TRIGONOMETRIC RATIO & IDENTITIES 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

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

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