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Mathematics LibreTexts

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(AB))cosA sinB = 12(sin(A+B)  sin(AB))cosA cosB = 12(cos(A+B) + cos(AB))sinA sinB = 12(cos(A+B)  cos(AB))

We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We see that
sin(A+B) + sin(AB) = (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(AB))) of such functions. We can go in the opposite direction, with the sum-to-product formulas:

sinA + sinB = 2sin12(A+B) cos12(AB)sinA  sinB = 2cos12(A+B) sin12(AB)cosA + cosB = 2cos12(A+B) cos12(AB)cosA  cosB = 2sin12(A+B) sin12(AB)

These formulas are just the product-to-sum formulas rewritten by using some clever substitutions: let x=12(A+B) and y=12(AB). Then x+y=A and xy=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(xy)= 212(cos(x+y) + cos(xy))= 2cosx cosy(by Equation 3.4.3)= 2cos12(A+B) cos12(AB) .


The proofs of the other sum-to-product formulas are similar (see Exercises 4-6).

Example 3.16

We are now in a position to prove Mollweide's equations, which we introduced in Section 2.3: For any triangle ABC,

abc = sin12(AB)cos12Canda+bc = cos12(AB)sin12C .

First, since C=212C, by the double-angle formula we have sinC=2sin12C cos12C. Thus,

abc = ac  bc = sinAsinC  sinBsinC(by the Law of Sines)= sinA  sinBsinC = sinA  sinB2sin12C cos12C= 2cos12(A+B) sin12(AB)2sin12C cos12C(by Equation 3.4.8)= cos12(180C) sin12(AB)sin12C cos12C(since A+B=180C)= cos(9012C) sin12(AB)sin12C cos12C= sin12(AB)cos12C(since cos(9012C)=sin12C) .

This proves the first equation. The proof of the other equation is similar (see Exercise 7).

Example 3.17

Using Mollweide's equations, we can prove the Law of Tangents: For any triangle ABC,

aba+b = tan12(AB)tan12(A+B) ,bcb+c = tan12(BC)tan12(B+C) ,cac+a = tan12(CA)tan12(C+A) .

We need only prove the first equation; the other two are obtained by cycling through the letters. We see that

aba+b = abca+bc = sin12(AB)cos12Ccos12(AB)sin12C(by Mollweide's equations)= sin12(AB)cos12(AB)sin12Ccos12C= tan12(AB)tan12C = tan12(AB)tan(9012(A+B))(since C=180(A+B))= tan12(AB)cot12(A+B)(since tan(9012(A+B))=cot12(A+B), see Section 1.5)= tan12(AB)tan12(A+B) .QED

Example 3.18

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(AB)= 1+2cos12(A+B) (cos12(A+B+2C)+cos12(AB))  , so= 1+2cos12(A+B)2cos12(A+C) cos12(B+C)  by Equation 3.4.9,


since 12(12(A+B+2C)+12(AB))=12(A+C) and 12(12(A+B+2C)12(AB))=12(B+C). Thus,}
cosA+cosB+cosC = 1+4cos(9012C) cos(9012B) cos(9012A)= 1+4sin12C sin12B sin12A  , so rearranging the order gives= 1+4sin12A sin12B sin12C .

Example 3.19

For any triangle ABC, show that sin12A sin12B sin12C18.

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(AB)) sin12C = 12(cos12(AB)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(AB) 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(AB)±cos212(AB)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(AB)8u  0u  18cos212(AB)  18sin12A sin12B sin12C  18 .

Example 3.20

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+418 = 32 .

Hence, 1 < cosA+cosB+cosC  32.

Example 3.21

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(θ1θ2) .

Solution

Multiply the top and bottom of t12p by sinθ1 sinθ2 to get:

t12p = 2n1 cosθ1n2 cosθ1 + n1 cosθ2sinθ1 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(2θ12θ2))(by formula 3.4.7)= 2cosθ1 sinθ2sin(θ1+θ2) cos(θ1θ2)

Example 3.22

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)= VmIm12(cos(2ωt+ϕ)+cos(ϕ))(by Equation 3.4.3)= 12VmImcosϕ + 12VmImcos(2ωt+ϕ)(since cos(ϕ)=cosϕ) .


This page titled 3.4: Other Identities is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Michael Corral via source content that was edited to the style and standards of the LibreTexts platform.

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