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+1 vote

Prove each identity.

39)   sin (-θ) cot(-θ) = cos θ

43)   csc (θ) + sin (-θ) =cos2 θ  sin θ

asked Feb 26, 2013 in TRIGONOMETRY by rockstar Apprentice

3 Answers

+1 vote

sin(–θ)cot(–θ)=cosθ

L .H.S sin(–θ)cot(–θ)

Even-Odd Identities:sin(–θ)= –sinθ and cot(–θ)= –cotθ

(–sinθ) (–cotθ)

Quotient Identities: cotθ=cosθ/sinθ

(–sinθ)(–cosθ/sinθ)

Cancel common terms.

(–)(–cosθ)

Product of two negative signs are positive.

cosθ=R.H.S

Therefore sin(–θ)cot(–θ)=cosθ.

answered Mar 4, 2013 by hussy Rookie
0 votes

 

L .H.S sin(–θ)cot(–θ)

Even-Odd Identities:sin(–θ)= –sinθ and cot(–θ)= –cotθ

(–sinθ) (–cotθ)    (Quotient Identities: cotθ=cosθ/sinθ)

(–sinθ)(–cosθ/sinθ)

Cancel common terms.

(–)(–cosθ)

Product of two negative signs are positive.

cosθ=R.H.S

Therefore sin(–θ)cot(–θ)=cosθ.

 

answered Mar 5, 2013 by Naren Answers Apprentice
0 votes

43).

Consider that, the trigonometric equation as csc (θ) + sin (- θ) = cos2 (θ) / sin (θ).

Left hand side identity : csc (θ) + sin (- θ).

Reciprocal identity : csc (θ) = 1/sin (θ).

csc (θ) + sin (- θ) = (1/ sin (θ)) + sin (- θ)

Even - Odd Identities : sin(– θ) = – sin (θ).

= (1/ sin (θ)) - sin (θ)

= (1 - sin2 θ) / sin (θ)

Pythagoras identity : sin2 θ + cos2 θ = 1.

= (cos2 θ) / sin (θ)

= Right hand side identity.

answered Jul 30, 2014 by lilly Expert

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