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.
These identities are also considered to be basic simplifications and are used to get rid of the 90° shift. By using them repeatedly they can be used to get rid of shifts of 180°, 270°, 360°, etc.
4. sin2(θ) + cos2(θ) = 1
1 − sin2(θ) = cos2(θ)or as
1 − cos2(θ) = sin2(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 + tan2(θ) = sec2(θ)Alternatively, if we divide both sides of identity (4) by sin2(θ) and simplify then it reads
1 + cot2(θ) = csc2(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 left-hand-side expression anywhere it appears by the right-hand-side expression.) The rest of the identities (identities 5 to 9, discussed next) are stated so that the left-hand-side is the combined form of some expression and the right-hand-side is the broken apart form. They can be applied in either direction depending on what we want to accomplish:
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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.
We see that it becomes part sin wave and part cos wave. It is interesting to compare this result with identity (3a), which shows that shifting a sin wave left by 90° turns it completely into a cos wave.
Use the fact that sin(γ) = h/c and then multiply (a + b) across to the other side.
Expand the right-hand-side and then use the two small triangles in the picture to replace each fraction by the appropriate trigonometric function.
We are done.
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) = cos2(α) − sin2(α)
= 1 − 2 sin2(α)
= 2 cos2(α) − 1
7c.
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Let β equal α and identities (7a), (7b), (7c) immediately pop out. Identity (7b) is written in 3 different forms. The second and third forms result from using Pythagoras’ theorem on the first form. They are the preferred forms because they only involve sin or only cos, and not both.
If we now change the name of the angle α to be A/2 then these become the so-called half angle identities:
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:
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Solution: Follow these steps:
Solution: Follow these steps:
Solution: Follow these steps:
where on the right-hand-side C and φ are given by
(b) Use the identity to express the sum of the two sinusoidal waveforms,
y1 = 3 sin(θ) and y2 = 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.)
(b) The left side of the identity can be considered to be the sum of two sinusoidal waveforms
with slightly different frequencies. Plot waveforms y1 and y2 on the same graph and compare them.
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 = I0 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 = V0 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.
We can change the product of trigonometric functions into a sum of trigonometric functions by using identity (9a) which says that sin(α)·sin(β) = ½[cos(α − β) − cos(α + β)]. If we make the replacements α→ωt+φ and β→ωt then identity (9a) reads
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.
This agrees with the result we derived in the previous example, which was that pavg = ( Ieffective )2 R where Ieffective = 0.7071 I0
pavg = 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 |