TRIGONOMETRY FORMULAE

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Trigonometry Formulae: Don’t try to remember just know the concept of derivation

With the help of sin (A + B), we can derive all the remaining trigonometric formulae. Let’s see, how …..

Compound Angle

(1) \(\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\)

(2) \(\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\)

(3) \(\cos \left( A+B \right)=\sin \left[ \frac{\pi }{2}-\left( A+B \right) \right]\)

\(=\sin \left[ \left( {}^{\pi }\!\!\diagup\!\!{}_{2}\;-A \right)-\left( B \right) \right]\)

\(=\sin \left( \frac{\pi }{2}-A \right)\cos B-\cos \left( \frac{\pi }{2}-A \right)\sin B\)

\(\therefore \,\,\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\)

(4) \(\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B\)

(5) \(\tan \left( A+B \right)=\frac{\sin \left( A+B \right)}{\cos \left( A+B \right)}\)

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

Dividing Numerator and Denominator by clip_image002[4]

\(\therefore \,\tan \left( A+B \right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)

(6) \(\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\)

Similarly,

(7) \(\cot \left( A+B \right)=\frac{\cot A\cot B-1}{\cot B+\cot A}\)

(8) \(\cot \left( A-B \right)=\frac{\cot A\cot B+1}{\cot B-\cot A}\)

Transformation Formulae:

By adding (1) and (2), we get

(9) \(2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)\)

By subtracting (2) from (1), we get

(10) \(2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)\)

By adding (3) and (4), we get

(11) \(\displaystyle 2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\)

By subtracting (4) from (3), we get

(12) \(2\sin A\sin B=-\cos \left( A+B \right)+\cos \left( A-B \right)\)

Now, consider \(A+B=C\) and \(A-B=D\)

Then, \(A=\frac{C+D}{2}\,\) and \(B=\frac{C-D}{2}\)

(13) From (9), \(\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}\)

(14) From (10), \(\sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)

(15) From (11), \(\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}\)

(16) From (12), \(\cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}\)

Multiply (1) and (2), we get

(17) \({{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right).\sin \left( A-B \right)\)

\({{\cos }^{2}}B-{{\cos }^{2}}A=\sin \left( A+B \right).\sin \left( A-B \right)\)

(18) \({{\cos }^{2}}B-{{\sin }^{2}}A=\cos \left( A+B \right).\cos \left( A-B \right)\)

\({{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right).\cos \left( A-B \right)\)

 

Multiple Angles

(19) \(\sin \left( 2A \right)=\sin \left( A+A \right)\)

\(=\sin A\cos A+\cos A\sin A\,\,\,\,\,\,\,\,\,\,\left\{ \text{from }\left( 1 \right) \right\}\)

\(\therefore \,\,\sin \left( 2A \right)=2\sin A\cos A\)

(20) \(\cos \left( 2A \right)=\cos \left( A+A \right)\)

\(=\cos A\cos A-\sin A\sin A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \text{from}\,\left( 3 \right) \right\}\)

\(\therefore \,\,\cos \left( 2A \right)={{\cos }^{2}}A-{{\sin }^{2}}A\)

\(=1-2{{\sin }^{2}}A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \text{from}\,\,\,{{\sin }^{2}}A+{{\cos }^{2}}A=1 \right\}\)

\(=2{{\cos }^{2}}A-1\)

(21) \(\tan \left( 2A \right)=\tan \left( A+A \right)\)

image

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

\(\therefore \tan \left( 2A \right)\,\,=\frac{2\tan A}{1-{{\tan }^{2}}A}\)

\(\sin \left( 2A \right)\,\,\,\,\,=\frac{2\tan A}{1+{{\tan }^{2}}A}\)

\(\cos \left( 2A \right)\,\,\,=\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\)

(22) \(\sin \left( 3A \right)=\sin \left( 2A+A \right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sin 2A\cos A+\cos 2A\sin A\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin A\cos A\cos A+\left( 1-2{{\sin }^{2}}A \right)\sin A\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin A\left( 1-{{\sin }^{2}}A \right)+\left( 1-2{{\sin }^{2}}A \right)\sin A\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin A-2{{\sin }^{3}}A+\sin A-2{{\sin }^{3}}A\)

\(\therefore \sin \left( 3A \right)=3\sin A-4{{\sin }^{3}}A\)

(23) \(\cos \left( 3A \right)=\cos \left( 2A+A \right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\cos 2A\cos A-\sin 2A\sin A\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( 2{{\cos }^{2}}A-1 \right)\cos A-2\sin A\cos A.\sin A\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( 2{{\cos }^{2}}A-1 \right)\cos A-2\cos A\left( 1-{{\cos }^{2}}A \right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2{{\cos }^{3}}A-\cos A-2\cos A+2{{\cos }^{3}}A\)

\(\therefore \cos \left( 3A \right)\,\,=4{{\cos }^{3}}A-3\cos A\)

(24) \(\tan \left( 3A \right)=\tan \left( 2A+A \right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\tan 2A+\tan A}{1-\tan 2A\tan A}\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\frac{2\tan A}{1-{{\tan }^{2}}A}+\tan A}{1-\frac{2\tan A}{1-{{\tan }^{2}}A}.\tan A}\)

\(\therefore \tan \left( 3A \right)=\frac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}\)

From (20), \(\cos 2A=1-2{{\sin }^{2}}A\)

And \(\cos 2A=2{{\cos }^{2}}A-1\)

(25) \(1-\cos 2A=2{{\sin }^{2}}A\)

(26) \(1+\cos 2A=2{{\cos }^{2}}A\)

(27) \(1-\sin 2A={{\sin }^{2}}A+{{\cos }^{2}}A-2\sin A\cos A\)

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(28) \(1+\sin 2A={{\left( \cos A+\sin A \right)}^{2}}\)

 

Half Angles

From (25), \(\sin A=\sqrt{\frac{1-\cos 2A}{2}}\)

(29) \(\sin \left( \frac{A}{2} \right)=\sqrt{\frac{1-\cos A}{2}}\)

(30) \(\cos \left( \frac{A}{2} \right)=\sqrt{\frac{1+\cos A}{2}}\)

(31) \(\tan \left( \frac{A}{2} \right)=\sqrt{\frac{1-\cos A}{1+\cos A}}\)

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