Trigonometry Examples

Example 1 : \(sin5x + sin2x – sinx\over {cos5x + 2cos3x + 2cos^x + cosx}\) is equal to –

Solution : L.H.S. = \(2sin2xcos3x + sin2x\over{2cos3x.cos2x + 2cos3x + 2cos^2x}\)

= \(sin2x[2cos3x+1]\over {2[cos3x(cos2x+1)+(cos^2x)]}\)

= \(sin2x[2cos3x+1]\over {2[cos3x(2cos^2x)+(cos^2x)]}\)

= \(sin2x[2cos3x+1]\over {2cos^2x(2cos3x+1)}\) = tanx

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Example 2 : Prove that : \(2cos2A+1\over {2cos2A-1}\) = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)

Solution : R.H.S. = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)

= (\(tan60^{\circ}+tanA\over {1-tan60^{\circ}tanA}\))(\(tan60^{\circ}-tanA\over {1+tan60^{\circ}tanA}\))

= (\(\sqrt{3}+tanA\over {1-\sqrt{3}tanA}\))(\(\sqrt{3}-tanA\over {1+\sqrt{3}tanA}\))

= \(3-tan^2A\over{1-3tan^2A}\) = \(3cos^2A-sin^2A\over {cos^2A-3sin^2A}\) = \(2cos^2A+cos^2A-2sin^2A+sin^2A\over {2cos^2A-2sin^2A-sin^2A-cos^2A}\)

= \(2(cos^2A-sin^2A)+cos^2A+sin^2A\over {2(cos^2A-sin^2A)-(sin^2A+cos^2A)}\)

= \(2cos2A+1\over {2cos2A-1}\) = R.H.S

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Example 3 : If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to-

Solution : cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B)+cos2C

= 2cos(\(3\pi\over 2\) – C)cos(A-B) + cos2C   \(\because\)   A + B + C = \(3\pi\over 2\)

= -2sinC cos(A-B) + 1 – 2\(sin^2C\) = 1 – 2sinC[cos(A-B)+sinC]

= 1 – 2sinC[cos(A-B) + sin(\(3\pi\over 2\)-(A+B))]

= 1 – 2sinC[cos(A-B)-cos(A+B)]

= 1 – 4sinA sinB sinC

Practice these given trigonometry examples to test your knowledge on concepts of trigonometry.