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Trigonometric formulas and identities

December 16, 2016 Mouctar Trigonometry 0

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Trigonometric formulas and identities

Since we have seen all the basic formulas, we are going to start listing them now and prepare for problem solving.

In some cases, We’ll add some extra steps to show how the new formulas are worked.

Interactive videos will be added to this page to show those formulas and their use in practice:

Right triangle trigonometry

For triangle ABC having $\angle C=90^{\circ}$

$\cos A=\frac{adjacent side}{hypotenuse}=\frac{1}{\sec A}$

$\sin A=\frac{opposite side}{hypotenuse}=\frac{1}{\csc A}$

$\tan A=\frac{opposite side}{adjacent \;side}=\frac{1}{\cot A}=\frac{\sin A}{\cos A}$

$\sec A=\frac{hypotenuse}{adjacent \; side}=\frac{1}{\cos A}$

$\csc A=\frac{hypotenuse}{opposite \;side}=\frac{1}{\sin A}$

$\cot A=\frac{adjacent side}{opposite \;side}=\frac{1}{\tan A}=\frac{\cos A}{\sin A}$

Please note that we know these relations. All we need is to divide all three side by the hypotenuse and we have the unit circle with $radius=1$ , the hypotenuse.

Law of cosines

For an oblique triangle:

${a}^2={b}^2+{c}^2-2 \cdot b \cdot c\cdot\cos{A}$

${b}^2={a}^2+{c}^2-2 \cdot a \cdot c\cdot\cos{B}$

${c}^2={a}^2+{b}^2-2 \cdot a \cdot b\cdot\cos{C}$

Law of sines

$\frac{\sin{A}}{a}=\frac{\sin{B}}{b}=\frac{\sin{C}}{c}$

Pythagorean identities

$\sin^{2} A+ \cos^{2} A=1$

We can easely show:

$1+\tan^{2} A= \sec^{2} A$

$1+\cot^{2} A= \csc^{2} A$

We may never need these, since we can easily reproduce from the unit circle.

Complementary angles: Cofunction Theorem

If $A$ and $B$ are complementary angles $sum=90^{\circ}$ :

$B=90^{\circ}-A$

$\sin A=\cos B$

$\sin B=\cos A$

$\sec A= \csc B$

$\sec B= \csc A$

$\tan A= \cot B$

$\tan B= \cot A$

Sum and difference formulas

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

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

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

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

We can use these to find:

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

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

Double angle formulas

From above when $B=A$

$\sin {2A}=2\sin {A} \cos{A}$

$\cos {2A}=\cos^{2}A-\sin^{2}A$

$\tan {2A}=\frac{2\tan A}{1-\tan^{2} A}$

Half-Angle formulas

From above when $B=A$

$\sin {\frac{A}{2}}=\pm \sqrt{\frac{1-\cos A}{2}}$

$\cos {\frac{A}{2}}=\pm \sqrt{\frac{1+\cos A}{2}}$

$\tan {\frac{A}{2}}=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}$

Product to sum formulas

From above :

$\sin {A} \cos{B}=\frac{1}{2}(\sin (A+B)+\sin (A-B) )$

$\cos {A} \sin{B}=\frac{1}{2}(\sin (A+B)-\sin (A-B) )$

$\cos {A} \cos{B}=\frac{1}{2}(\cos (A+B)+\cos (A-B) )$

$\sin {A} \sin{B}=\frac{1}{2}(\cos (A-B)-\cos (A+B) )$

Sum to product formulas

$\sin {A} + \sin{B}=2\sin {\frac{A+B}{2}}\cos {\frac{A-B}{2}}$

$\sin {A} - \sin{B}=2\cos {\frac{A+B}{2}}\sin {\frac{A-B}{2}}$

$\cos {A} + \cos{B}=2\cos {\frac{A+B}{2}}\cos {\frac{A-B}{2}}$

$\cos {A} - \cos{B}=-2\sin {\frac{A+B}{2}}\sin {\frac{A-B}{2}}$

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