PS: Note that all the angles are in degree.

Exercise 1

Knowing that \cos { \alpha } = \frac { 1 }{ 4 } , and that 270 < \alpha < 360, calculate the remaining trigonometric ratios of angle \alpha.

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Exercise 2

Knowing that \tan { \alpha } = 2, and that 180 < \alpha < 270, calculate the remaining trigonometric ratios of angle \alpha.

Exercise 3

Knowing that \sec { \alpha } = 2 and 0 < \alpha < \frac { \pi }{ 2 }, calculate the remaining trigonometric ratios of angle \alpha.

Exercise 4

Knowing that \csc { \alpha } = 3, calculate the remaining trigonometric ratios of angle \alpha.

Exercise 5

Prove the identities:

1\tan { \alpha } + \cot { \alpha } = \sec { \alpha } . \csc { \alpha }

2\cot^{ 2 }{ a } = \cos^{ 2 }{ a } + { (\cot { a } . \cos { a }) }^{ 2 }

3\frac { 1 }{ \sec^{ 2 }{ a }} = \sin^{ 2 }{ a } . \cos^{ 2 }{ a } + \cos^{ 4 }{ a }

4\cot { a } . \sec { a } = \csc { a }

5\sec^{ 2 }{ a } + \csc^{ 2 }{ a } = \frac { 1 }{ \sin^{ 2 }{ a } . \cos^{ 2 }{ a }}

Exercise 6

Simplify the fractions:

1 \frac { 1 + \tan^{ 2 }{ x }}{ 1 + \cot^{ 2 }{ x } }

2 \frac { \sec^{ 2 }{ a } - \cos^{ 2 }{ a }}{ \tan^{ 2 }{ a } }

3 \frac { \csc^{ 2 }{ a } - \sin^{ 2 }{ a } }{\csc^{ 2 }{ a } . (2 - \cos^{ 2 }{ a })}

Exercise 7

Prove the identities:

Part 1

\sin { b } . \cos { (a - b) } + \cos { b } . \sin { (a - b) } = \sin { a }

Part 2

\cot { (a + b) } = \frac { \cot { a } . \cot { b } - 1 }{ \cot { a } + \cot { b }}

Exercise 8

Simplify the fractions:

Part 1

\frac { \sin { 2x }}{ 1 + \cos { 2x }}

Part 2

\frac { \sin { 2a }}{ 1 - \cos^{ 2 }{ a }} . \frac { \sin { 2a }}{ \cos { a }}

Part 3

\frac { \sin { 3a } - \sin { 5a }}{ \cos { 3a } + \cos { 5a }}

Exercise 9

Calculate the trigonometric ratios of 15 (from the 45 and 30).

Exercise 10

Develop: \cos { (x+y+z) }.

Exercise 11

Calculate \sin { 3x }, depending on \sin { x }.

Exercise 12

Calculate \sin { x }, \cos { x }, and \tan { x }, in terms of \tan { \frac { x }{ 2 }}.

 

 

Solution of exercise 1

Knowing that \cos { \alpha } = \frac { 1 }{ 4 } , and that 270 < \alpha < 360, calculate the remaining trigonometric ratios of angle \alpha.

\sin { \alpha } = - \sqrt { 1 - { \frac { 1 }{ 4 } }^{ 2 } } = - \frac { \sqrt { 15 }}{ 4 } \qquad \csc { \alpha } = \frac { 4 \sqrt { 15 } }{ 15 }

\cos { \alpha } = \frac { 1 }{ 4 } \qquad \sec { \alpha } = 4

\tan { \alpha } = \frac { \frac { \sqrt { 15 }}{ 4 } }{ \frac { 1 }{ 4 } } = - \sqrt { 15 } \qquad \cot { \alpha } = - \frac { \sqrt { 15 } }{ 15 }

 

Solution of exercise 2

Knowing that \tan { \alpha } = 2, and that 180 < \alpha < 270, calculate the remaining trigonometric ratios of angle \alpha.

\sec { \alpha } = - \sqrt { 1 + 4 } = - \sqrt { 5 } \qquad \cos { \alpha } = - \frac { 1 }{ \sqrt { 5 } } = - \frac { \sqrt { 5 } }{ 5 }

\sin { \alpha } = 2 . (- \frac { \sqrt { 5 } }{ 5 }) = - \frac { 2 \sqrt { 5 } }{ 5 } \qquad \csc { \alpha } = - \frac { \sqrt { 5 } }{ 2 }

\tan { \alpha } = 2 \qquad \cot { \alpha } = \frac { 1 }{ 2 }

 

Solution of exercise 3

Knowing that \sec { \alpha } = 2 and 0 < \alpha < \frac { \pi }{ 2 }, calculate the remaining trigonometric ratios of angle \alpha.

\cos { \alpha } = \frac { 1 }{ 2 } \qquad \sec { \alpha } = 2

\sin { \alpha } = \sqrt { 1 - { (\frac { 1 }{ 2 }) }^{ 2 } } = \frac { \sqrt { 3 } }{ 2 } \qquad \csc { \alpha } = \frac { 2 \sqrt { 3 }}{ 3 }

\tan { \alpha } = \frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 1 }{ 2 } } = \sqrt { 3 } \qquad \cot { \alpha } = \frac { \sqrt { 3 } }{ 3 }

 

Solution of exercise 4

Knowing that \csc { \alpha } = 3, calculate the remaining trigonometric ratios of angle \alpha.

First quadrant:

\sin { \alpha } = \frac { 1 }{ 3 } \qquad \csc { \alpha } = 3

\cos { \alpha } = \sqrt { 1 - { (\frac { 1 }{ 3 }) }^{ 2 } } = \sqrt { \frac { 8 }{ 9 } } = \frac { 2 \sqrt { 2 } }{ 3 } \qquad \sec { \alpha } = \frac { 3 \sqrt { 2 } }{ 4 }

\tan { \alpha } = \frac { \frac { 1 }{ 3 } }{ \frac { 2 \sqrt { 2 } }{ 3 } } = \frac { 1 }{ 2 \sqrt { 2 } } = \frac { \sqrt { 2 } }{ 4 } \qquad \cot { \alpha } = 2 \sqrt { 2 }

 

Second quadrant:

\sin { \alpha } = \frac { 1 }{ 3 } \qquad \csc { \alpha } = 3

\cos { \alpha } = - \sqrt { 1 - { (\frac { 1 }{ 3 }) }^{ 2 } } = - \sqrt { \frac { 8 }{ 9 } } = - \frac { 2 \sqrt { 2 } }{ 3 } \qquad \sec { \alpha } = - \frac { 3 \sqrt { 2 } }{ 4 }

\tan { \alpha } = \frac { \frac { 1 }{ 3 } }{ - \frac { 2 \sqrt { 2 } }{ 3 } } = - \frac { 1 }{ 2 \sqrt { 2 } } = - \frac { \sqrt { 2 } }{ 4 } \qquad \cot { \alpha } = - 2 \sqrt { 2 }

 

Solution of exercise 5

Prove the identities:

1\tan { \alpha } + \cot { \alpha } = \sec { \alpha } . \csc { \alpha }

\tan { \alpha } + \cot { \alpha } = \frac { \sin { \alpha } }{ \cos { \alpha } } + \frac { \cos { \alpha } }{ \sin { \alpha }} = \frac { \sin^{ 2 }{ \alpha } + \cos^{ 2 }{ \alpha } }{ \cos { \alpha } . \sin { \alpha }}

= \frac { 1 }{ \cos { \alpha } . \sin { \alpha } } = \sec { \alpha } . \csc { \alpha }

2\cot^{ 2 }{ a } = \cos^{ 2 }{ a } + { (\cot { a } . \cos { a }) }^{ 2 }

\cos^{ 2 }{ a } + { (\cot { a } . \cos { a }) }^{ 2 } = \cos^{ 2 }{ a } . \csc^{ 2 }{ a } = \frac { \cos^{ 2 }{ a } }{ \sin^{ 2 }{ a } } = \cot^{ 2 }{ a }

3\frac { 1 }{ \sec^{ 2 }{ a }} = \sin^{ 2 }{ a } . \cos^{ 2 }{ a } + \cos^{ 4 }{ a }

\sin^{ 2 }{ a } . \cos^{ 2 }{ a } + \cos^{ 4 }{ a } = \cos^{ 2 }{ a }( \sin^{ 2 }{ a } + \cos^{ 2 }{ a } ) = \cos^{ 2 }{ a } = \frac { 1 }{ \sec^{ 2 }{ a } }

4\cot { a } . \sec { a } = \csc { a }

\cot { a } . \sec { a } = \frac { \cos { a } }{ \sin { a }} . \frac { 1 }{ \cos { a } } = \frac { 1 }{ \sin { a } } = \csc { a }

5\sec^{ 2 }{ a } + \csc^{ 2 }{ a } = \frac { 1 }{ \sin^{ 2 }{ a } . \cos^{ 2 }{ a }}

\sec^{ 2 }{ a } + \csc^{ 2 }{ a } = \frac { 1 }{ \cos^{ 2 }{ a } } + \frac { 1 }{ \sin^{ 2 }{ a } } = \frac { \sin^{ 2 }{ a } + \cos^{ 2 }{ a } }{ \sin^{ 2 }{ a } . \cos^{ 2 }{ a } } = \frac { 1 }{ \sin^{ 2 }{ a } . \cos^{ 2 }{ a }}

Solution of exercise 6

Simplify the fractions:

1 \frac { 1 + \tan^{ 2 }{ x }}{ 1 + \cot^{ 2 }{ x } }

\frac { 1 + \tan^{ 2 }{ x } }{ 1 + \cot^{ 2 }{ x } } = \frac { \sec^{ 2 }{ x } }{ \csc^{ 2 }{ x } } = \frac { \frac { 1 }{ \cos^{ 2 }{ x } } }{ \frac { 1 }{ \sin^{ 2 }{ x } } } = \tan^{ 2 }{ x }

 

2 \frac { \sec^{ 2 }{ a } - \cos^{ 2 }{ a }}{ \tan^{ 2 }{ a } }

\frac { \sec^{ 2 }{ a } - \cos^{ 2 }{ a }}{ \tan^{ 2 }{ a } } = \frac { \frac { 1 }{ \cos^{ 2 }{ a } } - \cos^{ 2 }{ a } }{ \tan^{ 2 }{ a } } =

= \frac { (1 - \cos^{ 2 }{ a } )(1 + \cos^{ 2 }{ a } ) }{ \sin^{ 2 }{ a }} = 1 + \cos^{ 2 }{ a }

 

3 \frac { \csc^{ 2 }{ a } - \sin^{ 2 }{ a } }{\csc^{ 2 }{ a } . (2 - \cos^{ 2 }{ a })}

\frac { \csc^{ 2 }{ a } - \sin^{ 2 }{ a } }{\csc^{ 2 }{ a } . (2 - \cos^{ 2 }{ a })} = \frac { 1 }{ 2 - \cos^{ 2 }{ a } } - \frac { \sin^{ 4 }{ a } }{ 2 - cos^{ 2 }{ a } } =

= \frac { 1 - \sin^{ 4 }{ a } }{ 1 + 1 - \cos^{ 2 }{ a } } = \frac { 1 - \sin^{ 4 }{ a } }{ 1 + \sin^{ 2 }{ a } } = \frac { (1 + \sin^{ 2 }{ a })(1 - \sin^{ 2 }{ a } ) }{ 1 + \sin^{ 2 }{ a } } = \cos^{ 2 }{ a }

 

Solution of exercise

Solution of exercise 7

Prove the identities:

1 \sin { b } . \cos { (a - b) } + \cos { b } . \sin { (a - b) } = \sin { a }

sin { [b + (a - b)] } = \sin { a }

 

2 \cot { (a + b) } = \frac { \cot { a } . \cot { b } - 1 }{ \cot { a } + \cot { b }}

\cot { (a + b) } = \frac { 1 }{ \tan { (a+b) }} = \frac { 1 }{ \frac { \tan { a } + \tan { b }}{1 - \tan { a } . \tan { b }} } = \frac { 1 - \tan { a } . \tan { b }}{ \tan { a } + \tan { b }} =

\frac { \frac { 1 }{ \tan { a } . \tan { b }} }{ \frac { \tan { a }}{ \tan { a } . \tan { b }} + \frac { \tan { b }}{ \tan { a } . \tan { b }}} = \frac { \cot { a } . \cot { b } - 1 }{ \cot { a } + \cot { b }}

 

Solution of exercise

Solution of exercise 8

Simplify the fractions:

1 \frac { \sin { 2x }}{ 1 + \cos { 2x }}

\frac { \sin { 2x }}{ 1 + \cos { 2x }} = \frac { 2 \sin { x } . \cos { x } }{ 1 + \cos^{ 2 }{ x } - \sin^{ 2 }{ x } } = \frac { 2 \sin { x } . \cos { x } }{ \cos^{ 2 }{ x } + \cos^{ 2 }{ x } } =\tan { x }

 

2 \frac { \sin { 2a }}{ 1 - \cos^{ 2 }{ a }} . \frac { \sin { 2a }}{ \cos { a }}

\frac { \sin { 2a } }{ 1 - \cos^{ 2 }{ a } } . \frac { \sin { 2a } }{ \cos { a } } = \frac { { (2 \sin { a } . \cos { a }) }^{ 2 } }{ \sin^{ 2 }{ a } . \cos { a } } = 4 \cos { a }

 

3 \frac { \sin { 3a } - \sin { 5a }}{ \cos { 3a } + \cos { 5a }}

\frac { \sin { 3a } - \sin { 5a }}{ \cos { 3a } + \cos { 5a }} = \frac { 2 \cos { 4a } . \sin { (-a) } }{ 2 \cos { 4a } . \cos { ( -a )}} = - \tan { a }

 

Solution of exercise

Solution of exercise 9

Calculate the trigonometric ratios of 15 (from the 45 and 30).

\sin { 15 } = \sin { (45 - 30) } = \sin { 45 } \cos { 30 } - \cos { 45 } \sin { 30 }

= \frac { \sqrt { 2 } }{ 2 } . \frac { \sqrt { 3 } }{ 2 } - \frac { \sqrt { 2 }}{ 2 } . \frac { 1 }{ 2 } = \frac { \sqrt { 2 } }{ 4 } ( \sqrt { 3 } - 1)

 

\cos { 15 } = \cos { (45 - 30) } = \cos { 45 } \cos { 30 } - \sin { 45 } \sin { 30 }

= \frac { \sqrt { 2 } }{ 2 } . \frac { \sqrt { 3 } }{ 2 } + \frac { \sqrt { 2 }}{ 2 } . \frac { 1 }{ 2 } = \frac { \sqrt { 2 } }{ 4 } ( \sqrt { 3 } + 1)

 

Solution of exercise

Solution of exercise 10

Develop: \cos { (x+y+z) }.

\cos { (x + y + z) } = \cos { [x + (y + z)] } =

= \cos { x } . \cos { (y + z) } - \sin { x } . \sin { (y + z) } =

= \cos { x } (\cos { y } . \cos { z } - \sin { y } . \sin { z }) - \sin { x } (\sin { y } . \cos { z } + \cos { y } . \sin { z }) =

= \cos { x } . \cos { y } . \cos { z } - \cos { x } . \sin { y } . \sin { z } - \sin { x } . \sin { y } . \cos { z } - \sin { x } . \cos { y } . \sin { z }

 

Solution of exercise 11

Calculate \sin { 3x }, depending on \sin { x }.

\sin { 3x } = \sin { (2x + x) } = \sin { 2x } . \cos { x } + \cos { 2x } . \sin { x } =

= (2 \sin { x } . \cos { x }) \cos { x } + (\cos^{ 2 }{ x } - \sin^{ 2 }{ x }) \sin { x }=

= 2 \sin { x } . \cos^{ 2 }{ x } + \cos^{ 2 }{ x } . \sin { x } - \sin^{ 3 }{ x } =

= 3 \sin { x } . \cos^{ 2 }{ x } - \sin^{ 3 }{ x } = 3 \sin { x }  (1 - \sin^{ 2 }{ x }) - \sin^{ 3 }{ x } =

= 3 \sin { x } - 4 \sin^{ 3 }{ x }

 

Solution of exercise 12

Calculate \sin { x }, \cos { x }, and \tan { x }, in terms of \tan { \frac { x }{ 2 }}.

\sin { x } = 2 \sin { \frac { x }{ 2 }} \cos { \frac { x }{ 2 } } = \frac { 2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } } = \frac { \frac { 2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } }{ \frac { \cos ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } +\frac { \sin ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } } = \frac { 2 \tan { \frac { x }{ 2 } }}{ 1 + \tan^{ 2 }{ \frac { x }{ 2 } } }

\cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } =\frac { \cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } } =\frac { \frac { \cos ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } -\frac { \sin ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } }{ \frac { \cos ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } +\frac { \sin ^{ 2 }{ \frac { x }{ 2 } } }{ \cos ^{ 2 }{ \frac { x }{ 2 } } } } =\frac { 1-\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } }

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

 

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Emma

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