3.4: Other Identities
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Though the identities in this section fall under the category of "other'', they are perhaps (along with cos2θ+sin2θ=1) the most widely used identities in practice. It is very common to encounter terms such as sinA+sinB or sinA cosB in calculations, so we will now derive identities for those expressions. First, we have what are often called the product-to-sum formulas:
sinA cosB = −12(sin(A+B) + sin(A−B))cosA sinB = −12(sin(A+B) − sin(A−B))cosA cosB = −12(cos(A+B) + cos(A−B))sinA sinB = −12(cos(A+B) − cos(A−B))
We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We see that
sin(A+B) + sin(A−B) = (sinA cosB + cosA sinB) + (sinA cosB − cosA sinB)= 2sinA cosB ,
so Equation 3.4.1 follows upon dividing both sides by 2. Notice how in each of the above identities a product (e.g. sinA cosB) of trigonometric functions is shown to be equivalent to a sum (e.g. 12(sin(A+B) + sin(A−B))) of such functions. We can go in the opposite direction, with the sum-to-product formulas:
sinA + sinB = −2sin12(A+B) cos12(A−B)sinA − sinB = −2cos12(A+B) sin12(A−B)cosA + cosB = −2cos12(A+B) cos12(A−B)cosA − cosB = −2sin12(A+B) sin12(A−B)
These formulas are just the product-to-sum formulas rewritten by using some clever substitutions: let x=12(A+B) and y=12(A−B). Then x+y=A and x−y=B. For example, to derive Equation 3.4.9, make the above substitutions in Equation 3.4.3 to get
cosA + cosB = cos(x+y) + cos(x−y)= 2⋅12(cos(x+y) + cos(x−y))= 2cosx cosy(by Equation 3.4.3)= 2cos12(A+B) cos12(A−B) .
The proofs of the other sum-to-product formulas are similar (see Exercises 4-6).
We are now in a position to prove Mollweide's equations, which we introduced in Section 2.3: For any triangle △ABC,
a−bc = sin12(A−B)cos12Canda+bc = cos12(A−B)sin12C .
First, since C=2⋅12C, by the double-angle formula we have sinC=2sin12C cos12C. Thus,
a−bc = ac − bc = sinAsinC − sinBsinC(by the Law of Sines)= sinA − sinBsinC = sinA − sinB2sin12C cos12C= 2cos12(A+B) sin12(A−B)2sin12C cos12C(by Equation 3.4.8)= cos12(180∘−C) sin12(A−B)sin12C cos12C(since A+B=180∘−C)= cos(90∘−12C) sin12(A−B)sin12C cos12C= sin12(A−B)cos12C(since cos(90∘−12C)=sin12C) .
This proves the first equation. The proof of the other equation is similar (see Exercise 7).
Using Mollweide's equations, we can prove the Law of Tangents: For any triangle △ABC,
a−ba+b = tan12(A−B)tan12(A+B) ,b−cb+c = tan12(B−C)tan12(B+C) ,c−ac+a = tan12(C−A)tan12(C+A) .
We need only prove the first equation; the other two are obtained by cycling through the letters. We see that
a−ba+b = a−bca+bc = sin12(A−B)cos12Ccos12(A−B)sin12C(by Mollweide's equations)= sin12(A−B)cos12(A−B)⋅sin12Ccos12C= tan12(A−B)⋅tan12C = tan12(A−B)⋅tan(90∘−12(A+B))(since C=180∘−(A+B))= tan12(A−B)⋅cot12(A+B)(since tan(90∘−12(A+B))=cot12(A+B), see Section 1.5)= tan12(A−B)tan12(A+B) .QED
For any triangle △ABC, show that
cosA + cosB + cosC = 1 + 4sin12A sin12B sin12C .
Solution
Since cos(A+B+C)=cos180∘=−1, we can rewrite the left side as
cosA+cosB+cosC = 1+(cos(A+B+C)+cosC)+(cosA+cosB) , so by Equation 3.4.9= 1+2cos12(A+B+2C) cos12(A+B)+2cos12(A+B) cos12(A−B)= 1+2cos12(A+B) (cos12(A+B+2C)+cos12(A−B)) , so= 1+2cos12(A+B)⋅2cos12(A+C) cos12(B+C) by Equation 3.4.9,
since 12(12(A+B+2C)+12(A−B))=12(A+C) and 12(12(A+B+2C)−12(A−B))=12(B+C). Thus,}
cosA+cosB+cosC = 1+4cos(90∘−12C) cos(90∘−12B) cos(90∘−12A)= 1+4sin12C sin12B sin12A , so rearranging the order gives= 1+4sin12A sin12B sin12C .
For any triangle △ABC, show that sin12A sin12B sin12C≤18.
Solution
Let u=sin12A sin12B sin12C. Apply Equation 3.4.4 to the first two terms in u to get
u = −12(cos12(A+B)−cos12(A−B)) sin12C = 12(cos12(A−B)−cos12(A+B)) cos12(A+B) ,
since sin12C=cos12(A+B), as we saw in Example 3.18. Multiply both sides by 2 to get
cos212(A+B) − cos12(A−B) cos12(A+B) + 2u = 0 ,
after rearranging the terms. Notice that the expression above is a quadratic equation in the term cos12(A+B). So by the quadratic formula,
cos12(A+B) = cos12(A−B)±√cos212(A−B)−4(1)(2u)2 ,
which has a real solution only if the quantity inside the square root is nonnegative. But we know that cos12(A+B) is a real number (and, hence, a solution exists), so we must have
cos212(A−B)−8u ≥ 0⇒u ≤ 18cos212(A−B) ≤ 18⇒sin12A sin12B sin12C ≤ 18 .
For any triangle △ABC, show that 1 < cosA+cosB+cosC ≤ 32.
Solution
Since 0∘<A,B,C<180∘, the sines of 12A, 12B, and 12C are all positive, so
cosA+cosB+cosC = 1+4sin12A sin12B sin12C > 1
by Example 3.18. Also, by Examples 3.18 and 3.19 we have
cosA+cosB+cosC = 1+4sin12A sin12B sin12C ≤ 1+4⋅18 = 32 .
Hence, 1 < cosA+cosB+cosC ≤ 32.
Recall Snell's law from Example 3.12 in Section 3.2: n1 sinθ1=n2 sinθ2. Use it to show that the p-polarization transmission Fresnel coefficient defined by
t12p = 2n1 cosθ1n2 cosθ1 + n1 cosθ2
can be written as:
t12p = 2cosθ1 sinθ2sin(θ1+θ2) cos(θ1−θ2) .
Solution
Multiply the top and bottom of t12p by sinθ1 sinθ2 to get:
t12p = 2n1 cosθ1n2 cosθ1 + n1 cosθ2⋅sinθ1 sinθ2sinθ1 sinθ2= 2(n1 sinθ1) cosθ1 sinθ2(n2 sinθ2) sinθ1 cosθ1 + (n1 sinθ1) sinθ2 cosθ2= 2cosθ1 sinθ2sinθ1 cosθ1 + sinθ2 cosθ2(by Snell's law)= 2cosθ1 sinθ212(sin2θ1 + sin2θ2)(by the double-angle formula)= 2cosθ1 sinθ212(2sin12(2θ1+2θ2) cos12(2θ1−2θ2))(by formula 3.4.7)= 2cosθ1 sinθ2sin(θ1+θ2) cos(θ1−θ2)
In an AC electrical circuit, the instantaneous power p(t) delivered to the entire circuit in the \emph{sinusoidal steady state} at time t is given by
p(t) = v(t)i(t) ,
where the voltage v(t) and current i(t) are given by
v(t) = Vmcosωt ,i(t) = Imcos(ωt+ϕ) ,
for some constants Vm, Im, ω, and ϕ. Show that the instantaneous power can be written as
p(t) = 12VmImcosϕ + 12VmImcos(2ωt+ϕ) .
Solution
By definition of p(t), we have
p(t) = VmImcosωt cos(ωt+ϕ)= VmIm⋅12(cos(2ωt+ϕ)+cos(−ϕ))(by Equation 3.4.3)= 12VmImcosϕ + 12VmImcos(2ωt+ϕ)(since cos(−ϕ)=cosϕ) .