Trigonometric Identities

Trigonometric Equation Solvers

Select an equation type below to solve it. Solutions will be found in the range $0^\circ \le A < 360^\circ$.

$b$:

$c$:

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$a$:

$c$:

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$a$:

$b$:

$c$:

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This solver demonstrates using $\cos^2 A = 1 - \sin^2 A$ to reduce the equation $A \cos^2 X + B \sin X + C = 0$ to a quadratic in $\sin X$.

(Coeff. of $\cos^2 X$, $A$):

(Coeff. of $\sin X$, $B$):

(Constant term, $C$):

Solution steps will appear here...

$\sin^2 \theta + \cos^2 \theta = 1$

$1 + \tan^2 \theta = \sec^2 \theta$

$1 + \cot^2 \theta = \csc^2 \theta$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{1}{\tan \theta}$

$\sec \theta = \frac{1}{\cos \theta}$

$\csc \theta = \frac{1}{\sin \theta}$

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

$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

$\sin(2A) = 2 \sin A \cos A$

$\cos(2A) = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$

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