DOUBLE ANGLE TRIG IDENTITIES PRACTICE PROBLEMS

Double angle formulas involving sin and cosine :

Problem 1 :

The expression 2sin 30° cos 30° has the same value as

a. sin 15°      b. cos 60°     c. sin 60°        d. cos 15°

Solution:

= 2sin 30° cos 30°

= 2 × (1/2) × (√3/2)

= √3/2

= sin 60°

So, option (C) is correct.

Problem 2 :

The expression csc A sin 2A is equivalent to

a. 2sin A        b. 2        c. 2cos A      d. 2cot A

Solution:

= csc A sin 2A

=1sin A×sin 2A=1sin A×2sinA cosA=2cosA

So, option (C) is correct.

Problem 3 :

If sin A = 2/3, find cos 2A.

Solution:

cos 2A=1-2sin2A=1-2232=1-249=1-89=9-89cos 2A=19

Problem 4 :

If sin A = 3/5, find cos 2A.

Solution:

cos 2A=1-2sin2A=1-2352=1-2925=1-1825=25-1825cos 2A=725

Problem 5 :

If θ is in Quadrant II and cos θ = -3/4, find an exact value for sin 2θ.

Solution:  

cos 𝜃=-34sin 𝜃=1-cos2𝜃=1--342=1-916=716sin 𝜃=74sin 2𝜃=2sin𝜃 cos𝜃=274-34=-378

Problem 6 :

Express cos 2A+sin2Acos A as a single trigonometric function for all values of A for which the fraction is defined.

Solution:

=cos 2A+sin2Acos A=2cos2A-1+sin2Acos A=2cos2A+-1+sin2Acos A=2cos2A-1-sin2Acos A=2cos2A-cos2Acos A=cos2Acos A=cos A

Verify each identity.

Problem 7 :

2sin x cos x - cos 2x = 2sin2x - 1 + sin 2x

Solution:

 2sin x cos x = sin 2x

cos 2x = 1 - 2sin2x

2sin x cos x - cos 2x = 2sin2x - 1 + sin 2x

Hence, it is verified.

Problem 8 :

cos2x + tan2x sin2x = sec2x sin2x + cos2x

Solution:

R.H.S :

= sec2x sin2x + cos2x

sec2x = 1 + tan2x

= (1 + tan2x) sin2x + 1 - 2sin2x

= sin2x + tan2x sin2x + 1 - 2sin2x

= tan2x sin2x + 1 - sin2x

1 - sin2x = cos2x

= tan2x sin2x + cos2x

L.H.S

Problem 9 :

2sin2xtan x=sin 2x

Solution:

2sin2xtan x=2sin2x×cos xsin x=2sin x cos x=sin 2x

Hence, it is verified.

Problem 10 :

1 + cos 2x -   tan 4x cos 4x = 2cos2x - sin 4x

Solution:

L.H.S :

= 1 + cos 2x - tan 4x cos 4x ----(1)

cos 2x = 2 cos2 x - 1

1 + cos 2x = 2 cos2 x

tan 4x = sin 4x / cos 4x

Applying these values in (1), we get

= 1 + cos 2x - (sin 4x / cos 4x) cos 4x

= 1 + cos 2x - sin 4x

= 2cos2x - sin 4x

R.H.S

Problem 11 :

tan xsin 2x=sec2x2

Solution:

tan xsin 2x=sin xcos xsin 2x=sin xcos x×1sin 2x=sin xcos x×12sin x cos x=1cos x×12cosx=12cos2x=sec2x2

Hence, it is verified.

Problem 12 :

Solution :

Problem 13 :

Solution :

Problem 14 :

Solution :

Problem 15 :

Solution :

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