Further Trigonometric Equations and Identities

Special Ratios:

 

If \tan { \cfrac { \theta }{ 2 } } =t then:

\tan { \theta } =\cfrac { 2t }{ 1-{ t }^{ 2 } }

\sin { \theta } =\cfrac { 2t }{ 1+{ t }^{ 2 } }

\cos { \theta } =\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }

 

Examples

 

1. Find the exact value of  \cfrac { 2\tan { 15 } }{ 1+\tan ^{ 2 }{ 15 } }.

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\cfrac { 2\tan { 15 } }{ 1+\tan ^{ 2 }{ 15 } } \\ \\ \sin { \theta } =\cfrac { 2t }{ 1+{ t }^{ 2 } } \quad where\quad t=\tan { \cfrac { \theta }{ 2 } } \\ \\ \therefore \quad \cfrac { 2\tan { 15 } }{ 1+\tan ^{ 2 }{ 15 } } =\sin { 30 } \\ \\ \qquad \sin { 30 } =\cfrac { 1 }{ 2 }

 

2. Prove that \cot { \cfrac { \theta }{ 2 } } -2\cot { \theta } =\tan { \cfrac { \theta }{ 2 } } .

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LHS=\cfrac { 1 }{ \tan { \cfrac { \theta }{ 2 } } } -\cfrac { 2 }{ \tan { \theta } } \\ \\ =\cfrac { 1 }{ t } -\cfrac { 2 }{ \cfrac { 2t }{ 1-{ t }^{ 2 } } } \quad where\quad t=\tan { \cfrac { \theta }{ 2 } } \\ \\ =\cfrac { 1 }{ t } -\cfrac { 2(1-{ t }^{ 2 }) }{ 2t } \\ \\ =\cfrac { 1 }{ t } -\cfrac { (1-{ t }^{ 2 }) }{ t } \\ \\ =\cfrac { 1-1+{ t }^{ 2 } }{ t } \\ \\ =\cfrac { { t }^{ 2 } }{ t } \\ \\ =\quad t\\ \\ \therefore \quad =\quad \tan { \cfrac { \theta }{ 2 } } \quad =\quad RHS

More Trigonometric Equations

a\sin { \theta } +b\cos { \theta } =r\sin { (\theta +\alpha ) } \\ \\ r=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad and\quad \tan { \alpha } =\cfrac { b }{ a }

 

Examples

 

1. Write \sqrt { 3 } \sin { x } +\cos { x } in the form r\sin { (x+\alpha ) } .

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Using the trigonometric equation:

a=\sqrt { 3 } ,\quad b=1\\ \\ r=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \\ \\ r=\sqrt { { \sqrt { 3 } }^{ 2 }+{ 1 }^{ 2 } } \\ \\ r=\sqrt { 3+1 } \\ \\ r=\sqrt { 4 } \\ \\ r=2\\ \\ \\ \tan { \alpha } =\cfrac { b }{ a } \\ \\ \qquad \quad =\cfrac { 1 }{ \sqrt { 3 } } \\ \\ \alpha =\tan ^{ -1 }{ \left( \cfrac { 1 }{ \sqrt { 3 } } \right) } \\ \\ \quad =30^{ \circ }\\ \\ \therefore \quad \sqrt { 3 } \sin { x } +\cos { x } =2\sin { (x+30^{ \circ }) }

 

2. Solve \sqrt { 3 } \sin { x } +\cos { x } =1\quad for\quad 0^{ \circ }\quad \le \quad x\quad \le \quad 360^{ \circ }.

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From the previous example, we know that a=\sqrt{3}b=1, r = 2 and \alpha = 30^\circ.

\therefore \quad \sqrt { 3 } \sin { x } +\cos { x } =2\sin { (x+30^{ \circ }) }

Solving:

\sqrt { 3 } \sin { x } +\cos { x } =1\qquad for\quad 0^{ \circ }\quad \le \quad x\quad \le \quad 360^{ \circ }\\ 2\sin { (x+30^{ \circ }) } =1\qquad \qquad for\quad 0^{ \circ }\quad \le \quad x+30^{ \circ }\quad \le \quad 360^{ \circ }\\ \\ \sin { (x+30^{ \circ }) } =\cfrac { 1 }{ 2 } \\ \\ x+30^{ \circ }=30^{ \circ },\quad 180^{ \circ }-30^{ \circ },\quad 360^{ \circ }+30^{ \circ }\\ x+30^{ \circ }=30^{ \circ },\quad 150^{ \circ },\quad 390^{ \circ }\\ x=0^{ \circ },\quad 120^{ \circ },\quad 360^{ \circ }