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.



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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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