SPECIAL ANGLES

Special angles are angles whose trigonometry function can be got without using tables or calculator. Example of such angles are 30°, 60° and 45°.

The triangle below will help you to get the sin, cos and tan of these angles. 

From fig 1, if we consider angle 30°, opp =1, adj = √3 and hyp = 2

Sin θ = opp/hyp

Sin 30 = ½

Cos 30 = adj/hyp =  \frac{\sqrt{3}}{2}

Tan 30 = 1/√3.

 

Considering angle 60, opp is √3, adj is 1 and hyp is 2

Sin 60 =  \frac{\sqrt{3}}{2}

Cos 60 = ½

Tan 60 = \frac{\sqrt{3}}{1} 

            = √3.

 

Considering the angle 45 in fig 2, opposite =1, adjacent =1 and hypotenuse = √2.

Sin 45 = 1/√2

Cos 45 = 1/√2

Tan 45 = 1/1

            = 1.

 

Master the two diagrams (fig 1 and fig 2) and how to put the values to each side correctly. This is very important when solving special angles questions.

 

    Example

Without using tables or calculator, find the value of the following.

   1. Cos 60°

   2. Sin 45°

   3. Tan 135°

   4. Cos 240°

   5. Cos 315°

   6. Sin 210°

 

         Solution

Once you see special angles questions telling you not to use tables, then draw your two triangles (fig1 and fig2 above.)

        1. Cos 60 = ½

 

        2. Sin 45 = 1/√2

 

        3. Tan 135

135 is not a special angle, so we will get its second value from the second quad. formula (180 – θ).

135 = (180 – 45)

 The 45 we subtracted to get 135 is the second value and it’s a special angle.

We will therefore look for the value of tan 45. from our fig 2,

Tan 45 = 1.

 

Our original angle 135 is on the second quadrant, tan is negative here. It means our answer will also take –.

∴ tan 135 = – 1.

 

** our final answer above is tan 135 and not 45 since the question is asking for tan 135 and not tan 45. We only used tan 45 to get our answer.

     

          4.  Cos 240°.

240 is a 3rd quadrant angle (180 + θ) and cos is negative here.

= – Cos (180 + 60)

= – cos 60

= – ½.

∴ cos 240 = – ½.

     

          5.  Cos 315°.

= cos (360 – 45)         cos is positive here

= cos 45

= 1/√2

∴ Cos 315 = 1/√2.

 

      6.  Sin 210°

= sin (180 + 30)       3rd quad. Sin is negative

= – sin 30

= – ½

∴ sin 240 = – ½.

 

              Exercise

Without using tables, find the value of the following

   1. Tan 120

   2. Cos 300

   3. Sin 225

   4. Sin 150

   5. Cos 210

   6. Tan 330

   7. Sin 315 + Cos 120

   8. 1 – 2 cos 15

   9. 2 sin 45 cos 60

  10. Tan 210 – sin 210

  11. \frac{\tan 300 + \tan 210}{1 - \tan 300 \tan210 }

   12. Cos2225 – tan2225 

   13. \frac{\sin 60 + \cos 30}{1 - \sin 30 \ tan 45} 

   14. \frac{1}{\cos 315} 

    15. Sin (– 150)

 

            Example 2

Without using table, evaluate the following.

        1. Sin (–240)

        2. Tan (–330)

        3. Cos (-120)

        4.   \frac{\sin ^2 60 + \tan ^2 315}{1 + 2 \sin 135\cos 300} leaving your answer in surd form

           

             SOLUTION

1     Sin (–240)

              = – Sin 240

              = – Sin (180 + 60)

              = – (– Sin 60)

The first – is the initial one from the question, the second – is because 240 belongs to the 3rd quad. And sin is negative here.

             = – (– \frac{\sqrt{3}}{2})

             =   \frac{\sqrt{3}}{2}

 

          2.  Tan (–330)

           = – tan 330

           = – tan (360 – 30)

           = – (– tan 30) 

330 belongs to 4th quad. And tan is negative here, this means

           = – (– 1/√3)

           = 1/√3.

 

          3. Cos (-120)

Unlike in sin and tan, cos of a negative number is same as that of a positive number i.e. cos (-210) = cos 210 and cos (- 150) = cos 150 etc.

               = cos 120

              = cos (180 – 60)        Notice there is no – in front of cos like in sin and tan.

             = – cos 60        the new here is because 120 is 2nd quad.

  

            = – ½

∴ cos 120 = – ½

     

      4.    \frac{\sin ^2 60 + \tan ^2 315}{1 + 2 \sin 135\cos 300} 

        first change all the angles (315, 135, 300) that are not special angles to special angles

            ** NOTE: tan2 315 means (tan 315)2  and sin2 60 means (sin 60)2 

             = \frac{(\sin 60) ^2 + (\tan 360 - 45)^2}{1 + 2 \sin (180 -45) \cos (360 - 60) } \\ \\ \\ = \ \ \ \frac{(\sin 60) ^2 + (- \tan 45)^2}{1 + 2 \sin 45 \cos 60 } \\

note: tan is negative in 4th quad. hence the –

= \frac{(\sin 60) ^2 + (- \tan 45)^2}{1 + 2 \sin 45 \cos 60 } \\ \\ \\ = \frac{(\frac{\sqrt{3}}{2}) ^2 + (-1)^2}{1 + 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} } \\

= \frac{\frac{3}{4} + 1}{1 + \frac{1}{\sqrt{2}} } \\

= \frac{\frac{3 + 4}{4}}{\frac{\sqrt{2} + 1}{\sqrt{2}} } \\ \\ = \frac{\frac{7}{4}}{\frac{\sqrt{2} + 1}{\sqrt{2}} } \\ \\ =\frac{7}{4} \div {\frac{\sqrt{2} + 1}{\sqrt{2}} } \\ \\

=\frac{7}{4} \times {\frac{\sqrt{2}}{\sqrt{2} + 1} } \\ \\ =\frac{7\sqrt{2}}{4\sqrt{2} + 4}

rationalise the denominator

= \frac{7\sqrt{2}}{4\sqrt{2} + 4} \times \frac{4 \sqrt{2} - 4}{4\sqrt{2} - 4} \\ \\ = \frac{7\sqrt{2} \times 4\sqrt{2} - 7\sqrt{2} \times 4 }{4\sqrt{2} \times 4\sqrt{2} - 4 \times 4} \\ \\ \\ = \frac{28 \times 2 -28\sqrt{2}}{16 \times 2 - 4 \times 4}

=\frac {28 (2 - \sqrt 2)}{16} \\ \\ =\frac {7 (2 - \sqrt 2)}{4}

 

          Exercise 

Evaluate the following without using tables

      1.   sin (–330)

      2.   tan (–315)

      3.   2 sin (–240)

      4.   cos (– 225)

      5.   tan2 (–300)

      6.   5 sin2 (–210) cos2(–300)

 

PRE: COMPLEMENTARY ANGLES

NEXT: TRIGONOMETRICAL EQUATIONS

PRE

NEXT

Leave a Reply

Your email address will not be published. Required fields are marked *