Trigonometric Equations
In this section we will discuss trigonometric equations.
An equation involving trigonometric ratios of an angle θ (say) is said to be a trigonometric equations, if it is satisfied for all values of θ for which the given trigonometric ratios are defined.
Some Trigonometric-equations (Identities) are as follows :
1) sin2θ + cos2θ= 1
2) sin2θ = 1 - cos2θ
3) cos2θ = 1 - sin 2θ
4) 1 + tan2θ = sec2θ
5) tan2θ = sec2θ - 1
6) sec2θ - tan2θ = 1
7) 1 + cot2θ = csc2θ
8) cot2θ = csc2θ - 1
9) csc2θ - cot2θ = 1
10) sec θ + tan θ = 1 / ( sec θ - tan θ)
11) csc θ + cot θ = 1 / ( csc θ - cot θ) |
These Trigonometric equations are true for any angle θ for which the trigonometric ratios are meaningful.
Some Solved Examples on Trigonometric Equations :
1) 1 / (1 + sin θ) + 1 / ( 1 – sin θ ) = 2 sec
2θ
Solution :
Consider 1 / (1 + sin θ) + 1 / ( 1 – sin θ )
LCM = (1 + sin θ)( 1 – sin θ )
= [1 (1 – sin θ) + 1 ( 1 + sin θ )] / (1 + sin θ)( 1 – sin θ)
= ( 1 – sin θ + 1 + sin θ ) / (1 + sin θ)( 1 – sin θ )
= 2 / (1 – sin
2θ ) [ use identity (a + b) ( a – b ) = a
2 - b
2]
= 2 / cos
2θ
= 2 sec
2θ
∴ 1 / (1 + sin θ) + 1 / ( 1 – sin θ ) = 2 sec
2θ
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2) Prove that : cos θ / ( 1- sin θ ) = ( 1 + sin θ) / cos θ
Solution :
Consider cos θ / ( 1- sin θ )
Multiply top (numerator) and bottom(denominator) by 1 + sin θ
= cos θ( 1 + sin θ) / ( 1- sin θ )( 1 + sin θ)
= cos θ ( 1 + sin θ) / ( 1- sin
2θ )
[ using this identity (a + b)(a – b ) = a
2- b
2]
= cos θ ( 1 + sin θ) / cos
2θ )
= (1 + sin θ) / cos θ
∴ cos θ / ( 1- sin θ ) = ( 1 + sin θ) / cos θ
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3) Cos
2 θ + 1 / ( 1 + cot
2 θ ) = 1
Solution :
Consider Cos
2 θ + 1 / ( 1 + cot
2 θ )
= Cos
2 θ + 1/ csc
2 θ [ since 1 + cot
2θ = csc
2θ]
= cos
2θ + sin
2θ [ since 1/ csc
2 θ = sin
2θ ]
= 1 [ since sin
2θ + cos
2θ= 1 ]
∴ Cos
2 θ + 1 / ( 1 + cot
2 θ ) = 1
Trigonometry
• SOHCAHTOA -Introduction to Trigonometry
• Trigonometric ratios and their Relation
• Trigonometry for specific angles
• Complementary angles in Trigonometry
• Trigonometric Equations
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