ত্রিকোণমিতির সূত্র

বিপরীত সূত্র (Reciprocal Identities):

\sin θ = \frac{1}{\csc θ}, \cos θ = \frac{1}{\sec θ}, \tan θ = \frac{1}{\cot θ}

\csc θ = \frac{1}{\sin θ}, \sec θ = \frac{1}{\cos θ}, \cot θ = \frac{1}{\tan θ}

\sin^{2} θ + \cos^{2} θ = 1

1 + \tan^{2} θ = \sec^{2} θ

1 + \cot^{2} θ = \csc^{2} θ

\tan θ = \frac{\sin θ}{\cos θ}, \cot θ = \frac{\cos θ}{\sin θ}

\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

\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}

\begin{aligned} \sin 2 A & = 2 \sin A \cos A \\ = \frac{2 \tan A}{1 + \tan^{2} A} \end{aligned}

\begin{aligned} \cos 2 A & = \cos^{2} A - \sin^{2} A \\ = 2 \cos^{2} A - 1 \\ = 1 - 2 \sin^{2} A \end{aligned}

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

\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ

\cos 3 θ = 4 \cos^{3} θ - 3 \cos θ

\tan 3 θ = \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ}

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

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

\begin{aligned} \tan \frac{A}{2} & = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} \\ = \frac{\sin A}{1 + \cos A} \\ = \frac{1 - \cos A}{\sin A} \end{aligned}

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

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

\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)]

\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}

\begin{aligned} \cos A - \cos B & = - 2 \sin \frac{A + B}{2} \sin \frac{A - B}{2} \\ = 2 \sin \frac{A + B}{2} \sin \frac{B - A}{2} \end{aligned}

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R

Info

এখানে,
$R =$ ত্রিভূজের পরিবৃত্তের ব্যাসার্ধ

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

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

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

\begin{aligned} Area & = \frac{1}{2} a b \sin C \\ = \frac{1}{2} b c \sin A \\ = \frac{1}{2} c a \sin B \end{aligned}

Area = \sqrt{s (s - a) (s - b) (s - c)}

Info

$s$ হচ্ছে অর্ধপরিসীমা
$s = \frac{a + b + c}{2}$

\sin \frac{A}{2} = \sqrt{\frac{(s - b) (s - c)}{b c}}

\sin \frac{B}{2} = \sqrt{\frac{(s - c) (s - a)}{c a}}

\sin \frac{C}{2} = \sqrt{\frac{(s - a) (s - b)}{a b}}

\cos \frac{A}{2} = \sqrt{\frac{(s - a)}{b c}}

\cos \frac{B}{2} = \sqrt{\frac{s (s - b)}{c a}}

\cos \frac{C}{2} = \sqrt{\frac{s (s - c)}{a b}}

\tan \frac{A}{2} = \sqrt{\frac{(s - b) (s - c)}{s (s - a)}} = \frac{(s - b) (s - c)}{Δ} = \frac{Δ}{s (s - a)}

\tan \frac{B}{2} = \sqrt{\frac{(s - c) (s - a)}{s (s - b)}} = \frac{(s - c) (s - a)}{Δ} = \frac{Δ}{s (s - b)}

\tan \frac{C}{2} = \sqrt{\frac{(s - a) (s - b)}{s (s - c)}} = \frac{(s - a) (s - b)}{Δ} = \frac{Δ}{s (s - c)}