Trigonometry · Higher Level

Trig Proofs

Proof 1

Prove $\cos(A-B) = \cos A\cos B + \sin A\sin B$

Find $|PQ|^2$

(i) Cosine Rule

(ii) Distance $|PQ| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

$a^2 = 1^2 + 1^2 - 2(1)(1)\cos(A-B)$

$a^2 = 2 - 2\cos(A-B)$

$|PQ|^2 = (x_2-x_1)^2 + (y_2-y_1)^2$

$a^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2$

$\quad = \cos^2 A - 2\cos A\cos B + \cos^2 B + \sin^2 A - 2\sin A\sin B + \sin^2 B$

$\cos^2 A + \sin^2 A = \cos^2 B + \sin^2 B = 1$

$a^2 = 2 - 2(\cos A\cos B + \sin A\sin B)$

$\cancel{2} - \cancel{2}\cos(A-B) = \cancel{2} - \cancel{2}(\cos A\cos B + \sin A\sin B)$

$\cos(A-B) = \cos A\cos B + \sin A\sin B$

Proof 2

Prove $\cos(A+B) = \cos A\cos B - \sin A\sin B$

$\cos(-A) = \cos A$

$\sin(-A) = -\sin A$

start $\cos(A-B) = \cos A\cos B + \sin A\sin B$

Replace $B$ with $-B$

$\cos(A-(-B)) = \cos A\cos(-B) + \sin A\sin(-B)$

$\cos(-B) = \cos B \quad$ and $\quad \sin(-B) = -\sin B$

$\cos(A+B) = \cos A\cos B - \sin A\sin B$

Proof 3

Prove $\sin(A+B) = \sin A\cos B + \cos A\sin B$

$\sin A = \dfrac{b}{a} \qquad \cos A = \dfrac{c}{a}$

$\cos(90-A) = \dfrac{b}{a} \qquad \sin(90-A) = \dfrac{c}{a}$

$\cos(90-A) = \sin A$

$\sin(90-A) = \cos A$

start $\cos(A-B) = \cos A\cos B + \sin A\sin B$

Replace $A$ with $90-A$

$\cos(90-A-B) = \cos(90-A)\cos B + \sin(90-A)\sin B$

$\cos(90-(A+B))$

$\sin(A+B) = \sin A\cos B + \cos A\sin B$

Proof 4

Prove $\sin(A-B) = \sin A\cos B - \cos A\sin B$

Replace $B$ with $-B$

$\sin(A+B) = \sin A\cos B + \cos A\sin B$

$\sin(A-B) = \sin A\cos(-B) + \cos A\sin(-B)$

$\quad\quad\quad\quad\;\; = \sin A\cos B - \cos A\sin B$

$\cos(-B) = \cos B$

$\sin(-B) = -\sin B$

Proof 5

Prove $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$

$\tan(A+B) = \dfrac{\sin(A+B)}{\cos(A+B)}$

$\quad\quad\quad\quad\;\;\; = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}$

$\quad\quad\quad\quad\;\;\; = \dfrac{\dfrac{\sin A\cos B}{\cos A\cos B} + \dfrac{\cos A\sin B}{\cos A\cos B}}{\dfrac{\cos A\cos B}{\cos A\cos B} - \dfrac{\sin A\sin B}{\cos A\cos B}}$

$\quad\quad\quad\quad\;\;\; = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$

Proof 6

Prove $\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A\tan B}$

Replace $B$ with $-B$

$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$

$\tan(A-B) = \dfrac{\tan A + \tan(-B)}{1 - \tan A\tan(-B)}$

$\tan(-B) = -\tan B$

$\quad\quad\quad\;\; = \dfrac{\tan A - \tan B}{1 + \tan A\tan B}$

Proof 7

Prove the Sine Rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$

Find $h$ twice

$\sin A = \dfrac{h}{b} \qquad\qquad \sin B = \dfrac{h}{a}$

$b\sin A = h \qquad\qquad\; a\sin B = h$

$h = h$

$\dfrac{a\sin B}{ab} = \dfrac{b\sin A}{ab}$

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$

Proof 8

Prove $a^2 = b^2 + c^2 - 2bc\cos A$

Find $h^2$ twice

$h^2 + x^2 = b^2$

$h^2 = b^2 - x^2$

$h^2 + (c-x)^2 = a^2$

$h^2 = a^2 - (c-x)^2$

$a^2 - (c-x)^2 = b^2 - x^2$

$a^2 = b^2 - x^2 + (c-x)^2$

$\quad = b^2 - \cancel{x^2} + c^2 - 2cx + \cancel{x^2}$

$\quad = b^2 + c^2 - 2cx$

$\cos A = \dfrac{x}{b}$

$b\cos A = x$

$a^2 = b^2 + c^2 - 2bc\cos A$