# Home > Trigonometry > Exact Values of Trigonometric Ratios

# Exact Values of Trigonometric Ratios To find the exact values of trigonometric ratios, consider an equilateral triangle ABC of sides 2 units.

Draw AD perpendicular to BC. AD is a perpendicular bisector (as AB = AC), and forms two isosceles triangles, with the altitude bisecting the third side in isosceles triangle. So BD = DC = 1 unit.

Now consider a right-angled triangle ABD. AD is perpendicular to DB, AB = 2 units and BD = 1 unit. AD2  =  AB2  –  BD2

=  22  –  12  =  4 – 1  =  3

AD = unit.

Using these values, we can get

sin 60° = = = ; and

sin 30°  = cos 60° = = = ; and

cos 30°  = tan 60° = = = ; and

tan 30°  = Now consider a right-angled isosceles triangle ABC with 1 unit of equal sides. AC  = AC = units.

sin 45°  = ;

cos 45°  = ; and

tan 45°  = =  1

Now let us look at some examples of calculating exact values of trigonometric ratios:

Example 1: Find the exact value of sin 30° + cos 60°

sin 30° + cos 60°  = = 1

Example 2: Find the exact value of sin 60° + sin 30°

sin 60° + sin 30° = Example 3: Find the exact value of 2sin 30°cos 30°

2sin 30°cos 30° = =  sin 60°

Example 4: Find the exact value of sin² 30° + cos² 30°

sin² 30° + cos² 30° = = =  1

Example 5: Find the exact value of 1 – 2sin² 30°

1 – 2sin² 30°  =  1 – 2 x =  1 – =  cos 60°

Example 6: Find the exact value of 2cos² 30° – 1

2cos² 30° – 1  =  2 x – 1

=  2 x – 1

= – 1

= =  cos 60°

Note  Examples 3, 4, 5, 6 can be given as trigonometric identities as stated below:

• 2 sin θ cos θ  =  sin 2θ
• sin² θ  +  cos² θ  = 1
• 1 – 2sin² θ  =  cos 2θ
• 2 cos² θ – 1  =  cos 2θ