sin(θ + 53.13°)The graph to the right shows the separate waveforms in green and blue and the resultant waveform in red. (Click here to see how the red waveform can be graphed.)
sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)In this case let α = 20t and β = 1t.
y = cos(t) · sin(20t).This graph is plotted below in red. We see that the factor cos(t) in this product plays the role of a slowly varying amplitude for the rapidly varying oscillation sin(20t). That is, we have a sinusoidal wave sin(20t), but that at certain times it has a small amplitude and at other times it has a large amplitude.
i = 0.7071 amperes
i = 1.0 sin(t) amperes
p = i^{ 2} R,where p is the power in watts, i is the current in amperes, and R is the resistance of the resistor in ohms. If the current changes with time then the power must be calculated at each instant in time with this formula. We say that the formula gives the instantaneous power.
p = i v,where i is the current in amperes at that instant and v is the voltage of the generator in volts. Suppose that the generator produces the AC current
i = I_{0} sin (ωt)and that the load is such that the voltage is out of phase with the current by angle φ. (The phase shift φ could be positive or negative.)
v = V_{0} sin (ωt+φ)Since i and v change with time, so does p. But as in the previous example, rather than the instantaneous power, we are interested in the average power.
sin(ωt+φ)·sin(ωt) = ½[cos(φ) − cos(2ωt+φ)]Substituting this into the formula for p gives or expanding, Because φ is a constant (remember it is the phase shift between v and i) the instantaneous power p is again a sinusoidal waveform with a DC component. The DC component gives the average power, Some special loads.
p_{avg} = 0.How can this be? The explanation is that inductors and capacitors absorb energy for one half of each cycle as they build up their magnetic and electric fields and then release the energy back to the generator in the other half of the cycle when the fields collapse.
Algebra Coach Exercises 