Free boundary problem

In trigonometry, it is common to use mnemonics to help remember trigonometric identities and the relationships between the various trigonometric functions. For example, the sine, cosine, and tangent ratios in a right triangle can be remembered by representing them as strings of letters, for instance SOH-CAH-TOA in English:

Sine = Opposite ÷ Hypotenuse

Cosine = Adjacent ÷ Hypotenuse

Tangent = Opposite ÷ Adjacent

One way to remember the letters is to sound them out phonetically (i.e. "SOH-CAH-TO-A"). Another method is to expand the letters into a sentence, such as "Some Old Hippie Caught Another Hippie Trippin' On Acid".

Communities exposed to Chinese dialect may choose to remember it as TOA-CAH-SOH, which also means 'big-footed woman' in Hokkien.

Mnemonic chart

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Trigonometric identities mnemonic

Reading across the central 1 in any direction gives reciprocal identities:

${\begin{array}{ccc}{1 \over \sin A}=\csc A&{\text{or}}&{1 \over \csc A}=\sin A\\\\{1 \over \tan A}=\cot A&{\text{or}}&{1 \over \cot A}=\tan A\\\\{1 \over \sec A}=\cos A&{\text{or}}&{1 \over \cos A}=\sec A\end{array}}$

Reading down any triangle gives the standard identities (starting at the top left and going clockwise):

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

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

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

Reading a function and dividing the two consecutive clockwise or counter clockwise neighbors gives these identities:

(Starting at tan and going clockwise)

$\tan A={\sin A \over \cos A}$

$\sin A={\cos A \over \cot A}$

$\cos A={\cot A \over \csc A}$

$\cot A={\csc A \over \sec A}$

$\csc A={\sec A \over \tan A}$

$\sec A={\tan A \over \sin A}$

(Starting at tan and going counter-clockwise)

$\tan A={\sec A \over \csc A}$

$\sec A={\csc A \over \cot A}$

$\csc A={\cot A \over \cos A}$

$\cot A={\cos A \over \sin A}$

$\cos A={\sin A \over \tan A}$

$\sin A={\tan A \over \sec A}$

Reading a function and multiplying the two nearest neighbors gives these identities (starting at tan and going clockwise):

$\tan A=\sin A\cdot \sec A\$

$\sin A=\cos A\cdot \tan A\$

$\cos A=\sin A\cdot \cot A\$

$\cot A=\cos A\cdot \csc A\$

$\csc A=\cot A\cdot \sec A\$

$\sec A=\csc A\cdot \tan A\$

References

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