Trigonometric Ratios of Standard Angles

Trigonometric Ratios of Standard Angles


Trigonometric Ratios of Standard Angles

 

In trigonometry, 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180° etc. are considered as the standard angles. We can derive the values of trigonometric ratios of standard angles geometrically.

We should remember those values for the calculations in trigonometry. Here is the geometrical procedure for the derivation of trigonometric ratios of standard angles.

 

 

Trigonometric ratios of 45°

 

Let OX be the initial line and trace out an XOP = 45°. From Q, a point on OP, draw QNOX. Then NQO = 90° – 45° = 45°. Therefore it is an isosceles triangle, i.e. ON = NQ. Let ON = NQ = a.

Right angled triangle for the Trigonometric ratios of 45°

From the right angled triangle ONQ, using Pythagoras relation, 

OQ2 = ON2 + NQ2

                  = a2 + a2

                  = 2a2

       OQ  = a√2

 

Now for right angled triangle ONQ with angle of reference NOQ = 45°,

sin45° = "NQ" /"OQ"  = "a" /("a" √("2" )) = "1" /√("2" )      cosec45° = "OQ" /"NQ"  = ("a" √("2" ))/"a"  = √("2" )      cos45° = "ON" /"OQ"  = "a" /("a" √("2" )) = "1" /√("2" )      sec45° = "OQ" /"ON"  = ("a" √("2" ))/"a"  = √("2" )      tan45° = "NQ" /"ON"  = "a" /"a" = 1     cot45° = "ON" /"NQ"  = "a" /"a"  = 1


Trigonometric ratios of 60° and 30°

 

Let ABC be a triangle in which AB = BC = CA and hence A = B = C = 60°. From A draw ADBC. Then BAD = CAD = 30° and BD = CD. Let AB = BC = CA = 2a. So, BD = CD = a.

Right angled triangle for trigonometric ratios of 60° and 30°

Now, from right angled DABD, 

          AD2 = AB2 – BD2

                  = 4a2 – a2

                  = 3a2

         AD = a√3

 

Now, in right angled DABD with angle of reference ABD = 60°,

Trigonometric ratios of 60°:   sin60° = "AD" /"AB"  = ("a" √("3" ))/"2a"  = √("3" )/"2"      cosec60° = "AB" /"AD"  = "2a" /("a" √("3" )) = "2" /√("3" )      cos60° = "BD" /"AB"  = "a" /"2a"  = "1" /"2"      sec60° = "AB" /"BD"  = "2a" /"a"  = 2      tan60° = "AD" /"BD"  = ("a" √("3" ))/a = √("3" )      cot60° = "BD" /"AD"  = "a" /("a" √("3" )) = "1" /√("3" )

And, with the angle of reference BAD = 30°,

Trigonometric ratios of 30°:    Sin30° = "BD" /"AB"  = "a" /"2a"  = "1" /"2"      cosec30° = "AB" /"BD"  = "2a" /"a"  = 2      cos30° = "AD" /"AB"  = ("a" √("3" ))/"2a"  = √("3" )/"2"     sec30° = "AB" /"AD"  = "2a" /("a" √("3" )) = "2" /√("3" )      tan30° = "BD" /"AD"  = "a" /("a" √("3" )) = "1" /√("3" )      cot30° = "AD" /"BD"  = ("a" √("3" ))/a = √("3" )


Trigonometric ratios of 0°

 

Let OX be the initial line and trace out an XOP = θ. Let Q be a point on OP. Draw QMOX.

Right angled triangle for the Trigonometric ratios of 0°

When θ is decreased and tends to be 0° keeping OM constant, then QM = 0 and OQ = OM.

 

Now, from right angled triangle MOQ with angle of reference θ = 0°,

Trigonometric ratios of 0°:  sin0° = "QM" /"OQ"  = "0" /"OQ"  = 0      cosec0° = "OQ" /"QM"  = "OQ" /"0"  = ∞ (undefined)     cos0°	= "OM" /"OQ"  = "OM" /"OM"  = 1      sec0° = "OQ" /"OM"  = "OM" /"OM"  = 1       tan0° = "QM" /"OM"  = "0" /"OM"  = 0        cot0°	= "OM" /"QM"  = "OM" /"0"  = ∞ (undefined)

Trigonometric ratios of 90°

 

Let OX be the initial line and trace out an XOP = θ. Let Q be a point on OP. Draw QMOX.

Right angled triangle for the Trigonometric ratios of 90°

When θ is increased and tends to be 90° keeping MQ constant, then OM = 0 and OQ = MQ.

 

Now, from right angled triangle MOQ with angle of reference θ = 90°,

Trigonometric ratios of 90°:    sin90° = "MQ" /"OQ"  = "MQ" /"MQ"  = 1     cosec90° = "OQ" /"MQ"  = "MQ" /"MQ"  = 1     cos90° = "OM" /"OQ"  = "0" /"OQ"  = 0     sec90° = "OQ" /"OM"  = "OQ" /"0"  = ∞ (undefined)     tan90° = "MQ" /"OM"  = "MQ" /"0"  = ∞ (undefined)     cot90° = "OM" /"MQ"  = "0" /"MQ"  = 0


Table of Trigonometric Ratios of Standard Angles


Table of Trigonometric Ratios of Standard Angles


Worked Out Examples

 

Example 1: Find the value of sin230° + sin245° + sin260°

 

Solution: Here,

sin230° + sin245° + sin260° = ("1" /"2" )^"2" + ("1" /√("2" ))^"2" + (√("3" )/"2" )^"2"  = "1" /"4" + "1" /"2" + "3" /"4"  = "1 + 2 + 3" /"4"  = "6" /"4"  = "3" /"2"   Ans.


Example 2: Find the value of:

Find the value of tan〖"π" /"3" 〗^"c" sin〖"π" /"3" 〗^"c" + sin〖"π" /"4" 〗^"c" cos〖"π" /"2" 〗^"c" + cos〖"π" /"2" 〗^"c" sin〖"π" /"3" 〗^"c"

Solution: Here,

tan〖"π" /"3" 〗^"c" sin〖"π" /"3" 〗^"c" + sin〖"π" /"4" 〗^"c" cos〖"π" /"2" 〗^"c" + cos〖"π" /"2" 〗^"c" sin〖"π" /"3" 〗^"c"  = tan60°sin60° + sin45°cos90° + cos90°sin60° = √("3" ) × √("3" )/"2"  + "1" /√("2" ) × 0 + 0 × √("3" )/"2"    = "3" /"2"   + 0 + 0   = "3" /"2"   Ans.

Example 3: Prove that:

Prove that: ("1 " +" tan30°" )/("1 " -" tan30°" ) = ("cos30° " +" sin30°" )/("cos30° " -" sin30°" )

Solution: Here,

LHS = ("1 " +" tan30°" )/("1 " -" tan30°" )          = (" 1 " +" "  "1" /√("3 " ))/("1 " -" "  "1" /√("3" ))            = ((√("3" ) " " +" 1" )/√("3" ))/((√("3" )  -" 1" )/√("3" ))            = (√("3" ) " " +" 1" )/(√("3" )  -" 1" ) RHS = ("cos30° " +" sin30°" )/("cos30° " -" sin30°" )          = ( √("3" )/"2"  " " +" "  "1" /"2 " )/(√("3" )/"2"  " " -" "  "1" /"2" )            = ((√("3" ) " " +" 1" )/√("3" ))/((√("3" )  -" 1" )/√("3" ))            = (√("3" ) " " +" 1" )/(√("3 " )-" 1" )  ∴ LHS = RHS.  Proved.


Example 4: Find the value of x: sin60°cos30° + x sin45°cos45° = sin30°cos60°

 

Solution: Here,

sin60°cos30° + x sin45°cos45° = sin30°cos60° or,	√("3" )/"2"   × √("3" )/"2"   + x "1" /√("2" ) × "1" /√("2" ) = "1" /"2"  × "1" /"2"  or,	"3" /"4"  + "x" /"2"  = "1" /"4"  or,	"x" /"2"  = "1" /"4"   – "3" /"4"  or,	"x" /"2"  = ("1 " -" 3" )/"4"  or,	"x" /"2"  = (-" 2" )/"4"  or,	"x" /"2"  = (-" 1" )/"2"  or,	2x = – 2   or,	x = – 1  Ans.


Example 5: In the given figure, the distance between house B and House C is 20m and BCA = 30°. Calculate the distance between house A and B.

 

Solution: Here,

ΔABC is a right angled triangle. BC = 20m, BCA = 30° and BAC = 90°

Example 5: Right angled triangle

          Sin30° = AB/BC

or,     ½ = AB/20m

or,     2AB = 20m

or,     AB = 20m/2

or,     AB = 10m

 

Hence, the distance between house A and B is 10m. 

If you have any question or problems regarding the Trigonometric Ratios of Standard Angles, you can ask here, in the comment section below.


 

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