Sums and differences as products (rad)
Products
If the sine and cosine sum and difference formulas are written down side-by-side it becomes apparent that useful results can be obtained by adding some of them them in pairs.
| $\sin\left(A+B\right)$sin(A+B) | $=$= | $\sin\left(A\right)\cos\left(B\right)+\sin\left(B\right)\cos\left(A\right)$sin(A)cos(B)+sin(B)cos(A) | (1) |
| $\sin\left(A-B\right)$sin(A−B) | $=$= | $\sin\left(A\right)\cos\left(B\right)-\sin\left(B\right)\cos\left(A\right)$sin(A)cos(B)−sin(B)cos(A) | (2) |
| $\cos\left(A+B\right)$cos(A+B) | $=$= | $\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)$cos(A)cos(B)−sin(A)sin(B) | (3) |
| $\cos\left(A-B\right)$cos(A−B) | $=$= | $\cos\left(A\right)\cos\left(B\right)+\sin\left(A\right)\sin\left(B\right)$cos(A)cos(B)+sin(A)sin(B) | (4) |
If we add (1) and (2), we have
| $\sin\left(A+B\right)+\sin\left(A-B\right)$sin(A+B)+sin(A−B) | $=$= | $2\sin\left(A\right)\cos\left(B\right)$2sin(A)cos(B) | (5) |
Similarly, from (3) and (4) we obtain, by addition,
| $\cos\left(A+B\right)+\cos\left(A-B\right)$cos(A+B)+cos(A−B) | $=$= | $2\cos\left(A\right)\cos\left(B\right)$2cos(A)cos(B) | (6) |
and by subtraction,
| $\cos\left(A-B\right)-\cos\left(A+B\right)$cos(A−B)−cos(A+B) | $=$= | $2\sin\left(A\right)\sin\left(B\right)$2sin(A)sin(B) | (7) |
Equations (5), (6) and (7) give the following three product formulas:
| $\sin\left(A\right)\cos\left(B\right)$sin(A)cos(B) | $=$= | $\frac{1}{2}\left(\sin\left(A+B\right)+\sin\left(A-B\right)\right)$12(sin(A+B)+sin(A−B)) | (5a) |
| $\cos\left(A\right)\cos\left(B\right)$cos(A)cos(B) | $=$= | $\frac{1}{2}\left(\cos\left(A+B\right)+\cos\left(A-B\right)\right)$12(cos(A+B)+cos(A−B)) | (6a) |
| $\sin\left(A\right)\sin\left(B\right)$sin(A)sin(B) | $=$= | $\frac{1}{2}\left(\cos\left(A-B\right)-\cos\left(A+B\right)\right)$12(cos(A−B)−cos(A+B)) | (7a) |
Sums
By re-writing (5a), (6a) and (7a) we can obtain formulas for the sums and differences of sines and cosines. To do this, we let $U=A+B$U=A+B and $V=A-B$V=A−B. Then, by solving these equations for $A$A and $B$B we get $A=\frac{U+V}{2}$A=U+V2 and $B=\frac{U-V}{2}$B=U−V2.
Thus, by substituting for $A$A and $B$B in the product formulas and rearranging slightly, we obtain:
| $\sin\left(U\right)+\sin\left(V\right)$sin(U)+sin(V) | $=$= | $2\sin\left(\frac{U+V}{2}\right)\cos\left(\frac{U-V}{2}\right)$2sin(U+V2)cos(U−V2) | (8) |
| $\cos\left(U\right)+\cos\left(V\right)$cos(U)+cos(V) | $=$= | $2\cos\left(\frac{U+V}{2}\right)\cos\left(\frac{U-V}{2}\right)$2cos(U+V2)cos(U−V2) | (9) |
| $\cos\left(V\right)-\cos\left(U\right)$cos(V)−cos(U) | $=$= | $2\sin\left(\frac{U+V}{2}\right)\sin\left(\frac{U-V}{2}\right)$2sin(U+V2)sin(U−V2) | (10) |
and from (8), using the fact that $-\sin\left(V\right)=\sin\left(-V\right)$−sin(V)=sin(−V), we can write
| $\sin\left(U\right)-\sin\left(V\right)$sin(U)−sin(V) | $=$= | $2\sin\left(\frac{U-V}{2}\right)\cos\left(\frac{U+V}{2}\right)$2sin(U−V2)cos(U+V2) | (11) |
Another type of sum, with a very useful simplification, occurs between different multiples of the sine and cosine of identical angles.
The expression $a\sin\left(\theta\right)+b\cos\left(\theta\right)$asin(θ)+bcos(θ) can be written in the form $r\sin\left(\theta+\alpha\right)$rsin(θ+α). The latter expands to $r\left(\sin\left(\theta\right)\cos\left(\alpha\right)+\cos\left(\theta\right)\sin\left(\alpha\right)\right)$r(sin(θ)cos(α)+cos(θ)sin(α)).
On comparing this with the original expression, we see that $a=r\cos\left(\alpha\right)$a=rcos(α) and $b=r\sin\left(\alpha\right)$b=rsin(α).
Hence, $r=\sqrt{a^2+b^2}$r=√a2+b2 and $\tan\left(\alpha\right)=\frac{b}{a}$tan(α)=ba. Then, using the notation $\tan^{-1}$tan−1 for the inverse tangent function, we can write
| $a\sin\left(\theta\right)+b\cos\left(\theta\right)$asin(θ)+bcos(θ) | $=$= | $\sqrt{a^2+b^2}\sin\left(\theta+\tan^{-1}\left(\frac{b}{a}\right)\right)$√a2+b2sin(θ+tan−1(ba)) | (12) |
Worked example
Express $\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)$cos(17π12)−cos(π4) more simply.
Do: Using (10), $\cos\left(V\right)-\cos\left(U\right)=2\sin\left(\frac{U+V}{2}\right)\sin\left(\frac{U-V}{2}\right)$cos(V)−cos(U)=2sin(U+V2)sin(U−V2), we have
$\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)=2\sin\left(\frac{\frac{17\pi}{12}+\frac{\pi}{4}}{2}\right)\sin\left(\frac{\frac{\pi}{4}-\frac{17\pi}{12}}{2}\right)$cos(17π12)−cos(π4)=2sin(17π12+π42)sin(π4−17π122)
That is,
| $\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)$cos(17π12)−cos(π4) | $=$= | $2\sin\left(\frac{5\pi}{6}\right)\sin\left(-\frac{7\pi}{12}\right)$2sin(5π6)sin(−7π12) |
| $=$= | $-2\sin\left(\frac{\pi}{6}\right)\sin\left(\frac{5\pi}{12}\right)$−2sin(π6)sin(5π12) | |
| $=$= | $-\sin\left(\frac{5\pi}{12}\right)$−sin(5π12) |
Using a half-angle formula, $\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1}{2}\left(1-\cos\left(\theta\right)\right)}$sin(θ2)=√12(1−cos(θ)) we can further simplify this to the exact value $-\frac{1}{2}\sqrt{2+\sqrt{3}}$−12√2+√3.
Practice questions
QUESTION 1
Express $\cos\left(3x+2y\right)\cos\left(x-y\right)$cos(3x+2y)cos(x−y) as a sum or difference of two trigonometric functions.
QUESTION 2
Express $\sin\left(6x\right)+\sin\left(4x\right)$sin(6x)+sin(4x) as a product of two trigonometric functions.
QUESTION 3
By expressing the left-hand side of the equation as a product, solve the equation $\sin5x+\sin x=0$sin5x+sinx=0 for $0\le x$0≤x$<=$<=$2\pi$2π.